Atomic Structure Questions and Answers

Q: The set of elements which has exceptional electronic configurations not obeying Aufbau principle is

The set is Cr, Cu, Ag & Ru.Each of these has a configuration that departs from the Aufbau order: Cr is 3d54s1 (not 3d44s2), Cu is 3d104s1, Ag is 4d105s1 and Ru is 4d75s1. Half-filled and filled d subshells, and the closeness of the (n-1)d and ns energies, explain these.Every other set contains a normal element: K (4s1), Ca (4s2) and Na (3s1) all follow the Aufbau order, which rules out the other four options.

Q: A 250 watt bulb emits monochromatic light of wavelength 198.78 nm. How many photons are emitted by the bulb per second?(Planck's constant = 6.626 x 10-34 J s and velocity of light = 3 x 108 ms-1).

The bulb emits 2.5 x \(10^{20}\) photons per second.Energy of one photon:\[ \begin{aligned} E &= \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{198.78 \times 10^{-9}} \\ &= 1.0 \times 10^{-18}\ \text{J} \end{aligned} \]A 250 W bulb emits 250 J each second, so the number of photons per second is\[ N = \frac{250}{1.0 \times 10^{-18}} = 2.5 \times 10^{20} \]The wavelength 198.78 nm is chosen so that \(hc/\lambda\) comes out exactly \(10^{-18}\) J. Leaving the wavelength in nanometres changes the power of ten.

Q: Which of the following statements are incorrect about boundary surface diagram of orbital?(i) A boundary surface is drawn in space for an orbital on which the value of probability density| Ψ |2 is constant.(ii) The boundary surface for | Ψ |2 and | Ψ | are not identical.(iii) The density of the dots represents the probability density of finding the electron in the region.(iv) In 1s and 2s orbitals, the probability of finding electron at a given distance is not equal in all the directions.(v) The 2p orbitals in the boundary surface diagram are not spherical.

The incorrect statements are (ii) and (iv).(ii) is wrong: |Ψ|2 is constant wherever |Ψ| is constant, so a boundary surface of constant |Ψ|2 is also one of constant |Ψ|; the two surfaces are identical. (iv) is wrong: 1s and 2s orbitals are spherically symmetric, so at a given distance from the nucleus the probability of finding the electron is the same in every direction.(i) is the definition of a boundary surface, and (v) is true since p orbitals are dumb-bell shaped. (iii) correctly describes the dot (charge cloud) representation, in which the density of dots shows the probability density. So the pair of incorrect statements is (ii) and (iv), and every option including (i) or (iii) is wrong.

Q: The radius of the first orbit He+ is

The radius is 26.45 pm.For a hydrogen-like ion, Bohr's model gives \(r_{n} = 52.9\frac{n^{2}}{Z}\) pm. For He+, \(Z = 2\), and the first orbit has \(n = 1\):\[ r_{1} = \frac{52.9 \times 1}{2} = 26.45 \text{ pm} \]52.9 pm is the hydrogen value, and 105.8 pm and 211.6 pm multiply by \(Z\) or \(Z^{2}\) instead of dividing.

Q: In Mosley experiment, the X-ray spectrum is obtained by plotting

Moseley plotted the square root of frequency against atomic number.Moseley found that the square root of the frequency of characteristic X-rays from an element rises in a straight line with its atomic number: \(\sqrt{\nu} = a(Z - b)\). The plot was a straight line against atomic number but not against atomic mass, which showed that atomic number is the fundamental property of an element and led to the modern periodic law.Plots against atomic mass were what failed to give a straight line, and wavelength or the square of frequency are not the quantities he used.

Q: The quantum number symbols are given in List-I and the informations about them are given in List-II. The correct match isList-IList-III) mla) Shape of orbitalII) msb) Size of orbitalIII) lc) Orientation of orbitalIV) nd) Orientation of electron spin

The correct match is I-c, II-d, III-a, IV-b.The magnetic quantum number ml gives the orientation of the orbital in space (c). The spin quantum number ms gives the orientation of the electron's spin (d). The azimuthal quantum number l gives the shape of the orbital (a). The principal quantum number n gives its size and energy (b).The common confusion is between ml and ms, and between n and l; each other option swaps one of these pairs.

Q: The transition energy of the electron from higher to lower level is 3.0 × 10-19 J. The wavelength of the emitted light is about(h = 6.625 × 10-34 J s, c = 3 × 108 m s-1)

The wavelength is about 660 nm.Use \(E = \frac{hc}{\lambda}\), so \(\lambda = \frac{hc}{E}\).\[ \begin{aligned} \lambda &= \frac{6.625 \times 10^{-34} \times 3 \times 10^{8}}{3.0 \times 10^{-19}} \\ &= 6.625 \times 10^{-7}\ \text{m} \approx 660\ \text{nm} \end{aligned} \]This is red light. 330 nm is half the answer, and 990 nm one and a half times it. Light of this wavelength is in the visible region; 330 nm and 310 nm would be ultraviolet.

Q: According to Hund's rule, most stable state for np2 configuration is ..........

The most stable state is 3P0.The np2 configuration gives the terms 3P, 1D and 1S. By Hund's first rule the term of highest multiplicity, 3P, lies lowest. It splits into J = 0, 1, 2 (L + S down to |L − S|).Hund's third rule picks the level: for a subshell less than half full the lowest J lies lowest. Two electrons in a p subshell that can hold six is less than half full, so 3P0 is the ground state, as in carbon. 3P2 would be the ground level of a more than half-filled shell (np4), 3P1 lies between them, and 1S0 is the highest of the three terms.

31.

The set of elements which has exceptional electronic configurations not obeying Aufbau principle is

A.K, Ca, Cr & Cu
B.Ca, Cr, Nb & Cu
C.Ca, Cr, Ag & Cu
D.Cr, Cu, Ag & Ru
E.Na, Rh, Cr & Cu

Exam Year(s)

KEAM - 2026,

Answer: D

The set is Cr, Cu, Ag & Ru.

Each of these has a configuration that departs from the Aufbau order: Cr is 3d⁵4s¹ (not 3d⁴4s²), Cu is 3d¹⁰4s¹, Ag is 4d¹⁰5s¹ and Ru is 4d⁷5s¹. Half-filled and filled d subshells, and the closeness of the (n-1)d and ns energies, explain these.

Every other set contains a normal element: K (4s¹), Ca (4s²) and Na (3s¹) all follow the Aufbau order, which rules out the other four options.

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32.

A 250 watt bulb emits monochromatic light of wavelength 198.78 nm. How many photons are emitted by the bulb per second?

(Planck's constant = 6.626 x 10^-34 J s and velocity of light = 3 x 10⁸ ms^-1).

A.2.5 x \(10^{20}\) \(\mathrm{s^{-1}}\)
B.2.5 x \(10^{18}\) \(\mathrm{s^{-1}}\)
C.2 x \(10^{20}\) \(\mathrm{s^{-1}}\)
D.2.2 x \(10^{16}\) \(\mathrm{s^{-1}}\)
E.2.6 x\(10^{20}\) \(\mathrm{s^{-1}}\)

Exam Year(s)

KEAM - 2026,

Answer: A

The bulb emits 2.5 x \(10^{20}\) photons per second.

Energy of one photon:

\[ \begin{aligned} E &= \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{198.78 \times 10^{-9}} \\ &= 1.0 \times 10^{-18}\ \text{J} \end{aligned} \]

A 250 W bulb emits 250 J each second, so the number of photons per second is

\[ N = \frac{250}{1.0 \times 10^{-18}} = 2.5 \times 10^{20} \]

The wavelength 198.78 nm is chosen so that \(hc/\lambda\) comes out exactly \(10^{-18}\) J. Leaving the wavelength in nanometres changes the power of ten.

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33.

Which of the following statements are incorrect about boundary surface diagram of orbital?

(i) A boundary surface is drawn in space for an orbital on which the value of probability density| Ψ |² is constant.
(ii) The boundary surface for | Ψ |² and | Ψ | are not identical.
(iii) The density of the dots represents the probability density of finding the electron in the region.
(iv) In 1s and 2s orbitals, the probability of finding electron at a given distance is not equal in all the directions.
(v) The 2p orbitals in the boundary surface diagram are not spherical.

A.(i), (iv)
B.(iii), (iv)
C.(ii), (iv)
D.(ii), (iii)
E.(i), (ii)

Exam Year(s)

KEAM - 2026,

Answer: C

The incorrect statements are (ii) and (iv).

(ii) is wrong: |Ψ|² is constant wherever |Ψ| is constant, so a boundary surface of constant |Ψ|² is also one of constant |Ψ|; the two surfaces are identical. (iv) is wrong: 1s and 2s orbitals are spherically symmetric, so at a given distance from the nucleus the probability of finding the electron is the same in every direction.

(i) is the definition of a boundary surface, and (v) is true since p orbitals are dumb-bell shaped. (iii) correctly describes the dot (charge cloud) representation, in which the density of dots shows the probability density. So the pair of incorrect statements is (ii) and (iv), and every option including (i) or (iii) is wrong.

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34.

The radius of the first orbit He⁺ is

A.52.9 pm
B.13.24 pm
C.211.6 pm
D.105.8 pm
E.26.45 pm

Exam Year(s)

KEAM - 2026,

Answer: E

The radius is 26.45 pm.

For a hydrogen-like ion, Bohr's model gives \(r_{n} = 52.9\frac{n^{2}}{Z}\) pm. For He⁺, \(Z = 2\), and the first orbit has \(n = 1\):

\[ r_{1} = \frac{52.9 \times 1}{2} = 26.45 \text{ pm} \]

52.9 pm is the hydrogen value, and 105.8 pm and 211.6 pm multiply by \(Z\) or \(Z^{2}\) instead of dividing.

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35.

In Mosley experiment, the X-ray spectrum is obtained by plotting

A.square root of frequency vs atomic mass
B.square root of frequency vs atomic number
C.square of frequency vs atomic mass
D.square root of wavelength vs atomic number
E.square root of wavelength vs atomic mass

Exam Year(s)

KEAM - 2026,

Answer: B

Moseley plotted the square root of frequency against atomic number.

Moseley found that the square root of the frequency of characteristic X-rays from an element rises in a straight line with its atomic number: \(\sqrt{\nu} = a(Z - b)\). The plot was a straight line against atomic number but not against atomic mass, which showed that atomic number is the fundamental property of an element and led to the modern periodic law.

Plots against atomic mass were what failed to give a straight line, and wavelength or the square of frequency are not the quantities he used.

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36.

The quantum number symbols are given in List-I and the informations about them are given in List-II. The correct match is

List-I	List-II
I) m_l	a) Shape of orbital
II) m_s	b) Size of orbital
III) l	c) Orientation of orbital
IV) n	d) Orientation of electron spin

A.I-d, II-a, III-c, IV-b
B.I-c, II-d, III-a, IV-b
C.I-b, II-c, III-d, IV-a
D.I-c, II-a, III-d, IV-b
E.I-b, II-d, III-a, IV-c

Exam Year(s)

KEAM - 2026,

Answer: B

The correct match is I-c, II-d, III-a, IV-b.

The magnetic quantum number m_l gives the orientation of the orbital in space (c). The spin quantum number m_s gives the orientation of the electron's spin (d). The azimuthal quantum number l gives the shape of the orbital (a). The principal quantum number n gives its size and energy (b).

The common confusion is between m_l and m_s, and between n and l; each other option swaps one of these pairs.

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37.

The transition energy of the electron from higher to lower level is 3.0 × 10^-19 J. The wavelength of the emitted light is about
(h = 6.625 × 10^-34 J s, c = 3 × 10⁸ m s^-1)

A.330 nm
B.660 nm
C.990 nm
D.310 nm
E.230 nm

Exam Year(s)

KEAM - 2026,

Answer: B

The wavelength is about 660 nm.

Use \(E = \frac{hc}{\lambda}\), so \(\lambda = \frac{hc}{E}\).

\[ \begin{aligned} \lambda &= \frac{6.625 \times 10^{-34} \times 3 \times 10^{8}}{3.0 \times 10^{-19}} \\ &= 6.625 \times 10^{-7}\ \text{m} \approx 660\ \text{nm} \end{aligned} \]

This is red light. 330 nm is half the answer, and 990 nm one and a half times it. Light of this wavelength is in the visible region; 330 nm and 310 nm would be ultraviolet.

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38.

According to Hund's rule, most stable state for np² configuration is ..........

A.\(^{1}S_{0}\)
B.\(^{3}P_{0}\)
C.\(^{3}P_{1}\)
D.\(^{3}P_{2}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The most stable state is ³P₀.

The np² configuration gives the terms ³P, ¹D and ¹S. By Hund's first rule the term of highest multiplicity, ³P, lies lowest. It splits into J = 0, 1, 2 (L + S down to |L − S|).

Hund's third rule picks the level: for a subshell less than half full the lowest J lies lowest. Two electrons in a p subshell that can hold six is less than half full, so ³P₀ is the ground state, as in carbon. ³P₂ would be the ground level of a more than half-filled shell (np⁴), ³P₁ lies between them, and ¹S₀ is the highest of the three terms.

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