basiv Test
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 Answered
 Review

Question 1 of 200
1. Question
0 pointsThe angular momentum of electron of H− atom is proportional to:
Correct
Angular momentum L is proportional to moment of inertia I and angular speed ω measured in radians per second.
Incorrect
From Bohr’s Model, Centripetal force of electron = Columbic force of attraction on electron by nucleus.
Unattempted
From Bohr’s Model, Centripetal force of electron = Columbic force of attraction on electron by nucleus.

Question 2 of 200
2. Question
0 pointsThe total energy in 1st orbit of hydrogen atom is given by:
Correct
Total energy of an electron in first orbit of hydrogen atom is – 13.6 eV.
Incorrect
This is the following solution:
Unattempted
This is the following solution:

Question 3 of 200
3. Question
0 pointsIf the radius of second Bohrs orbit of hydrogen atom is a_{0 }then the radius of third Bohrs orbit of Be^{3+} ion will be:
Correct
This option is incorrect.
Incorrect
This is the following solution:
Unattempted
This is the following solution:

Question 4 of 200
4. Question
0 pointsIf the total energy of an electron in the 1st shell of Hatom is –13.6 eV then its potential energy in the 1st excited state would be:
Correct
This option is incorrect.
Incorrect
This is the following solution:
Unattempted
This is the following solution:

Question 5 of 200
5. Question
0 pointsIf the speed of the electron in the first Bohr orbit of hydrogen atom be 'x', then the speed of electron in the third Bohr orbit is:
Correct
As speed is inversely propotional to the orbit number. This cannot be the answer.
Incorrect
This is the following solution:
Unattempted
This is the following solution:

Question 6 of 200
6. Question
0 pointsIf the radius of first Bohr orbit is 2x, then de Broglie wavelength of electron in 4th orbit is nearly:
Correct
This option is correct.
Incorrect
This is the following solution:
Unattempted
This is the following solution:

Question 7 of 200
7. Question
0 pointsIf the lowest energy Xrays have λ=3.055×10^{−8} m, estimate the minimum difference in energy between two Bohr's orbits such that an electronic transition would correspond to the emission of an Xray. Assuming that the electrons in other shells exert no influence, at what Z(minimum) would a transition from a second energy level to the first, result in the emission of an Xray?
Correct
This is correct option.
Incorrect
This is the following solution:
Unattempted
This is the following solution:

Question 8 of 200
8. Question
0 pointsPossible number of nodal planes present in the the M shell of Hlike species are:
Correct
Incorrect
This is the following solution:
Unattempted
This is the following solution:

Question 9 of 200
9. Question
0 pointsThe no. of radial nodes for 3P_{x} orbital is :
Correct
The total number of nodes present in a 3p orbital is two. Therefore, in a 3p orbital there is one angular node and one radial node.
Incorrect
Total Nodes=n1
and
Radial Nodes = n – 1 – ℓ
So, for 3Px n = 3 and ℓ = 1
So, the number of radial nodes = 3 – 1 1 = 1
Unattempted
Total Nodes=n1
and
Radial Nodes = n – 1 – ℓ
So, for 3Px n = 3 and ℓ = 1
So, the number of radial nodes = 3 – 1 1 = 1

Question 10 of 200
10. Question
0 pointsPhotoelectric emission is observed from a surface for frequencies ν_{1} and ν_{2} of incident radiations (ν_{1} >ν_{2}). If the maximum kinetic energy of photoelectrons in the two cases are in the ratio of 2 : 1, then threshold frequency v_{0} is given by:
Correct
This option is incorrect.
Incorrect
This is the following solution:
Unattempted
This is the following solution:

Question 11 of 200
11. Question
0 pointsThe spin angular momentum for the `s’ electron in Hatom is:
Correct
This option is correct.
Spin angular momentum =√S(S+1)h2π=√12(12+1)h2π=√3h4π
Incorrect
This is the following solution:
Unattempted
This is the following solution:

Question 12 of 200
12. Question
0 pointsIf a dye absorbs a photon of wavelength λ and reemits the absorbed energy into two photons of wavelengths λ_{1} and λ_{2} respectively. Then:
Correct
This option is incorrect.
Incorrect
This is the following solution:
Unattempted
This is the following solution:

Question 13 of 200
13. Question
0 pointsAn electron is allowed to move freely in a closed cubic box of length 10 cm. The minimum uncertainty in its velocity will be observed as :
Correct
This option is incorrect.
Incorrect
This is the following solution:
Unattempted
This is the following solution:

Question 14 of 200
14. Question
0 pointsIf a_{0} be the radius of first Bohr's orbit of Hatom, the deBroglie's wavelength of an electron revolving in the third Bohr's orbit will be:
Correct
This option is correct.
Incorrect
This is the following solution:
Unattempted
This is the following solution:

Question 15 of 200
15. Question
0 pointsThe probability of finding an electron residing in a p_{x} orbital is zero :
Correct
The yz plane is a twodimensional plane formed by the yaxis and zaxis. This option suggests that the probability of finding an electron residing in a px orbital is zero when considering positions that lie in this specific plane, because yz plane is the nodal point for px, and at the nodal point the probability of finding an electron is zero. Therefore it is the correct option.
Incorrect
The yz plane is a twodimensional plane formed by the yaxis and zaxis. This option suggests that the probability of finding an electron residing in a px orbital is zero when considering positions that lie in this specific plane, because yz plane is the nodal point for px, and at the nodal point the probability of finding an electron is zero._{The yz plane is a twodimensional plane formed by the yaxis and zaxis. This option suggests that the probability of finding an electron residing in a px orbital is zero when considering positions that lie in this specific plane, because yz plane is the nodal point for px, and at the nodal point the probability of finding an electron is zero.} The yz plane is a twodimensional plane formed by the yaxis and zaxis. This option suggests that the probability of finding an electron residing in a px orbital is zero when considering positions that lie in this specific plane, because yz plane is the nodal point for px, and at the nodal point the probability of finding an electron is zero.
Unattempted
The yz plane is a twodimensional plane formed by the yaxis and zaxis. This option suggests that the probability of finding an electron residing in a px orbital is zero when considering positions that lie in this specific plane, because yz plane is the nodal point for px, and at the nodal point the probability of finding an electron is zero._{The yz plane is a twodimensional plane formed by the yaxis and zaxis. This option suggests that the probability of finding an electron residing in a px orbital is zero when considering positions that lie in this specific plane, because yz plane is the nodal point for px, and at the nodal point the probability of finding an electron is zero.} The yz plane is a twodimensional plane formed by the yaxis and zaxis. This option suggests that the probability of finding an electron residing in a px orbital is zero when considering positions that lie in this specific plane, because yz plane is the nodal point for px, and at the nodal point the probability of finding an electron is zero.

Question 16 of 200
16. Question
0 pointsIn a single hydrogen atom, the electron is excited to its 6th orbit. The maximum no. of distinct lines possible, when it comes to the ground state is:
Correct
It is clear from the solution given below that this option is incorrect
The maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state is
2(n−n)(n−n+1) = 25×6 =15 lines.
Incorrect
The maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state is
2(n−n)(n−n+1) = 25×6 =15 lines.
Unattempted
The maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state is
2(n−n)(n−n+1) = 25×6 =15 lines.

Question 17 of 200
17. Question
0 pointsThe difference between the n^{th} and (n+1)^{th }Bohr’s radius of Hatom is equal to (n1)^{th} Bohr’s radius. Hence the value of 'n' is:
Correct
It is clear from the solution given below that the option is incorrect.
Since n=0 is not possible, the only possible value is n=4.
Incorrect
This is the following solution:
Unattempted
This is the following solution:

Question 18 of 200
18. Question
0 pointsConsider one He+ ion in excited state (n = 5). Which of the following observations will hold true as per the Bohr’s model?
Correct
A multitude of He+ will produce 10 possible emission spectral lines however a single He+ can produce maximum 4 lines . Therefore, this option is incorrect.
A multitude of He+ will produce 10 possible emission spectral lines however a single He+ can produce maximum 4 lines there for this option is incorrect.
Incorrect
This is the following solution:
Unattempted
This is the following solution:

Question 19 of 200
19. Question
0 pointsThe given diagram indicates the energy levels of certain atom.When an electron moves from 2E level to E level, a photon of wavelength λ is emitted.The wavelength of photon emitted during its transition from 4E/3 level to E level is:
Correct
After calculating the wavelength, it is clearly visible that this option is incorrect.
Incorrect
This is the following solution:
Unattempted
This is the following solution:

Question 20 of 200
20. Question
0 pointsTotal number of electrons in Cu^{2+} for which the summation of azimuthal quantum number and magnetic quantum number is zero is :
Correct
the electronic configuration of Cu+2 can be represented as:
1s2 2s2 2p6 3s2 3p6 3d9.
the total number of electrons for which the summation of the azimuthal quantum number and the magnetic quantum number is zero can be either 7 or 8.
Total electrons in S – subshell = 7Incorrect
This is the following solution:
Cu(29)=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{10},
l=0⇒S− subshell
Total electrons in S – subshell = 7Unattempted
This is the following solution:
Cu(29)=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{10},
l=0⇒S− subshell
Total electrons in S – subshell = 7 
Question 21 of 200
21. Question
0 pointsPhotons having energy equivalent to binding energy of 4th state of He^{+} ion is used on the metal surface of work function 1.4 eV. If electrons are further accelerated through the potential difference of 4V then the minimum value of deBroglie's wavelength associated with the electron is:
Correct
It is clearly visible from the solution given below that the option is incorrect.
Incorrect
This is the following solution:
Unattempted
This is the following solution:

Question 22 of 200
22. Question
0 pointsIf Rutherford would have used β − particles instead of α − particles in his Goldleaf experiment, which of the following observations would definitely not have been made:
Correct
β particles will converge while passing through atom as they bear negative charge. This option suggests that if Rutherford had used β particles instead of α particles in his Goldleaf experiment, the observation of β particles scattering beyond the gold foil would not have been made. This means that the β particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.β particles will converge while passing through atom as they bear negative charge. This option suggests that if Rutherford had used β particles instead of α particles in his Goldleaf experiment, the observation of β particles scattering beyond the gold foil would not have been made. This means that the β particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.
β particles will converge while passing through atom as they bear negative charge.This option suggests that if Rutherford had used β particles instead of α particles in his Goldleaf experiment, the observation of β particles scattering beyond the gold foil would not have been made. This means that the β particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.
This option suggests that if Rutherford had used β particles instead of α particles in his Goldleaf experiment, the observation of β particles scattering beyond the gold foil would not have been made. This means that the β particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.This option suggests that if Rutherford had used β particles instead of α particles in his Goldleaf experiment, the observation of β particles scattering beyond the gold foil would not have been made. This means that the β particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.
Incorrect
β particles will converge while passing through atom as they bear negative charge. if Rutherford had used β particles instead of α particles in his Goldleaf experiment, the observation of β particles scattering beyond the gold foil would not have been made. This means that the β particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.
This is the following questions:β particles will converge while passing through atom as they bear negative charge. if Rutherford had used β particles instead of α particles in his Goldleaf experiment, the observation of β particles scattering beyond the gold foil would not have been made. This means that the β particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.
β particles will converge while passing through atom as they bear negative charge. if Rutherford had used β particles instead of α particles in his Goldleaf experiment, the observation of β particles scattering beyond the gold foil would not have been made. This means that the β particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.β particles will converge while passing through atom as they bear negative charge.
β particles will converge while passing through atom as they bear negative charge. if Rutherford had used β particles instead of α particles in his Goldleaf experiment, the observation of β particles scattering beyond the gold foil would not have been made. This means that the β particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.iment, the observation of β particles scattering beyond the gold foil would not have been made. This means that the β particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.
This option suggests that if Rutherford had used β particles instead of α particles in his Goldleaf experiment, the observation of β particles scattering beyond the gold foil would not have been made. This means that the β particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.This option suggests that if Rutherford had used β particles instead of α particles in his Goldleaf experiment, the observation of β particles scattering beyond the gold foil would not have been made. This means that the β particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.
Unattempted
β particles will converge while passing through atom as they bear negative charge. if Rutherford had used β particles instead of α particles in his Goldleaf experiment, the observation of β particles scattering beyond the gold foil would not have been made. This means that the β particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.
This is the following questions:β particles will converge while passing through atom as they bear negative charge. if Rutherford had used β particles instead of α particles in his Goldleaf experiment, the observation of β particles scattering beyond the gold foil would not have been made. This means that the β particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.
β particles will converge while passing through atom as they bear negative charge. if Rutherford had used β particles instead of α particles in his Goldleaf experiment, the observation of β particles scattering beyond the gold foil would not have been made. This means that the β particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.β particles will converge while passing through atom as they bear negative charge.
β particles will converge while passing through atom as they bear negative charge. if Rutherford had used β particles instead of α particles in his Goldleaf experiment, the observation of β particles scattering beyond the gold foil would not have been made. This means that the β particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.iment, the observation of β particles scattering beyond the gold foil would not have been made. This means that the β particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.
This option suggests that if Rutherford had used β particles instead of α particles in his Goldleaf experiment, the observation of β particles scattering beyond the gold foil would not have been made. This means that the β particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.This option suggests that if Rutherford had used β particles instead of α particles in his Goldleaf experiment, the observation of β particles scattering beyond the gold foil would not have been made. This means that the β particles would not have experienced significant deflection or scattering as they passed through the gold foil. Thus this option is correct.

Question 23 of 200
23. Question
0 pointsA 1kW radio transmitter operates at a frequency of 880 KHz. How many quantas per second does it emit?
Correct
This option does not agree with solution and hence is incorrect.
Incorrect
This is the following solution:
Unattempted
This is the following solution:

Question 24 of 200
24. Question
0 pointsThe ratio of the e/m values of a proton and an α − particle is:
Correct
For a proton (H), e =1.6×1019 C and m=1.67×1027 Kg
(e/m) = For an alpha particle (He+2), e=2×1.6×1019 C and m= 4×1.67×1027 Kg
Dividing e/m of proton by that of He+2 we get 4/2 = 2/1.
Incorrect
This is the following solution:
Unattempted
This is the following solution:

Question 25 of 200
25. Question
0 pointsFor sets of values of quantum numbers (n, l, m and s) are given below. Which of these does not provide a permissible solution of the wave equation?
Correct
None
Incorrect
n = l is not permissible.
Value of l is always less than the value of n.Unattempted
n = l is not permissible.
Value of l is always less than the value of n. 
Question 26 of 200
26. Question
0 pointsThe angular momentum of an electron in a Bohr’s orbit of He^{+} is 3.1652×10^{34} kg.m^{2} / sec.The wave number of the spectral line emitted when an electron falls from this level to the first excited state is:
Correct
Not correct because it doesn't match the calculated value for the wave number based on the given angular momentum.
Incorrect
This is the following solution:
Unattempted
This is the following solution:

Question 27 of 200
27. Question
0 pointsThe electronic configuration of Fe ^{3+} is:
Correct
This configuration suggests Fe3+ loses only 2 electrons, which is incorrect.
Incorrect
Fe(Z=26)=[Ar]3d^{6}As^{2}
Fe^{3+} (23 electrons) = [Ar]3d^{5}
Unattempted
Fe(Z=26)=[Ar]3d^{6}As^{2}
Fe^{3+} (23 electrons) = [Ar]3d^{5}

Question 28 of 200
28. Question
0 pointsThe value of the magnetic moment of a particular ion is 2.83 Bohr magneton. The ion is :
Correct
None
Incorrect
This is the following solution:
Unattempted
This is the following solution:

Question 29 of 200
29. Question
0 pointsThe ratio of the radii of the first three Bohr orbits is:
Correct
Incorrect, doesn't follow the square of quantum numbers.
Incorrect
The radii of the Bohr orbits are proportional to the square of the principal quantum numbers (n^{2}). The first three Bohr orbits correspond to n = 1, 2, and 3. Therefore, their radii ratio would be 1^{2} : 2^{2} : 3^{2} = 1 : 4 : 9.
Unattempted
The radii of the Bohr orbits are proportional to the square of the principal quantum numbers (n^{2}). The first three Bohr orbits correspond to n = 1, 2, and 3. Therefore, their radii ratio would be 1^{2} : 2^{2} : 3^{2} = 1 : 4 : 9.

Question 30 of 200
30. Question
0 pointsAtomic number of chromium is 24, then Cr^{3+ }will be:
Correct
Diamagnetic substances have all their electrons paired up, resulting in no net magnetic moment. Cr3+ has unpaired electrons, so it's not diamagnetic.
Incorrect
Chromium (Cr) with an atomic number of 24 has an electronic configuration of 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵. When it loses three electrons to form Cr3+ (3+ charge), its electronic configuration becomes 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³. In this configuration, it has unpaired electrons in the 3d subshell, making it paramagnetic.
Unattempted
Chromium (Cr) with an atomic number of 24 has an electronic configuration of 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵. When it loses three electrons to form Cr3+ (3+ charge), its electronic configuration becomes 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³. In this configuration, it has unpaired electrons in the 3d subshell, making it paramagnetic.

Question 31 of 200
31. Question
0 pointsFind the incorrect set of quantum number.
Correct
Incorrect.
If the angular momentum quantum number of an electron =3, which means (n1) values are possible for l, so the possible values of l = 0,1,2. The possible values of m will be ± l that are 2,1,0,1,2, and s will always be +1/2. The value of m will never be more than l, thus this is not a valid set of quantum numbers
Incorrect
If the angular momentum quantum number of an electron =3, which means (n1) values are possible for l, so the possible values of l = 0,1,2. The possible values of m will be ± l that are 2,1,0,1,2, and s will always be +1/2. The value of m will never be more than l, thus this is not a valid set of quantum numbers.
Unattempted
If the angular momentum quantum number of an electron =3, which means (n1) values are possible for l, so the possible values of l = 0,1,2. The possible values of m will be ± l that are 2,1,0,1,2, and s will always be +1/2. The value of m will never be more than l, thus this is not a valid set of quantum numbers.

Question 32 of 200
32. Question
0 pointsIf the speed of electron in the Bohr's first orbit of hydrogen is x, the speed in the third orbit is:
Correct
This is the correct speed for the electron in the third orbit according to the model.
Incorrect
According to Bohr's model, the speed of an electron in an orbit is inversely proportional to the principal quantum number (n). The formula for the speed (v) of an electron in the nth orbit is given by:
v ∝ 1/n
If the speed of an electron in the first orbit (n = 1) is x, then the speed in the third orbit (n = 3) will be: v3 = v1 x (1/n3) = x * (1/3) = x/3
Unattempted
According to Bohr's model, the speed of an electron in an orbit is inversely proportional to the principal quantum number (n). The formula for the speed (v) of an electron in the nth orbit is given by:
v ∝ 1/n
If the speed of an electron in the first orbit (n = 1) is x, then the speed in the third orbit (n = 3) will be: v3 = v1 x (1/n3) = x * (1/3) = x/3

Question 33 of 200
33. Question
0 pointsChoose the correct set of quantum numbers for the Ne atom’s electron of the second excited state:
Correct
Incorrect principal quantum number for the second excited state. Other quantum numbers are not in line with the correct level of excitation.
Incorrect
Principal Quantum Number (n): 3 (Third energy level)
Azimuthal Quantum Number (ℓ): 0 (s orbital)
Magnetic Quantum Number (mℓ): 0 (One of the orientation states within the s orbital)
Spin Quantum Number (ms): +1/2 (Spin orientation)
Unattempted
Principal Quantum Number (n): 3 (Third energy level)
Azimuthal Quantum Number (ℓ): 0 (s orbital)
Magnetic Quantum Number (mℓ): 0 (One of the orientation states within the s orbital)
Spin Quantum Number (ms): +1/2 (Spin orientation)

Question 34 of 200
34. Question
0 pointsSelect the one among the following transitions in hydrogen atom, which gives an absorption line of lowest frequency:
Correct
This transition is of higher energy compared to option C, resulting in a higher frequency absorption line.
Incorrect
Transition from n = 4 to n = 6 involves the largest energy change, resulting in the absorption of a photon with the lowest frequency and longest wavelength among the given options. This is because the energy of a photon is inversely proportional to its wavelength, and larger energy gaps correspond to lowerfrequency transitions.
Unattempted
Transition from n = 4 to n = 6 involves the largest energy change, resulting in the absorption of a photon with the lowest frequency and longest wavelength among the given options. This is because the energy of a photon is inversely proportional to its wavelength, and larger energy gaps correspond to lowerfrequency transitions.

Question 35 of 200
35. Question
0 pointsAn electron is moving with a kinetic energy of 4.55×10^{25}J. What will be the de Broglie wave length for this electron?
Correct
This value is too small for an electron's de Broglie wavelength based on the given kinetic energy.
Incorrect
The de Broglie wavelength (λ) of a particle is given by the formula λ = h / p, where h is the Planck constant and p is the momentum of the particle. For an electron, the momentum can be related to its kinetic energy (K) as p = √(2mK), where m is the mass of the electron. Given the kinetic energy K = 4.55×1025 J, the mass of an electron (m), and the Planck constant (h), the calculated de Broglie wavelength is indeed 7.28×107 m.
Unattempted
The de Broglie wavelength (λ) of a particle is given by the formula λ = h / p, where h is the Planck constant and p is the momentum of the particle. For an electron, the momentum can be related to its kinetic energy (K) as p = √(2mK), where m is the mass of the electron. Given the kinetic energy K = 4.55×1025 J, the mass of an electron (m), and the Planck constant (h), the calculated de Broglie wavelength is indeed 7.28×107 m.

Question 36 of 200
36. Question
0 pointsSuppose 10^{17}J of energy is needed by the interior of human eye to see an object. How many photons of green light (λ = 550 nm) are needed to generate this minimum amount of energy?
Correct
Incorrect option.
Incorrect
This is the following solution:
Unattempted
This is the following solution:

Question 37 of 200
37. Question
0 pointsThreshold frequency of a metal is f_{o}. When light of frequency v = 2f_{o} is incident on the metal plate, maximum velocity of e– emitted is v1, when frequency of incident radiation is 5f_{o}, maximum velocity of emitted e– is v2. then the ratio of v1/v2 is :
Correct
Incorrect
This is the following solution:
Unattempted
This is the following solution:

Question 38 of 200
38. Question
0 pointsNumber of waves in the third Bohr’s orbit of hydrogen will be:
Correct
Incorrect
This is the following solution :
Unattempted
This is the following solution :

Question 39 of 200
39. Question
0 pointsThe binding energy of an electron in the ground state of the He atom is equal to 24.6 eV. The energy required to remove both the electrons from the atom will be :
Correct
To remove both electrons from a helium atom, we need to consider the ionization energy of helium. The ionization energy is the energy required to completely remove an electron from an atom or ion. The first ionization energy of helium is the energy required to remove one electron from a neutral helium atom, and is equal to 24.6 eV, as given in the problem statement. To remove both electrons from the helium atom, we need to consider the second ionization energy, which is the energy required to remove the second electron from a singly ionized helium ion (He+). The second ionization energy of helium is greater than the first, because the positive charge of the ion makes it more difficult to remove the second electron.The second ionization energy of helium is 54.4 eV. Therefore, the total energy required to remove both electrons from a neutral helium atom is:
24.6 eV + 54.4 eV = 79 eV
So the energy required to remove both electrons from a helium atom is 79 eV.
Incorrect
This is the following solution :
Unattempted
This is the following solution :

Question 40 of 200
40. Question
0 pointsThe spectral line of the shortest wavelength in Balmer series of atomic hydrogen will be :
Correct
The spectral line of the shortest wavelength in Balmer series of atomic hydrogen is shown in picture below.
Incorrect
Shortest wavelength means maximum energy. Therefore, the electronic transition involved should be :
Unattempted
Shortest wavelength means maximum energy. Therefore, the electronic transition involved should be :

Question 41 of 200
41. Question
0 pointsFor a d – electron, the orbital angular momentum is:
Correct
Incorrect
This is the folllowing solution:
Unattempted
This is the folllowing solution:

Question 42 of 200
42. Question
0 pointsThe electronic configuration of an element is 1s^{2} , 2s^{2} , 2p^{6} , 3s^{2} , 3p^{6} , 3d^{5} , 4s¹.This represents its:
Correct
The electronic configuration 1s2, 2s2, 2p6, 3s2, 3p6, 3d5, 4s1 belongs to the element chromium (Cr) which has atomic number 24.
Incorrect
The given electronic configuration is ground state for chromium. Hence, (B) is the correct answer.
Unattempted
The given electronic configuration is ground state for chromium. Hence, (B) is the correct answer.

Question 43 of 200
43. Question
0 pointsWhich of the following pairs of ions have the same electronic configuration?
Correct
Cr3+ and Fe3+ do not have the same electronic configuration.
The electronic configuration of Cr3+ is 1s2 2s2 2p6 3s2 3p6 3d3, which is the same as the electronic configuration of a neutral chromium (Cr) atom with three electrons removed.
The electronic configuration of Fe3+ is 1s2 2s2 2p6 3s2 3p6 3d5, which is the same as the electronic configuration of a neutral iron (Fe) atom with three electrons removed.
Therefore, Cr3+ and Fe3+ have different electronic configurations.
Incorrect
Fe³+and Mn²+ have same electronic configuration. Hence (B) is the correct answer.
Unattempted
Fe³+and Mn²+ have same electronic configuration. Hence (B) is the correct answer.

Question 44 of 200
44. Question
0 pointsPrincipal, azimuthal and magnetic quantum numbers are respectively related to :
Correct
Principal, azimuthal and magnetic quantum numbers are respectively related to size,shape and orientation
Incorrect
Principal, azimuthal and magnetic quantum numbers are respectively related to size,shape and orientation
Hence, (B) is the correct answer.Unattempted
Principal, azimuthal and magnetic quantum numbers are respectively related to size,shape and orientation
Hence, (B) is the correct answer. 
Question 45 of 200
45. Question
0 pointsNumber of waves in third Bohr’s orbit of hydrogen will be:
Correct
The number of waves in the third Bohr's orbit of hydrogen will be 3.The third Bohr's orbit of hydrogen has a principal quantum number of n=3. The number of waves in a circular orbit is equal to the circumference of the orbit divided by the wavelength of the wave.
Incorrect
This is the following solution:
Unattempted
This is the following solution:

Question 46 of 200
46. Question
0 pointsThe mass of proton is
Correct
Mass of protons is 1.6726 X 10^{27} kg
Incorrect
Mass of protons is 1.6726 X 10^{27} kg
Unattempted
Mass of protons is 1.6726 X 10^{27} kg

Question 47 of 200
47. Question
0 pointsWhat will be the effect on proton when passes through the electric field?
Correct
It will be deflected towards the cathode not the anode.
Incorrect
Protons are positively charged particles which show deflection towards cathode in electric field
Unattempted
Protons are positively charged particles which show deflection towards cathode in electric field

Question 48 of 200
48. Question
0 pointsPositive rays are produced
Correct
There is no evidence of production of positive rays by this method.
Incorrect
Positive rays are produced by bombardment of cathode rays on gas molecules in discharge tube.
Unattempted
Positive rays are produced by bombardment of cathode rays on gas molecules in discharge tube.

Question 49 of 200
49. Question
0 pointsPressure in gas discharge tube was kept
Correct
It was kept at 0.01 torr.
Incorrect
At low pressure (0.01 torr) there is large spaces between gas molecules and there will be less hinderance for cathode rays to travel from cathode to anode.
Unattempted
At low pressure (0.01 torr) there is large spaces between gas molecules and there will be less hinderance for cathode rays to travel from cathode to anode.

Question 50 of 200
50. Question
0 pointsPositive rays give flash on
Correct
Positive rays produce flashes upon ZnS plate
Incorrect
Positive rays produce flashes upon ZnS plate
Unattempted
Positive rays produce flashes upon ZnS plate

Question 51 of 200
51. Question
0 pointsThe relationship between energy of a photon of light and its frequency is given by:
Correct
This states that all matter, like light, should behave as particles and waves. It does not give the relationship between energy of a photon of light and its frequency.
Incorrect
According to Planck's quantum theory, E a v or E = hv
Unattempted
According to Planck's quantum theory, E a v or E = hv

Question 52 of 200
52. Question
0 pointsThe velocity of the photon
Correct
Velocity of photon is independent of wavelength
c= f x λ
Product of frequency and wavelength remains constant quantity because of inverse relationship between frequency and wavelength
Incorrect
Velocity of photon is independent of wavelength
c= f x λ
Product of frequency and wavelength remains constant quantity because of inverse relationship between frequency and wavelength
Unattempted
Velocity of photon is independent of wavelength
c= f x λ
Product of frequency and wavelength remains constant quantity because of inverse relationship between frequency and wavelength

Question 53 of 200
53. Question
0 pointsPlanck's theory says energy is emitted:
Correct
This is the exact opposite of Plank's theory.
Incorrect
According to Planck's quantum theory, there is no continuous emission or absorption of energy.
Unattempted
According to Planck's quantum theory, there is no continuous emission or absorption of energy.

Question 54 of 200
54. Question
0 pointsPlanck's equation is:
Correct
This is Einstein's equation that relates energy and mass.
Incorrect
Explanation is:
Unattempted
Explanation is:

Question 55 of 200
55. Question
0 pointsQuantum number which is not derived from Schrodinger's wave equation
Correct
This can be derived from Schrodinger's wave equation
Incorrect
Quantum number which is not derived from Schrodinger wave equation is spin quantum number. It was given by Uhlenbeck and Goudsmith
Unattempted
Quantum number which is not derived from Schrodinger wave equation is spin quantum number. It was given by Uhlenbeck and Goudsmith

Question 56 of 200
56. Question
0 pointsA nodal plane in an orbital is the plane where electron density is
Correct
Electron density is 0 in the nodal plane
Incorrect
Nodal plane is the area where the probability of finding of electron is zero
Unattempted
Nodal plane is the area where the probability of finding of electron is zero

Question 57 of 200
57. Question
0 pointsThe number of degenerate orbitals in a subshell having sausage shape are:
Correct
It is incorrect because there is more than one degenerate orbital in a subshell with a sausage shape.
Incorrect
dsubshell has five degenerate orbitals with sausage shape.
Unattempted
dsubshell has five degenerate orbitals with sausage shape.

Question 58 of 200
58. Question
0 pointsQuantum number values for 3p orbital are
Correct
This combination represents the 3s orbital, not the 3p orbital. The principal quantum number (n) denotes the energy level or shell, while the azimuthal quantum number (ℓ) indicates the shape of the orbital. The s orbitals have a spherical shape, which corresponds to ℓ = 0.
The correct option is b) n = 3, ℓ = 1.
The quantum numbers describe various properties of an electron's orbital in an atom. In the case of the 3p orbital, the correct quantum number values are n = 3 and ℓ = 1.
Let's examine why the other options are incorrect:
a) n = 3, ℓ = 0: This combination represents the 3s orbital, not the 3p orbital. The principal quantum number (n) denotes the energy level or shell, while the azimuthal quantum number (ℓ) indicates the shape of the orbital. The s orbitals have a spherical shape, which corresponds to ℓ = 0.
c) n = 2, ℓ = 1: This combination represents the 2p orbital, not the 3p orbital. The principal quantum number (n) represents the energy level, and in this case, it corresponds to the second shell. However, the 2p orbital belongs to the second shell, not the third shell.
d) n = 2, ℓ = 3: This combination is invalid because the azimuthal quantum number (ℓ) can only have integer values ranging from 0 to (n – 1). Since n = 2, the maximum value of ℓ can be 1, not 3. Therefore, this combination does not correspond to any valid orbital.
To summarize, the correct quantum number values for the 3p orbital are n = 3 and ℓ = 1, as stated in option b.
Incorrect
The correct option is b) n = 3, ℓ = 1.
The quantum numbers describe various properties of an electron's orbital in an atom. In the case of the 3p orbital, the correct quantum number values are n = 3 and ℓ = 1.
Let's examine why the other options are incorrect:
a) n = 3, ℓ = 0: This combination represents the 3s orbital, not the 3p orbital. The principal quantum number (n) denotes the energy level or shell, while the azimuthal quantum number (ℓ) indicates the shape of the orbital. The s orbitals have a spherical shape, which corresponds to ℓ = 0.
c) n = 2, ℓ = 1: This combination represents the 2p orbital, not the 3p orbital. The principal quantum number (n) represents the energy level, and in this case, it corresponds to the second shell. However, the 2p orbital belongs to the second shell, not the third shell.
d) n = 2, ℓ = 3: This combination is invalid because the azimuthal quantum number (ℓ) can only have integer values ranging from 0 to (n – 1). Since n = 2, the maximum value of ℓ can be 1, not 3. Therefore, this combination does not correspond to any valid orbital.
To summarize, the correct quantum number values for the 3p orbital are n = 3 and ℓ = 1, as stated in option b.
Unattempted
The correct option is b) n = 3, ℓ = 1.
The quantum numbers describe various properties of an electron's orbital in an atom. In the case of the 3p orbital, the correct quantum number values are n = 3 and ℓ = 1.
Let's examine why the other options are incorrect:
a) n = 3, ℓ = 0: This combination represents the 3s orbital, not the 3p orbital. The principal quantum number (n) denotes the energy level or shell, while the azimuthal quantum number (ℓ) indicates the shape of the orbital. The s orbitals have a spherical shape, which corresponds to ℓ = 0.
c) n = 2, ℓ = 1: This combination represents the 2p orbital, not the 3p orbital. The principal quantum number (n) represents the energy level, and in this case, it corresponds to the second shell. However, the 2p orbital belongs to the second shell, not the third shell.
d) n = 2, ℓ = 3: This combination is invalid because the azimuthal quantum number (ℓ) can only have integer values ranging from 0 to (n – 1). Since n = 2, the maximum value of ℓ can be 1, not 3. Therefore, this combination does not correspond to any valid orbital.
To summarize, the correct quantum number values for the 3p orbital are n = 3 and ℓ = 1, as stated in option b.

Question 59 of 200
59. Question
0 pointsMaximum number of electrons that can be accommodated in psubshell
Correct
It is incorrect because 2 is the maximum number of electrons that can be accommodated in a single p orbital, not the entire psubshell.
Note: Number of electrons in a subshell = 2(2ℓ + 1).
The correct option is b) 6.The psubshell can accommodate a maximum of 6 electrons. This is because the psubshell consists of three orbitals (px, py, and pz), and each orbital can hold a maximum of 2 electrons according to the Pauli exclusion principle.
Let's examine why the other options are incorrect:
a) 2 is incorrect because 2 is the maximum number of electrons that can be accommodated in a single p orbital, not the entire psubshell.
c) 10 is incorrect because 10 is the maximum number of electrons that can be accommodated in a dsubshell, not the psubshell.
d) 14 is incorrect because 14 is the maximum number of electrons that can be accommodated in an fsubshell, not the psubshell.
Incorrect
Note: Number of electrons in a subshell = 2(2ℓ + 1).
The correct option is b) 6.
The psubshell can accommodate a maximum of 6 electrons. This is because the psubshell consists of three orbitals (px, py, and pz), and each orbital can hold a maximum of 2 electrons according to the Pauli exclusion principle.
Let's examine why the other options are incorrect:
a) 2 is incorrect because 2 is the maximum number of electrons that can be accommodated in a single p orbital, not the entire psubshell.
c) 10 is incorrect because 10 is the maximum number of electrons that can be accommodated in a dsubshell, not the psubshell.
d) 14 is incorrect because 14 is the maximum number of electrons that can be accommodated in an fsubshell, not the psubshell.
Unattempted
Note: Number of electrons in a subshell = 2(2ℓ + 1).
The correct option is b) 6.
The psubshell can accommodate a maximum of 6 electrons. This is because the psubshell consists of three orbitals (px, py, and pz), and each orbital can hold a maximum of 2 electrons according to the Pauli exclusion principle.
Let's examine why the other options are incorrect:
a) 2 is incorrect because 2 is the maximum number of electrons that can be accommodated in a single p orbital, not the entire psubshell.
c) 10 is incorrect because 10 is the maximum number of electrons that can be accommodated in a dsubshell, not the psubshell.
d) 14 is incorrect because 14 is the maximum number of electrons that can be accommodated in an fsubshell, not the psubshell.

Question 60 of 200
60. Question
0 pointsWhich subshell has highest energy?
Correct
The energy of an electron in an atom depends on its principal quantum number (n), azimuthal quantum number (ℓ), and magnetic quantum number (m).The general trend is that as the principal quantum number increases, the energy of the subshell also increases. Additionally, within a given principal quantum number, the energy increases as the azimuthal quantum number increases.
The correct option is a) n=5, ℓ=3, m=+1.
The energy of an electron in an atom depends on its principal quantum number (n), azimuthal quantum number (ℓ), and magnetic quantum number (m). The general trend is that as the principal quantum number increases, the energy of the subshell also increases. Additionally, within a given principal quantum number, the energy increases as the azimuthal quantum number increases.
Let's analyze the other options to understand why they are incorrect:
b) n=5, ℓ=2, m=+2: This subshell has a lower energy compared to option a. The azimuthal quantum number (ℓ) is lower, which indicates a lower energy level. Therefore, option b has a lower energy than option a.
c) n=4, ℓ=3, m=0: This subshell has a lower energy compared to option a. The principal quantum number (n) is lower, which indicates a lower energy level. Therefore, option c has a lower energy than option a.
d) n=4, ℓ=0, m=+1: This subshell has a lower energy compared to option a. The azimuthal quantum number (ℓ) is lower, which indicates a lower energy level. Therefore, option d has a lower energy than option a.
In summary, option a has the highest energy among the given choices because it has the highest principal quantum number (n=5) and highest azimuthal quantum number (ℓ=3).
Incorrect
The correct option is a) n=5, ℓ=3, m=+1.
The energy of an electron in an atom depends on its principal quantum number (n), azimuthal quantum number (ℓ), and magnetic quantum number (m). The general trend is that as the principal quantum number increases, the energy of the subshell also increases. Additionally, within a given principal quantum number, the energy increases as the azimuthal quantum number increases.
Let's analyze the other options to understand why they are incorrect:
b) n=5, ℓ=2, m=+2: This subshell has a lower energy compared to option a. The azimuthal quantum number (ℓ) is lower, which indicates a lower energy level. Therefore, option b has a lower energy than option a.
c) n=4, ℓ=3, m=0: This subshell has a lower energy compared to option a. The principal quantum number (n) is lower, which indicates a lower energy level. Therefore, option c has a lower energy than option a.
d) n=4, ℓ=0, m=+1: This subshell has a lower energy compared to option a. The azimuthal quantum number (ℓ) is lower, which indicates a lower energy level. Therefore, option d has a lower energy than option a.
In summary, option a has the highest energy among the given choices because it has the highest principal quantum number (n=5) and highest azimuthal quantum number (ℓ=3).
Unattempted
The correct option is a) n=5, ℓ=3, m=+1.
The energy of an electron in an atom depends on its principal quantum number (n), azimuthal quantum number (ℓ), and magnetic quantum number (m). The general trend is that as the principal quantum number increases, the energy of the subshell also increases. Additionally, within a given principal quantum number, the energy increases as the azimuthal quantum number increases.
Let's analyze the other options to understand why they are incorrect:
b) n=5, ℓ=2, m=+2: This subshell has a lower energy compared to option a. The azimuthal quantum number (ℓ) is lower, which indicates a lower energy level. Therefore, option b has a lower energy than option a.
c) n=4, ℓ=3, m=0: This subshell has a lower energy compared to option a. The principal quantum number (n) is lower, which indicates a lower energy level. Therefore, option c has a lower energy than option a.
d) n=4, ℓ=0, m=+1: This subshell has a lower energy compared to option a. The azimuthal quantum number (ℓ) is lower, which indicates a lower energy level. Therefore, option d has a lower energy than option a.
In summary, option a has the highest energy among the given choices because it has the highest principal quantum number (n=5) and highest azimuthal quantum number (ℓ=3).

Question 61 of 200
61. Question
0 pointsMagnetic quantum number values for the dsubshell are:
Correct
This option is incorrect because the total value of the magnetic quantum number (m) for the dsubshell is 5, not 2.
The correct option is c) 5.
The magnetic quantum number (m) represents the orientation of an orbital within a subshell. For the dsubshell, which has an azimuthal quantum number (ℓ) of 2, the possible values of the magnetic quantum number range from 2 to +2.
Now let's examine why the other options are incorrect:
a) 2: This option is incorrect because the total value of the magnetic quantum number (m) for the dsubshell is 5, not 2.
b) 3: This option is incorrect because the total value of the magnetic quantum number (m) for the dsubshell is 5, not 3.
d) 7: This option is incorrect because the total value of the magnetic quantum number (m) for the dsubshell is 5, not 7.
Therefore, the correct answer is c) 5, as it correctly represents the range of values for the magnetic quantum number in the dsubshell (2, 1, 0, 1, 2).
Incorrect
The correct option is c) 5.
The magnetic quantum number (m) represents the orientation of an orbital within a subshell. For the dsubshell, which has an azimuthal quantum number (ℓ) of 2, the possible values of the magnetic quantum number range from 2 to +2.
Now let's examine why the other options are incorrect:
a) 2: This option is incorrect because the total value of the magnetic quantum number (m) for the dsubshell is 5, not 2.
b) 3: This option is incorrect because the total value of the magnetic quantum number (m) for the dsubshell is 5, not 3.
d) 7: This option is incorrect because the total value of the magnetic quantum number (m) for the dsubshell is 5, not 7.
Therefore, the correct answer is c) 5, as it correctly represents the range of values for the magnetic quantum number in the dsubshell (2, 1, 0, 1, 2).
Unattempted
The correct option is c) 5.
The magnetic quantum number (m) represents the orientation of an orbital within a subshell. For the dsubshell, which has an azimuthal quantum number (ℓ) of 2, the possible values of the magnetic quantum number range from 2 to +2.
Now let's examine why the other options are incorrect:
a) 2: This option is incorrect because the total value of the magnetic quantum number (m) for the dsubshell is 5, not 2.
b) 3: This option is incorrect because the total value of the magnetic quantum number (m) for the dsubshell is 5, not 3.
d) 7: This option is incorrect because the total value of the magnetic quantum number (m) for the dsubshell is 5, not 7.
Therefore, the correct answer is c) 5, as it correctly represents the range of values for the magnetic quantum number in the dsubshell (2, 1, 0, 1, 2).

Question 62 of 200
62. Question
0 pointsA porbital has__________ energy than the sorbital of same principal quantum number
Correct
Lower is incorrect because the porbital has higher energy than the sorbital.
Note: Energy of subshell α(n+ ℓ), Energy order s
The correct option is b) Higher.
An sorbital has lower energy than a porbital of the same principal quantum number.
Lower is incorrect because, as mentioned above, the porbital has higher energy than the sorbital.Equal is incorrect because the energy of an sorbital is lower than that of a porbital. They are not equal in energy.
Vary orbit to orbit is incorrect because the comparison is between the sorbital and the porbital of the same principal quantum number. While energy levels may vary between different orbitals within the same subshell (such as between px, py, and pz orbitals), the energy of the sorbital is consistently lower than the energy of the porbital.
This difference in energy arises from the different shapes and orientations of the orbitals. The sorbital is spherically symmetric and has a lower energy, while the porbitals have a dumbbell shape and are oriented along the x, y, and z axes, resulting in higher energy
Incorrect
Note: Energy of subshell α(n+ ℓ), Energy order s
The correct option is b) Higher.
An sorbital has lower energy than a porbital of the same principal quantum number.
Lower is incorrect because, as mentioned above, the porbital has higher energy than the sorbital.
Equal is incorrect because the energy of an sorbital is lower than that of a porbital. They are not equal in energy.
Vary orbit to orbit is incorrect because the comparison is between the sorbital and the porbital of the same principal quantum number. While energy levels may vary between different orbitals within the same subshell (such as between px, py, and pz orbitals), the energy of the sorbital is consistently lower than the energy of the porbital.
This difference in energy arises from the different shapes and orientations of the orbitals. The sorbital is spherically symmetric and has a lower energy, while the porbitals have a dumbbell shape and are oriented along the x, y, and z axes, resulting in higher energy.
Unattempted
Note: Energy of subshell α(n+ ℓ), Energy order s
The correct option is b) Higher.
An sorbital has lower energy than a porbital of the same principal quantum number.
Lower is incorrect because, as mentioned above, the porbital has higher energy than the sorbital.
Equal is incorrect because the energy of an sorbital is lower than that of a porbital. They are not equal in energy.
Vary orbit to orbit is incorrect because the comparison is between the sorbital and the porbital of the same principal quantum number. While energy levels may vary between different orbitals within the same subshell (such as between px, py, and pz orbitals), the energy of the sorbital is consistently lower than the energy of the porbital.
This difference in energy arises from the different shapes and orientations of the orbitals. The sorbital is spherically symmetric and has a lower energy, while the porbitals have a dumbbell shape and are oriented along the x, y, and z axes, resulting in higher energy.

Question 63 of 200
63. Question
0 pointsWhich set of quantum number represents 29^{th} electron of Cu atom
Correct
It represents the 29th electron being in the 4s orbital of the copper atom.
The correct option is a) n=4, ℓ=0, m=0, s=+½.
The quantum numbers provide specific information about the electron's energy level, orbital shape, orientation, and spin. Let's analyze why the other options are incorrect:
Option b) n=3, ℓ=2, m=0, s=+½: This set of quantum numbers represents the 29th electron being in the 3d orbital. However, copper (Cu) has 29 electrons, and the filling order of electrons follows the Aufbau principle. The 4s orbital is filled before the 3d orbital. Therefore, the 29th electron should occupy the 4s orbital, not the 3d orbital.
Option c) n=4, ℓ=1, m=0, s=+½: This set of quantum numbers represents the 29th electron being in the 4p orbital. However, as mentioned above, the 29th electron occupies the 4s orbital in copper, not the 4p orbital.
Option d) n=3, ℓ=0, m=0, s=+½: This set of quantum numbers represents the 29th electron being in the 3s orbital. However, copper has 29 electrons, and the 3s orbital can accommodate only 2 electrons. Therefore, the 29th electron cannot be in the 3s orbital.
Hence, option a) n=4, ℓ=0, m=0, s=+½ is correct. It represents the 29th electron being in the 4s orbital of the copper atom.
Incorrect
The correct option is a) n=4, ℓ=0, m=0, s=+½.
The quantum numbers provide specific information about the electron's energy level, orbital shape, orientation, and spin. Let's analyze why the other options are incorrect:
Option b) n=3, ℓ=2, m=0, s=+½: This set of quantum numbers represents the 29th electron being in the 3d orbital. However, copper (Cu) has 29 electrons, and the filling order of electrons follows the Aufbau principle. The 4s orbital is filled before the 3d orbital. Therefore, the 29th electron should occupy the 4s orbital, not the 3d orbital.
Option c) n=4, ℓ=1, m=0, s=+½: This set of quantum numbers represents the 29th electron being in the 4p orbital. However, as mentioned above, the 29th electron occupies the 4s orbital in copper, not the 4p orbital.
Option d) n=3, ℓ=0, m=0, s=+½: This set of quantum numbers represents the 29th electron being in the 3s orbital. However, copper has 29 electrons, and the 3s orbital can accommodate only 2 electrons. Therefore, the 29th electron cannot be in the 3s orbital.
Hence, option a) n=4, ℓ=0, m=0, s=+½ is correct. It represents the 29th electron being in the 4s orbital of the copper atom.
Unattempted
The correct option is a) n=4, ℓ=0, m=0, s=+½.
The quantum numbers provide specific information about the electron's energy level, orbital shape, orientation, and spin. Let's analyze why the other options are incorrect:
Option b) n=3, ℓ=2, m=0, s=+½: This set of quantum numbers represents the 29th electron being in the 3d orbital. However, copper (Cu) has 29 electrons, and the filling order of electrons follows the Aufbau principle. The 4s orbital is filled before the 3d orbital. Therefore, the 29th electron should occupy the 4s orbital, not the 3d orbital.
Option c) n=4, ℓ=1, m=0, s=+½: This set of quantum numbers represents the 29th electron being in the 4p orbital. However, as mentioned above, the 29th electron occupies the 4s orbital in copper, not the 4p orbital.
Option d) n=3, ℓ=0, m=0, s=+½: This set of quantum numbers represents the 29th electron being in the 3s orbital. However, copper has 29 electrons, and the 3s orbital can accommodate only 2 electrons. Therefore, the 29th electron cannot be in the 3s orbital.
Hence, option a) n=4, ℓ=0, m=0, s=+½ is correct. It represents the 29th electron being in the 4s orbital of the copper atom.

Question 64 of 200
64. Question
0 pointsThe dorbital which is bilobed with collar is represented as
Correct
The dxy orbital is characterized by lobes along the x and y axes, without the presence of a collar. It does not exhibit the bilobed structure with a collar mentioned in the question.
The correct option is c) dz2
.The dorbital which is bilobed with a collar is represented by the dz2 orbital.
Option a) dxy: The dxy orbital is characterized by lobes along the x and y axes, without the presence of a collar. It does not exhibit the bilobed structure with a collar mentioned in the question.
Option b) dyz: The dyz orbital is characterized by lobes along the y and z axes, without the presence of a collar. It also does not possess the bilobed structure with a collar mentioned in the question.
Option d) dz2y2
: The dz2y2
orbital is characterized by two lobes along the zaxis and a nodal plane along the yaxis. It does not exhibit the specific bilobed structure with a collar mentioned in the question.The dz2 orbital, on the other hand, has two lobes along the zaxis and a ringshaped collar in the xy plane. This collar is responsible for the “bilobed with a collar” description mentioned in the question.
Incorrect
The correct option is c) dz^{2}.
The dorbital which is bilobed with a collar is represented by the dz^{2} orbital.
Option a) dxy: The dxy orbital is characterized by lobes along the x and y axes, without the presence of a collar. It does not exhibit the bilobed structure with a collar mentioned in the question.
Option b) dyz: The dyz orbital is characterized by lobes along the y and z axes, without the presence of a collar. It also does not possess the bilobed structure with a collar mentioned in the question.
Option d) dz^{2}y^{2}: The dz^{2}y^{2} orbital is characterized by two lobes along the zaxis and a nodal plane along the yaxis. It does not exhibit the specific bilobed structure with a collar mentioned in the question.
The dz2 orbital, on the other hand, has two lobes along the zaxis and a ringshaped collar in the xy plane. This collar is responsible for the “bilobed with a collar” description mentioned in the question.
Unattempted
The correct option is c) dz^{2}.
The dorbital which is bilobed with a collar is represented by the dz^{2} orbital.
Option a) dxy: The dxy orbital is characterized by lobes along the x and y axes, without the presence of a collar. It does not exhibit the bilobed structure with a collar mentioned in the question.
Option b) dyz: The dyz orbital is characterized by lobes along the y and z axes, without the presence of a collar. It also does not possess the bilobed structure with a collar mentioned in the question.
Option d) dz^{2}y^{2}: The dz^{2}y^{2} orbital is characterized by two lobes along the zaxis and a nodal plane along the yaxis. It does not exhibit the specific bilobed structure with a collar mentioned in the question.
The dz2 orbital, on the other hand, has two lobes along the zaxis and a ringshaped collar in the xy plane. This collar is responsible for the “bilobed with a collar” description mentioned in the question.

Question 65 of 200
65. Question
0 points____________is/are isoelectronic with K^{+}
Correct
Phosphorus (P) normally has 15 electrons. By gaining three electrons, it forms the P3 ion, which has a configuration of 2, 8, 8, 2. This configuration is isoelectronic with K+ (2, 8, 8)
Incorrect
The correct option is d) All of these.
Isoelectronic species are atoms or ions that have the same number of electrons. K+ is a potassium ion that has lost one electron, resulting in a configuration of 2, 8, 8. Let's analyze each option:
a) P3: Phosphorus (P) normally has 15 electrons. By gaining three electrons, it forms the P3 ion, which has a configuration of 2, 8, 8, 2. This configuration is isoelectronic with K+ (2, 8, 8).
b) Si4: Silicon (Si) normally has 14 electrons. By gaining four electrons, it forms the Si4 ion, which has a configuration of 2, 8, 8, 4. This configuration is isoelectronic with K+ (2, 8, 8).
c) S2: Sulfur (S) normally has 16 electrons. By gaining two electrons, it forms the S2 ion, which has a configuration of 2, 8, 8, 6. This configuration is isoelectronic with K+ (2, 8, 8).
Therefore, all three options (a, b, and c) are correct because P3, Si4, and S2 are isoelectronic with K+. They all have the same number of electrons as K+ and share the same electron configuration of 2, 8, 8.
So, the correct option is d) All of these.
Unattempted
The correct option is d) All of these.
Isoelectronic species are atoms or ions that have the same number of electrons. K+ is a potassium ion that has lost one electron, resulting in a configuration of 2, 8, 8. Let's analyze each option:
a) P3: Phosphorus (P) normally has 15 electrons. By gaining three electrons, it forms the P3 ion, which has a configuration of 2, 8, 8, 2. This configuration is isoelectronic with K+ (2, 8, 8).
b) Si4: Silicon (Si) normally has 14 electrons. By gaining four electrons, it forms the Si4 ion, which has a configuration of 2, 8, 8, 4. This configuration is isoelectronic with K+ (2, 8, 8).
c) S2: Sulfur (S) normally has 16 electrons. By gaining two electrons, it forms the S2 ion, which has a configuration of 2, 8, 8, 6. This configuration is isoelectronic with K+ (2, 8, 8).
Therefore, all three options (a, b, and c) are correct because P3, Si4, and S2 are isoelectronic with K+. They all have the same number of electrons as K+ and share the same electron configuration of 2, 8, 8.
So, the correct option is d) All of these.

Question 66 of 200
66. Question
0 pointsThe atomic number of an element is 26. How many electrons are present in M Shell of this element in ground state?
Correct
11 does not represent the correct number of electrons in the Mshell of iron.
Incorrect
The correct option is b) 14.
To determine the number of electrons present in the Mshell of an element with an atomic number of 26, we need to determine the electron configuration of the element.
The atomic number 26 corresponds to the element iron (Fe). The electron configuration of iron is:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶
In the ground state, all the lower energy levels (shells) will be filled before the Mshell. Therefore, the Mshell represents the 3rd shell (n=3). The Mshell consists of the 3s, 3p, and 3d subshells.
The 3s subshell can hold a maximum of 2 electrons, the 3p subshell can hold a maximum of 6 electrons, and the 3d subshell can hold a maximum of 10 electrons.
To determine the number of electrons in the Mshell, we need to add up the electrons in the 3s, 3p, and 3d subshells.
3s² + 3p⁶ + 3d⁶ = 2 + 6 + 6 = 14
Therefore, in the Mshell of iron (Fe) in its ground state, there are 14 electrons.
Hence, the correct option is b) 14. The other options (a, c, d) are incorrect because they do not represent the correct number of electrons in the Mshell of iron.
Unattempted
The correct option is b) 14.
To determine the number of electrons present in the Mshell of an element with an atomic number of 26, we need to determine the electron configuration of the element.
The atomic number 26 corresponds to the element iron (Fe). The electron configuration of iron is:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶
In the ground state, all the lower energy levels (shells) will be filled before the Mshell. Therefore, the Mshell represents the 3rd shell (n=3). The Mshell consists of the 3s, 3p, and 3d subshells.
The 3s subshell can hold a maximum of 2 electrons, the 3p subshell can hold a maximum of 6 electrons, and the 3d subshell can hold a maximum of 10 electrons.
To determine the number of electrons in the Mshell, we need to add up the electrons in the 3s, 3p, and 3d subshells.
3s² + 3p⁶ + 3d⁶ = 2 + 6 + 6 = 14
Therefore, in the Mshell of iron (Fe) in its ground state, there are 14 electrons.
Hence, the correct option is b) 14. The other options (a, c, d) are incorrect because they do not represent the correct number of electrons in the Mshell of iron.

Question 67 of 200
67. Question
0 pointsThe electronic configuration of an element is 1s^{2}, 2s^{2}, 2P^{1}_{x}, 2P^{1}_{y}, 2P^{1}_{z}.This represesnts a/an:
Correct
The ground state of an atom is its most stable and lowest energy electron configuration. In this state, electrons fill the available orbitals starting from the lowest energy level (closest to the nucleus) and following the Pauli exclusion principle and Hund's rule. The given electronic configuration “1s^{2}, 2s^{2}, 2P^{1}_{x}, 2P^{1}_{y}, 2P^{1}_{z}” follows this pattern, with electrons filling the 1s and 2s orbitals completely and then populating the three available 2p orbitals with a total of 5 electrons. This arrangement represents the ground state of a nitrogen atom.
Incorrect
The ground state of an atom is its most stable and lowest energy electron configuration. In this state, electrons fill the available orbitals starting from the lowest energy level (closest to the nucleus) and following the Pauli exclusion principle and Hund's rule. The given electronic configuration “1s^{2}, 2s^{2}, 2P^{1}_{x}, 2P^{1}_{y}, 2P^{1}_{z}” follows this pattern, with electrons filling the 1s and 2s orbitals completely and then populating the three available 2p orbitals with a total of 5 electrons. This arrangement represents the ground state of a nitrogen atom.
Unattempted
The ground state of an atom is its most stable and lowest energy electron configuration. In this state, electrons fill the available orbitals starting from the lowest energy level (closest to the nucleus) and following the Pauli exclusion principle and Hund's rule. The given electronic configuration “1s^{2}, 2s^{2}, 2P^{1}_{x}, 2P^{1}_{y}, 2P^{1}_{z}” follows this pattern, with electrons filling the 1s and 2s orbitals completely and then populating the three available 2p orbitals with a total of 5 electrons. This arrangement represents the ground state of a nitrogen atom.

Question 68 of 200
68. Question
0 pointsThe correct electronic configuration of _{29}Cu is
Correct
Incorrect
Incorrect
Electronic configuration of _{29}Cu = [Ar], 4s^{1}, 3d^{10}
Unattempted
Electronic configuration of _{29}Cu = [Ar], 4s^{1}, 3d^{10}

Question 69 of 200
69. Question
0 pointsWhich one of the following ion has similar electronic configuration like Ar?
Correct
Scandium (Sc) has an atomic number of 21. If it loses 2 electrons to become Sc^{+2}, its electronic configuration becomes: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d^{1} 4s⁰. This configuration is not similar to argon's electronic configuration, which is 1s² 2s² 2p⁶ 3s² 3p⁶. Therefore, Sc+2 is not the correct answer.
Incorrect
Titanium (Ti) has an atomic number of 22. If it loses 4 electrons to become Ti+4, its electronic configuration becomes: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁰ 4s⁰. This configuration is identical to argon's electronic configuration (1s² 2s² 2p⁶ 3s² 3p⁶). Therefore, Ti+4 is the correct answer as it has a similar electronic configuration to argon.
Unattempted
Titanium (Ti) has an atomic number of 22. If it loses 4 electrons to become Ti+4, its electronic configuration becomes: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁰ 4s⁰. This configuration is identical to argon's electronic configuration (1s² 2s² 2p⁶ 3s² 3p⁶). Therefore, Ti+4 is the correct answer as it has a similar electronic configuration to argon.

Question 70 of 200
70. Question
0 pointsIn electronic configuration of Cu and Cr,which of following is true?
Correct
This option is true for copper (Cu) as its 3d orbital is fully filled with 10 electrons, but it is not true for chromium (Cr) as its 3d orbital is not fully filled(half filled)
Incorrect
Option C is true for both copper and chromium. Both copper and chromium have electrons in their sorbitals: Cu has 1 electron in the 4s orbital, and Cr has 1 electron in the 4s orbital.
Unattempted
Option C is true for both copper and chromium. Both copper and chromium have electrons in their sorbitals: Cu has 1 electron in the 4s orbital, and Cr has 1 electron in the 4s orbital.

Question 71 of 200
71. Question
0 pointsWhich of the following violates Hund's rule?
Correct
This option does not violate Hund's rule. Electrons are first singly placed in the 2p orbitals before they start pairing up.
Incorrect
Option D violates Hund's rule. According to Hund's rule, electrons preferentially occupy degenerate orbitals singly before pairing up. In this option, the 2p orbitals (2p^{2}_{x}, 2p^{0}_{y}, 2p^{0}_{z}) should be singly occupied first before pairing, but two electrons are placed in the 2p^{2}_{x} orbital before any are placed in the 2p^{0}_{y} and 2p^{0}_{z} orbitals. This arrangement violates Hund's rule.
Unattempted
Option D violates Hund's rule. According to Hund's rule, electrons preferentially occupy degenerate orbitals singly before pairing up. In this option, the 2p orbitals (2p^{2}_{x}, 2p^{0}_{y}, 2p^{0}_{z}) should be singly occupied first before pairing, but two electrons are placed in the 2p^{2}_{x} orbital before any are placed in the 2p^{0}_{y} and 2p^{0}_{z} orbitals. This arrangement violates Hund's rule.

Question 72 of 200
72. Question
0 pointsWhich one of the following represents most stable configuration of three electrons for ground state of an element in group VA
Correct
This option suggests that two electrons are in one orbital (np^{2}_{x}), one electron is in a different orbital (np^{1}_{y}), and no electrons are in another orbital (np^{0}_{Z}). This arrangement is not following the expected ground state electron filling pattern.
Incorrect
Halffilled and completely filled arbitas are relatively more stable.
Option B suggests that each of the three electrons is in a different orbital (np^{1}_{x}, np^{1}_{Y}, np^{1}_{Z}). This arrangement follows Hund's rule and is a valid ground state configuration for three electrons in group VA.Unattempted
Halffilled and completely filled arbitas are relatively more stable.
Option B suggests that each of the three electrons is in a different orbital (np^{1}_{x}, np^{1}_{Y}, np^{1}_{Z}). This arrangement follows Hund's rule and is a valid ground state configuration for three electrons in group VA. 
Question 73 of 200
73. Question
0 pointsWhich one of the following electronic configuration represents an element that will form a simple ion with 3 charge?
Correct
This configuration suggests that the element has a total of 9 valence electrons. To form a 3 charged ion, the element would need to gain 3 electrons. However, this would result in a configuration of 1s² 2s² 2p⁶ 3s² 3p⁴, which is not the expected configuration.
Incorrect
This configuration accurately represents the electronic configuration of phosphorus (P). It has 5 valence electrons in the 3s and 3p orbitals. To form a 3 charged ion, phosphorus would gain 3 electrons, resulting in a stable electron configuration by completing its octet: 1s² 2s² 2p⁶ 3s² 3p⁶.
Unattempted
This configuration accurately represents the electronic configuration of phosphorus (P). It has 5 valence electrons in the 3s and 3p orbitals. To form a 3 charged ion, phosphorus would gain 3 electrons, resulting in a stable electron configuration by completing its octet: 1s² 2s² 2p⁶ 3s² 3p⁶.

Question 74 of 200
74. Question
0 pointsA specie Z has following electronic configuration 1s^{2},2s^{2},2P_{x}^{2},2p_{y}^{2},2p_{y}^{2},3s^{2},3p_{x}^{2},3p_{y}^{1},3p_{z}^{1.} What could Z be
Correct
The electronic configuration of argon is [Ne]3s² 3p⁶. This does not match the given configuration, so Ar is not the correct answer.
Incorrect
The electronic configuration of sulfur is [Ne]3s² 3p⁴. The given configuration matches the electron distribution of sulfur (S), with 16 electrons and the distribution 1s² 2s² 2px² 2py² 2pz² 3s² 3px² 3py¹ 3pz¹. Therefore, S is the correct answer.
Unattempted
The electronic configuration of sulfur is [Ne]3s² 3p⁴. The given configuration matches the electron distribution of sulfur (S), with 16 electrons and the distribution 1s² 2s² 2px² 2py² 2pz² 3s² 3px² 3py¹ 3pz¹. Therefore, S is the correct answer.

Question 75 of 200
75. Question
0 pointsThe total number of lobes in all the orbitals of a dsubshell are:
Correct
This option suggests that the total number of lobes in all the d orbitals is only 2. However, each d orbital has more than 2 lobes due to their complex shapes. This answer is not accurate.
Incorrect
Option D suggests that the total number of lobes in all the d orbitals is 18. Each d orbital has a specific number of lobes: dxy, dxz, dyz, and dx²y² each have 4 lobes, while dz² has 2. Adding these up gives a total of 18 lobes.
Unattempted
Option D suggests that the total number of lobes in all the d orbitals is 18. Each d orbital has a specific number of lobes: dxy, dxz, dyz, and dx²y² each have 4 lobes, while dz² has 2. Adding these up gives a total of 18 lobes.

Question 76 of 200
76. Question
0 points“Two electrons in the same orbital should have opposite spins” according to:
Correct
Aufbau's principle states that electrons fill the lowest energy orbitals first. It explains the order in which electrons occupy different orbitals but does not directly address the spin of electrons in the same orbital.
Incorrect
Pauli's exclusion principle states that no two electrons in an atom can have the same set of four quantum numbers, and furthermore, electrons in the same orbital must have opposite spins. This principle is precisely related to the statement given.
Unattempted
Pauli's exclusion principle states that no two electrons in an atom can have the same set of four quantum numbers, and furthermore, electrons in the same orbital must have opposite spins. This principle is precisely related to the statement given.

Question 77 of 200
77. Question
0 pointsIf more than one degenerate orbitals are available then electrons go one by one with same spin. According to which rule:
Correct
If more than one degenerate orbitals are available then electrons go one by one with same spin. is called hund's rule The rule that electrons go one by one with the same spin in degenerate orbitals is called Hund's rule of maximum multiplicity. This rule states that electrons will occupy degenerate orbitals singly before they pair up, and that they will have the same spin in each orbital. This is because electrons have a property called spin, which can be either up or down. The Pauli exclusion principle states that no two electrons in an atom can have the same four quantum numbers. This means that if two electrons are in the same orbital, they must have different spins.
Incorrect
Option A is correct.
If more than one degenerate orbitals are available then electrons go one by one with same spin is called hund's rule.The rule states that electrons go one by one with the same spin in degenerate orbitals and is called Hund's rule of maximum multiplicity. This rule states that electrons will occupy degenerate orbitals singly before they pair up, and that they will have the same spin in each orbital. This is because electrons have a property called spin, which can be either up or down.The Pauli exclusion principle states that no two electrons in an atom can have the same four quantum numbers. This means that if two electrons are in the same orbital, they must have different(opposite) spins.Unattempted
Option A is correct.
If more than one degenerate orbitals are available then electrons go one by one with same spin is called hund's rule.The rule states that electrons go one by one with the same spin in degenerate orbitals and is called Hund's rule of maximum multiplicity. This rule states that electrons will occupy degenerate orbitals singly before they pair up, and that they will have the same spin in each orbital. This is because electrons have a property called spin, which can be either up or down.The Pauli exclusion principle states that no two electrons in an atom can have the same four quantum numbers. This means that if two electrons are in the same orbital, they must have different(opposite) spins. 
Question 78 of 200
78. Question
0 pointsThe e/m value for the canal rays is maximum for
Correct
The e/m value for canal rays is maximum for hydrogen. This is because hydrogen has the lowest mass of all the elements, so the e/m value will be the highest. The other elements you mentioned have higher masses, so their e/m values will be lower. Canal rays are also known as anode rays. They are positively charged ions that are produced when a gas is ionized in a discharge tube. The e/m value of canal rays is a measure of the chargetomass ratio of the ions. The higher the e/m value, the lighter the ions are.The e/m value for canal rays is maximum for hydrogen, and it is equal to 1.
Incorrect
Option A is correct.
The e/m value for canal rays is maximum for hydrogen. This is because hydrogen has the lowest mass of all the elements, so the e/m value will be the highest. The other elements you mentioned have higher masses, so their e/m values will be lower. Canal rays are also known as anode rays. They are positively charged ions that are produced when a gas is ionized in a discharge tube. The e/m value of canal rays is a measure of the chargetomass ratio of the ions. The higher the e/m value, the lighter the ions are.The e/m value for canal rays is maximum for hydrogen, and it is equal to 1.Unattempted
Option A is correct.
The e/m value for canal rays is maximum for hydrogen. This is because hydrogen has the lowest mass of all the elements, so the e/m value will be the highest. The other elements you mentioned have higher masses, so their e/m values will be lower. Canal rays are also known as anode rays. They are positively charged ions that are produced when a gas is ionized in a discharge tube. The e/m value of canal rays is a measure of the chargetomass ratio of the ions. The higher the e/m value, the lighter the ions are.The e/m value for canal rays is maximum for hydrogen, and it is equal to 1. 
Question 79 of 200
79. Question
0 pointsCorrect order of energy in the given subshells is:
Correct
Option C is correct.
The correct order of energy in the given subshells is 5s>3d>4s>3p.
The energy of an electron subshell is determined by two factors: the principal quantum number (n) and the angular momentum quantum number (l). The principal quantum number tells us the energy level of the subshell, and the angular momentum quantum number tells us the shape of the subshell. In general, subshells with higher principal quantum numbers have higher energy than subshells with lower principal quantum numbers. So, 5s has a higher energy than 4s, and 3d has a higher energy than 3p. However, the angular momentum quantum number also plays a role in determining the energy of a subshell. Subshells with the same principal quantum number but different angular momentum quantum numbers have different energies, with subshells with higher angular momentum quantum numbers having higher energy. So, 3d has a higher energy than 3p, even though they have the same principal quantum number.
Therefore, the correct order of energy in the given subshells is 5s>3d>4s>3p. Option C is correct.
The correct order of energy in the given subshells is 5s>3d>4s>3p.
The energy of an electron subshell is determined by two factors: the principal quantum number (n) and the angular momentum quantum number (l). The principal quantum number tells us the energy level of the subshell, and the angular momentum quantum number tells us the shape of the subshell. In general, subshells with higher principal quantum numbers have higher energy than subshells with lower principal quantum numbers. So, 5s has a higher energy than 4s, and 3d has a higher energy than 3p. However, the angular momentum quantum number also plays a role in determining the energy of a subshell. Subshells with the same principal quantum number but different angular momentum quantum numbers have different energies, with subshells with higher angular momentum quantum numbers having higher energy. So, 3d has a higher energy than 3p, even though they have the same principal quantum number.
Therefore, the correct order of energy in the given subshells is 5s>3d>4s>3p.Incorrect
Option C is correct.
The correct order of energy in the given subshells is 5s>3d>4s>3p.
The energy of an electron subshell is determined by two factors: the principal quantum number (n) and the angular momentum quantum number (l). The principal quantum number tells us the energy level of the subshell, and the angular momentum quantum number tells us the shape of the subshell. In general, subshells with higher principal quantum numbers have higher energy than subshells with lower principal quantum numbers. So, 5s has a higher energy than 4s, and 3d has a higher energy than 3p. However, the angular momentum quantum number also plays a role in determining the energy of a subshell. Subshells with the same principal quantum number but different angular momentum quantum numbers have different energies, with subshells with higher angular momentum quantum numbers having higher energy. So, 3d has a higher energy than 3p, even though they have the same principal quantum number.
Therefore, the correct order of energy in the given subshells is 5s>3d>4s>3p.Unattempted
Option C is correct.
The correct order of energy in the given subshells is 5s>3d>4s>3p.
The energy of an electron subshell is determined by two factors: the principal quantum number (n) and the angular momentum quantum number (l). The principal quantum number tells us the energy level of the subshell, and the angular momentum quantum number tells us the shape of the subshell. In general, subshells with higher principal quantum numbers have higher energy than subshells with lower principal quantum numbers. So, 5s has a higher energy than 4s, and 3d has a higher energy than 3p. However, the angular momentum quantum number also plays a role in determining the energy of a subshell. Subshells with the same principal quantum number but different angular momentum quantum numbers have different energies, with subshells with higher angular momentum quantum numbers having higher energy. So, 3d has a higher energy than 3p, even though they have the same principal quantum number.
Therefore, the correct order of energy in the given subshells is 5s>3d>4s>3p. 
Question 80 of 200
80. Question
0 pointsMaximum numbers of electrons in a subshell is given by:
Correct
The maximum number of electrons in a subshell is given by 2(2l + 1). This formula is based on the Pauli exclusion principle, which states that no two electrons in an atom can have the same four quantum numbers. The four quantum numbers are the principal quantum number (n), the angular momentum quantum number (l), the magnetic quantum number (ml), and the spin quantum number (ms). The angular momentum quantum number (l) determines the shape of the orbital, and it can have values from 0 to n – 1. The maximum number of electrons in a subshell is therefore 2(2l + 1), where l is the angular momentum quantum number.
No. of orbitals in a subshell = (2ℓ +1)No. of electrons in a subshell = 2(2ℓ +I)Incorrect
Option B is correct.
The maximum number of electrons in a subshell is given by 2(2l + 1). This formula is based on the Pauli exclusion principle, which states that no two electrons in an atom can have the same four quantum numbers. The four quantum numbers are the principal quantum number (n), the angular momentum quantum number (l), the magnetic quantum number (ml), and the spin quantum number (ms). The angular momentum quantum number (l) determines the shape of the orbital, and it can have values from 0 to n – 1. The maximum number of electrons in a subshell is therefore 2(2l + 1), where l is the angular momentum quantum number.
No. of orbitals in a subshell = (2ℓ +1)
No. of electrons in a subshell = 2(2ℓ +I)Unattempted
Option B is correct.
The maximum number of electrons in a subshell is given by 2(2l + 1). This formula is based on the Pauli exclusion principle, which states that no two electrons in an atom can have the same four quantum numbers. The four quantum numbers are the principal quantum number (n), the angular momentum quantum number (l), the magnetic quantum number (ml), and the spin quantum number (ms). The angular momentum quantum number (l) determines the shape of the orbital, and it can have values from 0 to n – 1. The maximum number of electrons in a subshell is therefore 2(2l + 1), where l is the angular momentum quantum number.
No. of orbitals in a subshell = (2ℓ +1)
No. of electrons in a subshell = 2(2ℓ +I) 
Question 81 of 200
81. Question
0 pointsThe set of orbitals having n + / = 5 are:
Correct
The 2p orbital has n = 2 and l = 1, so n + l = 3.
The set of orbitals having n + l = 5 are 5s, 4p, 3d. The principal quantum number (n) tells us the energy level of the orbital, and the angular momentum quantum number (l) tells us the shape of the orbital. The sum of n + l is called the total angular momentum quantum number.
For the 5s orbital, n = 5 and l = 0, so n + l = 5.
For the 4p orbital, n = 4 and l = 1, so n + l = 5.
For the 3d orbital, n = 3 and l = 2, so n + l = 5.
The other sets of orbitals mentioned do not have a total angular momentum quantum number of 5.Incorrect
Option D is correct.
The set of orbitals having n + l = 5 are 5s, 4p, 3d. The principal quantum number (n) tells us the energy level of the orbital, and the angular momentum quantum number (l) tells us the shape of the orbital. The sum of n + l is called the total angular momentum quantum number.
For the 5s orbital, n = 5 and l = 0, so n + l = 5.
For the 4p orbital, n = 4 and l = 1, so n + l = 5.
For the 3d orbital, n = 3 and l = 2, so n + l = 5.
The other sets of orbitals mentioned do not have a total angular momentum quantum number of 5.Unattempted
Option D is correct.
The set of orbitals having n + l = 5 are 5s, 4p, 3d. The principal quantum number (n) tells us the energy level of the orbital, and the angular momentum quantum number (l) tells us the shape of the orbital. The sum of n + l is called the total angular momentum quantum number.
For the 5s orbital, n = 5 and l = 0, so n + l = 5.
For the 4p orbital, n = 4 and l = 1, so n + l = 5.
For the 3d orbital, n = 3 and l = 2, so n + l = 5.
The other sets of orbitals mentioned do not have a total angular momentum quantum number of 5. 
Question 82 of 200
82. Question
0 pointsThe ratio of e/m values of a proton and an a—particle is:
Correct
The ratio of e/m values of a proton and an alpha particle is 2:1.
The charge of a proton is +1, and its mass is 1 u. The charge of an alpha particle is +2, and its mass is 4 u. The e/m value of a proton is therefore 1/1 = 1, and the e/m value of an alpha particle is 2/4 = 0.5. The ratio of these two values is 2/0.5 = 2:1.
Therefore, the answer is 2:1.Incorrect
Option A is correct.
The ratio of e/m values of a proton and an alpha particle is 2:1.
The charge of a proton is +1, and its mass is 1 u. The charge of an alpha particle is +2, and its mass is 4 u. The e/m value of a proton is therefore 1/1 = 1, and the e/m value of an alpha particle is 2/4 = 0.5. The ratio of these two values is 2/0.5 = 2:1.
Therefore, the answer is 2:1.Unattempted
Option A is correct.
The ratio of e/m values of a proton and an alpha particle is 2:1.
The charge of a proton is +1, and its mass is 1 u. The charge of an alpha particle is +2, and its mass is 4 u. The e/m value of a proton is therefore 1/1 = 1, and the e/m value of an alpha particle is 2/4 = 0.5. The ratio of these two values is 2/0.5 = 2:1.
Therefore, the answer is 2:1. 
Question 83 of 200
83. Question
0 pointsTwo electrons present in same orbital can be distinguished by:
Correct
The principal quantum number (n) is a quantum number that describes the energy level of an electron in an atom. It is the first of a set of quantum numbers that describe the unique quantum state of an electron (the others being the azimuthal quantum number ℓ, the magnetic quantum number ml, and the spin quantum number ms). It is denoted by the symbol n.
Incorrect
Option C is correct.
Two electrons present in the same orbital can be distinguished by their spin quantum number. The spin quantum number can have two values, +1/2 and 1/2. This means that there are two possible ways for two electrons to occupy the same orbital, with opposite spins.
The principal quantum number, azimuthal quantum number, and magnetic quantum number all describe the spatial distribution of an electron in an atom. They do not tell us anything about the spin of the electron. Therefore, the answer is spin quantum number.The principal quantum number, azimuthal quantum number, and magnetic quantum number all describe the spatial distribution of an electron in an atom. They do not tell us anything about the spin of the electron. Therefore, the answer is spin quantum numberUnattempted
Option C is correct.
Two electrons present in the same orbital can be distinguished by their spin quantum number. The spin quantum number can have two values, +1/2 and 1/2. This means that there are two possible ways for two electrons to occupy the same orbital, with opposite spins.
The principal quantum number, azimuthal quantum number, and magnetic quantum number all describe the spatial distribution of an electron in an atom. They do not tell us anything about the spin of the electron. Therefore, the answer is spin quantum number.The principal quantum number, azimuthal quantum number, and magnetic quantum number all describe the spatial distribution of an electron in an atom. They do not tell us anything about the spin of the electron. Therefore, the answer is spin quantum number 
Question 84 of 200
84. Question
0 pointsAn atom has 35 nucleons and has atomic number equal to 17. The number of electrons with n = 2, I = 0 in it is:
Correct
The number of electrons with n = 2, l = 0 in an atom with 35 nucleons and an atomic number of 17 is 2. The principal quantum number (n) tells us the energy level of the electron, and the angular momentum quantum number (l) tells us the shape of the orbital. The value of n = 2 tells us that the electron is in the second energy level, and the value of l = 0 tells us that the orbital is an s orbital. The s orbital in the second energy level can hold up to 2 electrons, so the number of electrons with n = 2, l = 0 in an atom with 35 nucleons and an atomic number of 17 is 2.
Here is the electron configuration of chlorine, which has an atomic number of 17:
1s2 2s2 2p6 3s2 3p5
The electrons with n = 2, l = 0 are the 2s2 electrons. There are 2 of these electrons in chlorine, so the answer is 2.Incorrect
Option A is correct.
The number of electrons with n = 2, l = 0 in an atom with 35 nucleons and an atomic number of 17 is 2. The principal quantum number (n) tells us the energy level of the electron, and the angular momentum quantum number (l) tells us the shape of the orbital. The value of n = 2 tells us that the electron is in the second energy level, and the value of l = 0 tells us that the orbital is an s orbital. The s orbital in the second energy level can hold up to 2 electrons, so the number of electrons with n = 2, l = 0 in an atom with 35 nucleons and an atomic number of 17 is 2.
Here is the electron configuration of chlorine, which has an atomic number of 17:
1s2 2s2 2p6 3s2 3p5
The electrons with n = 2, l = 0 are the 2s2 electrons. There are 2 of these electrons in chlorine, so the answer is 2.Unattempted
Option A is correct.
The number of electrons with n = 2, l = 0 in an atom with 35 nucleons and an atomic number of 17 is 2. The principal quantum number (n) tells us the energy level of the electron, and the angular momentum quantum number (l) tells us the shape of the orbital. The value of n = 2 tells us that the electron is in the second energy level, and the value of l = 0 tells us that the orbital is an s orbital. The s orbital in the second energy level can hold up to 2 electrons, so the number of electrons with n = 2, l = 0 in an atom with 35 nucleons and an atomic number of 17 is 2.
Here is the electron configuration of chlorine, which has an atomic number of 17:
1s2 2s2 2p6 3s2 3p5
The electrons with n = 2, l = 0 are the 2s2 electrons. There are 2 of these electrons in chlorine, so the answer is 2. 
Question 85 of 200
85. Question
0 pointsThe e/m ratio of cathode rays is:
Correct
The e/m ratio of cathode rays is 1.76 x 10^11 C/kg. This means that the chargetomass ratio of the particles that make up cathode rays is 1.76 x 10^11 coulombs per kilogram. Cathode rays are beams of electrons that are produced when a high voltage is applied to a gas in a vacuum tube. The electrons are accelerated by the electric field, and they travel in straight lines through the gas. The e/m ratio of cathode rays was first measured by J.J. Thomson in 1897. Thomson's experiment showed that the e/m ratio of cathode rays is independent of the gas that is used in the vacuum tube. This led Thomson to conclude that the particles that make up cathode rays are a fundamental particle of matter, which he called the electron. The e/m ratio of cathode rays is a very important quantity in physics. It is used to determine the mass of the electron, and it is also used to study the properties of electrons.
Incorrect
Option B is correct.
The e/m ratio of cathode rays is 1.76 x 10^11 C/kg. This means that the chargetomass ratio of the particles that make up cathode rays is 1.76 x 10^11 coulombs per kilogram. Cathode rays are beams of electrons that are produced when a high voltage is applied to a gas in a vacuum tube. The electrons are accelerated by the electric field, and they travel in straight lines through the gas. The e/m ratio of cathode rays was first measured by J.J. Thomson in 1897. Thomson's experiment showed that the e/m ratio of cathode rays is independent of the gas that is used in the vacuum tube. This led Thomson to conclude that the particles that make up cathode rays are a fundamental particle of matter, which he called the electron. The e/m ratio of cathode rays is a very important quantity in physics. It is used to determine the mass of the electron, and it is also used to study the properties of electrons.Unattempted
Option B is correct.
The e/m ratio of cathode rays is 1.76 x 10^11 C/kg. This means that the chargetomass ratio of the particles that make up cathode rays is 1.76 x 10^11 coulombs per kilogram. Cathode rays are beams of electrons that are produced when a high voltage is applied to a gas in a vacuum tube. The electrons are accelerated by the electric field, and they travel in straight lines through the gas. The e/m ratio of cathode rays was first measured by J.J. Thomson in 1897. Thomson's experiment showed that the e/m ratio of cathode rays is independent of the gas that is used in the vacuum tube. This led Thomson to conclude that the particles that make up cathode rays are a fundamental particle of matter, which he called the electron. The e/m ratio of cathode rays is a very important quantity in physics. It is used to determine the mass of the electron, and it is also used to study the properties of electrons. 
Question 86 of 200
86. Question
0 pointsThe wave number (li) of a radiation with X, = 2 X 10^{8} nm will be,
Correct
The wave number (ν) of a radiation with wavelength λ = 2 x 10^8 nm will be 0.5 x 10^8 nm^1. The wave number is defined as the reciprocal of the wavelength. In other words, it is the number of waves per unit distance. The unit for wave number is usually expressed in cm^1, but it can also be expressed in nm^1. The wavelength of a wave is the distance between two consecutive peaks or troughs of the wave. In the case of electromagnetic radiation, the wavelength is the distance between two consecutive crests of the wave. The wave number of a radiation can be calculated using the following formula: ν = 1/λ where:
ν is the wave number
λ is the wavelength
In this case, the wavelength is given as 2 x 10^8 nm. So, the wave number can be calculated as follows:
ν = 1/λ = 1/(2 x 10^8 nm) = 0.5 x 10^8 nm^1
Therefore, the wave number of a radiation with wavelength λ = 2 x 10^8 nm will be 0.5 x 10^8 nm^1.Incorrect
Option Ais correct.
The wave number (ν) of a radiation with wavelength λ = 2 x 10^8 nm will be 0.5 x 10^8 nm^1. The wave number is defined as the reciprocal of the wavelength. In other words, it is the number of waves per unit distance. The unit for wave number is usually expressed in cm^1, but it can also be expressed in nm^1. The wavelength of a wave is the distance between two consecutive peaks or troughs of the wave. In the case of electromagnetic radiation, the wavelength is the distance between two consecutive crests of the wave. The wave number of a radiation can be calculated using the following formula: ν = 1/λ where:
ν is the wave number
λ is the wavelength
In this case, the wavelength is given as 2 x 10^8 nm. So, the wave number can be calculated as follows:
ν = 1/λ = 1/(2 x 10^8 nm) = 0.5 x 10^8 nm^1
Therefore, the wave number of a radiation with wavelength λ = 2 x 10^8 nm will be 0.5 x 10^8 nm^1.Unattempted
Option Ais correct.
The wave number (ν) of a radiation with wavelength λ = 2 x 10^8 nm will be 0.5 x 10^8 nm^1. The wave number is defined as the reciprocal of the wavelength. In other words, it is the number of waves per unit distance. The unit for wave number is usually expressed in cm^1, but it can also be expressed in nm^1. The wavelength of a wave is the distance between two consecutive peaks or troughs of the wave. In the case of electromagnetic radiation, the wavelength is the distance between two consecutive crests of the wave. The wave number of a radiation can be calculated using the following formula: ν = 1/λ where:
ν is the wave number
λ is the wavelength
In this case, the wavelength is given as 2 x 10^8 nm. So, the wave number can be calculated as follows:
ν = 1/λ = 1/(2 x 10^8 nm) = 0.5 x 10^8 nm^1
Therefore, the wave number of a radiation with wavelength λ = 2 x 10^8 nm will be 0.5 x 10^8 nm^1. 
Question 87 of 200
87. Question
0 pointsIf Aufbau rule is not followed in filling of the subshells, then the block of which will change in the periodic table?
Correct
This is the correct option . If Aufbau rule is not followed then its last electron will be in 3d rather than 4s , changing its block.
According to Aufbau's principle the elcetrons should be filled itno the orbitals according to increasing energy levels. In K electrons first fill into 4s then 3d. Not obeying this rule will change K's block from s to d.
Incorrect
According to Aufbau's principle the elcetrons should be filled itno the orbitals according to increasing energy levels. In K electrons first fill into 4s then 3d. Not obeying this rule will change K's block from s to d.
Unattempted
According to Aufbau's principle the elcetrons should be filled itno the orbitals according to increasing energy levels. In K electrons first fill into 4s then 3d. Not obeying this rule will change K's block from s to d.

Question 88 of 200
88. Question
0 pointsPositive rays produce in discharge tube from:
Correct
This is incorrect. Positive rays are formed when high speed electrons collide with the gas molecules to knock of their electrons making them positively charged.
High speed electrons collide with the gas molecules and knock off electrons from them making them positively charged . These ions then travel towards the cathode. Hence, option B is correct.
Incorrect
High speed electrons collide with the gas molecules and knock off electrons from them making them positively charged . These ions then travel towards the cathode. Hence, option B is correct.
FTB Pg 24
Unattempted
High speed electrons collide with the gas molecules and knock off electrons from them making them positively charged . These ions then travel towards the cathode. Hence, option B is correct.
FTB Pg 24

Question 89 of 200
89. Question
0 pointsWhich quantum number describes the energy of electron
Correct
This just shows the spin of electron.
D is correct beacuse it shows the energy level of electrons , which is often referred as the energy of electron.
Incorrect
D is correct beacuse it shows the energy level of electrons , which is often referred as the energy of electron.
Unattempted
D is correct beacuse it shows the energy level of electrons , which is often referred as the energy of electron.

Question 90 of 200
90. Question
0 pointsMaximum number of electrons that can be accommodated in N shell:
Correct
It is incorrect. N is the fourth shell and can accomodate a maximum number of 32e.
Incorrect
Option D is correct . N is the fourth shell and can accomodate a maximum number of 32e. We can find this by using formula 2n^{2}
Unattempted
Option D is correct . N is the fourth shell and can accomodate a maximum number of 32e. We can find this by using formula 2n^{2}

Question 91 of 200
91. Question
0 pointsWhich of the following quantum number has negative quantum number value?
Correct
It is incorrect . Principal Quantum number has only positive values as it indicates energy levels.
Incorrect
Option C is correct , Magnetic quantum number has positive as well as negative value for its quantum number.
Unattempted
Option C is correct , Magnetic quantum number has positive as well as negative value for its quantum number.

Question 92 of 200
92. Question
0 pointsIn ratio, positive rays is_________________than that of cathode rays
Correct
Positive rays are formed when highspeed electrons collide with gas molecules and knock off their electrons making them positively charge, but not all gas molecules are ionized. Hence, the ratio of positive rays is less than cathode rays.
Incorrect
Positive rays are formed when high speed electrons collide with gas molecules and knock off there electrons making them positively charge, but not all gas molecules are ionized. Hence, the ratio of positive rays is less than cathode rays.
Unattempted
Positive rays are formed when high speed electrons collide with gas molecules and knock off there electrons making them positively charge, but not all gas molecules are ionized. Hence, the ratio of positive rays is less than cathode rays.

Question 93 of 200
93. Question
0 pointsFor an electron in Kshell, the correct four quantum numbers are represented by;
Correct
L should be 0 for K shell.
Incorrect
The correct four quantum numbers for an electron in the Kshell are:
 Principal quantum number (n) = 1
 Azimuthal quantum number (l) = 0
 Magnetic quantum number (m) = 0
 Spin quantum number (s) = +1/2 or 1/2
The principal quantum number (n) determines the energy level and distance from the nucleus, and for the Kshell, it has a value of 1. The azimuthal quantum number (l) determines the shape of the electron's orbital, and for the Kshell, it has a value of 0, indicating that the electron is in an s orbital. The magnetic quantum number (m) specifies the orientation of the orbital in space, and for an s orbital, it can only have a value of 0. The spin quantum number (s) indicates the direction of the electron's spin, and it can have a value of +1/2 or 1/2.
Unattempted
The correct four quantum numbers for an electron in the Kshell are:
 Principal quantum number (n) = 1
 Azimuthal quantum number (l) = 0
 Magnetic quantum number (m) = 0
 Spin quantum number (s) = +1/2 or 1/2
The principal quantum number (n) determines the energy level and distance from the nucleus, and for the Kshell, it has a value of 1. The azimuthal quantum number (l) determines the shape of the electron's orbital, and for the Kshell, it has a value of 0, indicating that the electron is in an s orbital. The magnetic quantum number (m) specifies the orientation of the orbital in space, and for an s orbital, it can only have a value of 0. The spin quantum number (s) indicates the direction of the electron's spin, and it can have a value of +1/2 or 1/2.

Question 94 of 200
94. Question
0 pointsAll are electromagnetic in nature except,
Correct
They are electromagnetic in nature.
Cathode rays are not electromagnetic in nature, and are made of high speed electrons released from cathode.
Incorrect
Cathode rays are not electromagnetic in nature, and are made of high speed electrons released from cathode.
Unattempted
Cathode rays are not electromagnetic in nature, and are made of high speed electrons released from cathode.

Question 95 of 200
95. Question
0 pointsIndicate the correct order of sub shell w.r.t penetration power.
Correct
Smaller the size of subshell and lower the enrgy level, higher will be the penetration power.
Smaller the size of subshell and lower the enrgy level, higher will be the penetration power.
Incorrect
Smaller the size of subshell and lower the enrgy level, higher will be the penetration power.
Unattempted
Smaller the size of subshell and lower the enrgy level, higher will be the penetration power.

Question 96 of 200
96. Question
0 pointsAn electron moving in orbit having principal quantum number 4 & azimuthal quantum number 2, then the subshell will be
Correct
Azimuthal quantum number 2 shows that electron is in D subshell.
Principal Quantum number shows the orbit is 4 and Azimuthal quantum number 2 shows that electron is in D subshell. So, option C is correct.
Incorrect
Principal Quantum number shows the orbit is 4 and Azimuthal quantum number 2 shows that electron is in D subshell. So, option C is correct.
Unattempted
Principal Quantum number shows the orbit is 4 and Azimuthal quantum number 2 shows that electron is in D subshell. So, option C is correct.

Question 97 of 200
97. Question
0 pointsNitrogen has three unpaired electrons in outermost shells according to
Correct
None
Incorrect
Correct option is C
Thus, nitrogen contains three unpaired electrons following Hund's rule of maximum multiplic
According to this rule, electron pairing in any orbital (s, p, d or f) does not take place until each orbital of the a subshell contains one electron with paralell spin.
The electronic configuration of nitrogen is 1s2 2s2 2p3.Nitrogen is Trivalent.
FTB Pg42Unattempted
Correct option is C
Thus, nitrogen contains three unpaired electrons following Hund's rule of maximum multiplic
According to this rule, electron pairing in any orbital (s, p, d or f) does not take place until each orbital of the a subshell contains one electron with paralell spin.
The electronic configuration of nitrogen is 1s2 2s2 2p3.Nitrogen is Trivalent.
FTB Pg42 
Question 98 of 200
98. Question
0 pointsShape of orbital is determined by_____________quantum number is
Correct
It represents the energy level or shell in which the electron resides. The value of 'n' can be any positive integer starting from 1, and it determines the overall size and energy of the orbital.
Incorrect
The shape of an orbital is described by the azimuthal quantum number or orbital angular momentum quantum number (l).
The azimuthal quantum number defines the shape of the orbital and determines the subshell to which the orbital belongs. It specifies the angular momentum of the electron in the orbital and governs the threedimensional shape of the orbital.
Unattempted
The shape of an orbital is described by the azimuthal quantum number or orbital angular momentum quantum number (l).
The azimuthal quantum number defines the shape of the orbital and determines the subshell to which the orbital belongs. It specifies the angular momentum of the electron in the orbital and governs the threedimensional shape of the orbital.

Question 99 of 200
99. Question
0 pointsA quantum number which is not obtained by solving Schrodinger wave equation is:
Correct
Spin Quantum Number is not derived from Schrodinger’s wave equation.
Principal, azimuthal and magnetic quantum number can be obtained from Schinodiger's wave equation.
Incorrect
Spin Quantum Number is not derived from Schrodinger’s wave equation.
Unattempted
Spin Quantum Number is not derived from Schrodinger’s wave equation.

Question 100 of 200
100. Question
0 pointsAt infinite distance from nucleus, potential energy of electron is:
Correct
At infinite distance, potential energy can't be positive or negative.
Incorrect
correct option is B
The energy of an electron is considered as zero at infinite distance from the nucleus as electron come nearer it gets bounded from the nucleus so the energy is negative. Energy is inversely proportional to the distance so, as the distance decreases energy increases to a greater negative value.
Unattempted
correct option is B
The energy of an electron is considered as zero at infinite distance from the nucleus as electron come nearer it gets bounded from the nucleus so the energy is negative. Energy is inversely proportional to the distance so, as the distance decreases energy increases to a greater negative value.

Question 101 of 200
101. Question
0 pointsThe energy associated with quantum of radiation is:
Correct
According to Planck’s quantum theory, The energy of the radiation absorbed or emitted is directly proportional to the frequency of the radiation,
Incorrect
According to Planck’s quantum theory, The energy of the radiation absorbed or emitted is directly proportional to the frequency of the radiation
The energy of radiation is expressed in terms of frequency as,
E=hν
Where,
E is the energy of tradition;
h is the Planck’s constant (6.626×10−34 J.s);
ν is the frequency of radiation.
FTB Pg 35
Unattempted
According to Planck’s quantum theory, The energy of the radiation absorbed or emitted is directly proportional to the frequency of the radiation
The energy of radiation is expressed in terms of frequency as,
E=hν
Where,
E is the energy of tradition;
h is the Planck’s constant (6.626×10−34 J.s);
ν is the frequency of radiation.
FTB Pg 35

Question 102 of 200
102. Question
0 pointsWhich of the following has the maximum number of unpaired electrons?
Correct
None
Incorrect
Fe^{+2} has 5 unpaired e^{–}
Unattempted
Fe^{+2} has 5 unpaired e^{–}

Question 103 of 200
103. Question
0 pointsTotal number of delectrons in an element of atomic number 26 is:
Correct
This option is incorrect as the total number of d electrons in an element of atomic number 26 is 6.
Incorrect
The total number of delectrons in an atom is determined by its atomic number. For an atom with an atomic number of 26, which corresponds to the element iron (Fe), the electron configuration can be determined as follows: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶.
To find the total number of delectrons, we look at the electrons in the 3d subshell, which is 6 delectrons. So, the total number of delectrons in an atom of atomic number 26 (iron) is 6.
Unattempted
The total number of delectrons in an atom is determined by its atomic number. For an atom with an atomic number of 26, which corresponds to the element iron (Fe), the electron configuration can be determined as follows: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶.
To find the total number of delectrons, we look at the electrons in the 3d subshell, which is 6 delectrons. So, the total number of delectrons in an atom of atomic number 26 (iron) is 6.

Question 104 of 200
104. Question
0 pointsWhich one of the following has the same number of electrons as an alpha particle?
Correct
In a neutral hydrogen atom (H), there is one electron while an alpha particle contains zero electrons. Hence, this option is incorrect.
Incorrect
Hydrogen ion (H^{+}). It contains the same number of electrons as an alpha particle, i.e. 0 electrons. It's worth noting that alpha particles contain the same number of protons as Helium , i.e. 2 protons. There are no electrons in both H^{+} and alpha particle.
Unattempted
Hydrogen ion (H^{+}). It contains the same number of electrons as an alpha particle, i.e. 0 electrons. It's worth noting that alpha particles contain the same number of protons as Helium , i.e. 2 protons. There are no electrons in both H^{+} and alpha particle.

Question 105 of 200
105. Question
0 pointsWhich of the following orbital is not possible:
Correct
2p orbital is possible. The 2p orbitals are electron orbitals associated with the second principal energy level (n = 2) in an atom. The azimuthal quantum number (l) for the 2p orbitals is 1, and the magnetic quantum number (m) can take on values from 1, 0 or +1.
In the case of the 2p orbitals, there are three different orbitals, often referred to as 2px, 2py, and 2pz. These orbitals have different orientations in space.
Incorrect
The Azimuthal quantum number (l) specifies the shape of an orbital in an atom.The azimuthal quantum number can take on integer values from 0 to (n – 1), where n is the principal quantum number.
However, the azimuthal quantum number (l) cannot be equal to 2 in a 2d orbital because the values of l must follow certain restrictions based on the principal quantum number (n). The allowed values of l for a given n are from 0 to (n – 1).
For example:
 When n = 1, the only allowed value for l is 0.
 When n = 2, the allowed values for l are 0 and 1.
 When n = 3, the allowed values for l are 0, 1 and 2.
In the case of n = 2, the only possible values of l are 0 and 1. These correspond to the s and p orbitals respectively and therefore 2d orbital is not possible.
Unattempted
The Azimuthal quantum number (l) specifies the shape of an orbital in an atom.The azimuthal quantum number can take on integer values from 0 to (n – 1), where n is the principal quantum number.
However, the azimuthal quantum number (l) cannot be equal to 2 in a 2d orbital because the values of l must follow certain restrictions based on the principal quantum number (n). The allowed values of l for a given n are from 0 to (n – 1).
For example:
 When n = 1, the only allowed value for l is 0.
 When n = 2, the allowed values for l are 0 and 1.
 When n = 3, the allowed values for l are 0, 1 and 2.
In the case of n = 2, the only possible values of l are 0 and 1. These correspond to the s and p orbitals respectively and therefore 2d orbital is not possible.

Question 106 of 200
106. Question
0 pointsThe number of electrons in M shell of an element with atomic number 24 is:
Correct
This option is incorrect as the number of electrons in M shell are 13 according to the electron configuration of the element.
Incorrect
Electronic configuration of Atomic number 24 (Chromium) is :1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}4s^{1}
Mshell is the third shell, i.e. n = 3
Number of electrons in third shell are 2+6+5 = 13
Hence, the correct option is 13.
Unattempted
Electronic configuration of Atomic number 24 (Chromium) is :1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}4s^{1}
Mshell is the third shell, i.e. n = 3
Number of electrons in third shell are 2+6+5 = 13
Hence, the correct option is 13.

Question 107 of 200
107. Question
0 pointsThe maximum number of electrons with n = 3 and I = 2 is:
Correct
The respective orbital where n = 3 and l = 2 will be '3d' because 3d orbitals contain five subshells and the principal quantum number is 3. We know that 3d orbital can accommodate 10 electrons (each subshell can accommodate two electrons). Therefore, the number of electrons in n = 3, l = 2 are 10. Hence this option is correct.
Incorrect
The respective orbital where n = 3, l = 2 will be '3d' because 3d orbitals contain five subshells and the principal quantum number is 3. We know that 3d orbital can accommodate 10 electrons (each subshell can accommodate two electrons). Therefore the number of electrons in n = 3, l = 2 are 10.
Unattempted
The respective orbital where n = 3, l = 2 will be '3d' because 3d orbitals contain five subshells and the principal quantum number is 3. We know that 3d orbital can accommodate 10 electrons (each subshell can accommodate two electrons). Therefore the number of electrons in n = 3, l = 2 are 10.

Question 108 of 200
108. Question
0 pointsBohr's theory cannot be applied on:
Correct
Bohr's theory is only applicable to hydrogenlike species containing one electron only. As H contains 1 electron, so the Bohr's theory is applicable on it.
Incorrect
Bohr's theory is only applicable to hydrogenlike species containing one electron only. As H, He^{1+} and Li^{2+} all have one electron each, hence Bohr's theory is applicable on them. But H^{+ }has 0 electrons and that is the reason Bohr's theory is not appliable on it.
Unattempted
Bohr's theory is only applicable to hydrogenlike species containing one electron only. As H, He^{1+} and Li^{2+} all have one electron each, hence Bohr's theory is applicable on them. But H^{+ }has 0 electrons and that is the reason Bohr's theory is not appliable on it.

Question 109 of 200
109. Question
0 pointsWhich of the following set of quantum numbers is not possible?
Correct
 “n” represents the principal quantum number, which determines the energy level or shell of an electron. In this case, n=3 means the electron is in the third energy level.
 “l” represents the azimuthal quantum number or the orbital angular momentum quantum number. It determines the shape of the orbital. For l=2, the orbital shape is a d orbital.
 “m” represents the magnetic quantum number, which specifies the orientation of the orbital in space. For m=0, the orbital is oriented along the zaxis.
 “s” represents the spin quantum number, which describes the spin state of the electron. For s=1/2, it means the electron has a spindown orientation.
This set of quantum numbers is possible.
Incorrect
n=3 represents the third energy level, l=3 indicates an f orbital, m=3 means the f orbital is oriented in a specific direction, s=+1/2 represents a spinup electron.
This set of quantum numbers is not possible.
In the third energy level, the maximum value of azimuthal quantum number (l) is 2, so l=3 is invalid. Based on the explanations above, option B (n=3, l=3, m=3, s=+1/2) is the correct answer as it contains an invalid value for the azimuthal quantum number (l). This option would have been correct if the value of l was smaller than the value of n i.e. less than 3.
Unattempted
n=3 represents the third energy level, l=3 indicates an f orbital, m=3 means the f orbital is oriented in a specific direction, s=+1/2 represents a spinup electron.
This set of quantum numbers is not possible.
In the third energy level, the maximum value of azimuthal quantum number (l) is 2, so l=3 is invalid. Based on the explanations above, option B (n=3, l=3, m=3, s=+1/2) is the correct answer as it contains an invalid value for the azimuthal quantum number (l). This option would have been correct if the value of l was smaller than the value of n i.e. less than 3.

Question 110 of 200
110. Question
0 pointsThe electronic configuration of four elements is given. Which of these elements has the highest first ionization energy?
Correct
The given element has highest ionization energy as it has half filled p orbital which is comparatively more stable than partially filled p orbital. Hence, greater amount of energy is required to remove an electron from this element.
Incorrect
Option A is correct. The given element in option A has highest ionization energy as it has half filled p orbital which is comparatively more stable than partially filled p orbital. Hence, greater amount of energy is required to remove an electron from this element. It is extra stable configuration.
Option B is incorrect (refer to the explanation in option B)
Option C is incorrect (refer to the explanation in option C)
Option D is also incorrect (refer to the explanation in option D)
Unattempted
Option A is correct. The given element in option A has highest ionization energy as it has half filled p orbital which is comparatively more stable than partially filled p orbital. Hence, greater amount of energy is required to remove an electron from this element. It is extra stable configuration.
Option B is incorrect (refer to the explanation in option B)
Option C is incorrect (refer to the explanation in option C)
Option D is also incorrect (refer to the explanation in option D)

Question 111 of 200
111. Question
0 pointsThe magnetic quantum number (m) of a free gaseous atom is associated with:
Correct
The effective volume of an orbital is not determined by magnetic quantum number (m). Instead, it is related to the principal quantum number (n) which determines the size and energy of the orbital.
Incorrect
In quantum mechanics, the magnetic quantum number (m) specifies the orientation or magnetic component of an atomic orbital within a given energy level. It determines the number of orbitals in a subshell and the spatial orientation of these orbitals.
The value of (m) can range from l to +l, including zero (where 'l' is azimuthal quantum number). It represents the different possible orientations of an orbital in space. For example, for an s orbital (l = 0), the magnetic quantum number (m) is always zero. For p orbital (l = 1), (m) can take the values 1, 0 or +1, representing the three different orientations along the x, y, and z axes.
In summary, the magnetic quantum number (m) of a free gaseous atom is associated with the spatial orientation or magnetic component of atomic orbitals.
Unattempted
In quantum mechanics, the magnetic quantum number (m) specifies the orientation or magnetic component of an atomic orbital within a given energy level. It determines the number of orbitals in a subshell and the spatial orientation of these orbitals.
The value of (m) can range from l to +l, including zero (where 'l' is azimuthal quantum number). It represents the different possible orientations of an orbital in space. For example, for an s orbital (l = 0), the magnetic quantum number (m) is always zero. For p orbital (l = 1), (m) can take the values 1, 0 or +1, representing the three different orientations along the x, y, and z axes.
In summary, the magnetic quantum number (m) of a free gaseous atom is associated with the spatial orientation or magnetic component of atomic orbitals.

Question 112 of 200
112. Question
0 pointsThe atomic number of an element is 18. The number of electron pairs in its valence shell is:
Correct
The electronic configuration of an element with atomic number 18 is given by: 2,8,8
Hence, it is clear that there are 4 electron pairs (8 electrons) in its valence shell. This option is incorrect.Incorrect
The electronic configuration of an element with atomic number 18 is given by: 2,8,8
Hence, it is clear that there are 4 electron pairs (8 electrons) in its valence shell.All other options are incorrect.
Unattempted
The electronic configuration of an element with atomic number 18 is given by: 2,8,8
Hence, it is clear that there are 4 electron pairs (8 electrons) in its valence shell.All other options are incorrect.

Question 113 of 200
113. Question
0 pointsThe correct representation for 'd' orbital is:
Correct
This option is incorrect.
d block elements have a general electronic configuration of (n1)d ^{110}ns^{12}. These elements can find stability in halffilled orbitals as well as fully filled d orbitals. An example of this is the electronic configuration of chromium, which has halffilled d and s orbitals in its configuration : [Ar]3d^{5}4s^{1}.
Incorrect
d block elements have a general electronic configuration of (n1)d ^{110}ns^{12}. These elements can find stability in halffilled orbitals as well as fully filled d orbitals. An example of this is the electronic configuration of chromium, which has halffilled d and s orbitals in its configuration: [Ar]3d^{5}4s^{1}.
Unattempted
d block elements have a general electronic configuration of (n1)d ^{110}ns^{12}. These elements can find stability in halffilled orbitals as well as fully filled d orbitals. An example of this is the electronic configuration of chromium, which has halffilled d and s orbitals in its configuration: [Ar]3d^{5}4s^{1}.

Question 114 of 200
114. Question
0 pointsNumber of unpaired electrons in chromium (Cr) are:
Correct
Chromium (Cr) has 6 unpaired electrons. Its electronic configuration is: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
In the 3d orbital, there are 5 unpaired electrons. Unpaired electrons are those electrons in an atom that have their spins in the same direction and are located in different orbitals with parallel spins. In this case, the 5 unpaired electrons are in the 3d orbital, making chromium a paramagnetic element. Hence this option is incorrect.
Incorrect
Chromium (Cr) has 6 unpaired electrons. Its electronic configuration is:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
In the 3d orbital, there are 5 unpaired electrons. Unpaired electrons are those electrons in an atom that have their spins in the same direction and are located in different orbitals with parallel spins. In this case, the 5 unpaired electrons are in the 3d orbital, making chromium a paramagnetic element.
Unattempted
Chromium (Cr) has 6 unpaired electrons. Its electronic configuration is:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
In the 3d orbital, there are 5 unpaired electrons. Unpaired electrons are those electrons in an atom that have their spins in the same direction and are located in different orbitals with parallel spins. In this case, the 5 unpaired electrons are in the 3d orbital, making chromium a paramagnetic element.

Question 115 of 200
115. Question
0 pointsThe frequency of a green light is 6 x 10^{14 }Hz, its wavelength is: