Atomic Structure Questions and Answers

Q: The total orbital angular momentum quantum number L and spin quantum S of the term symbol 4G is :

L and S are 4 and 3/2.In a term symbol 2S+1L, the letter gives L (S = 0, P = 1, D = 2, F = 3, G = 4) and the left superscript gives the multiplicity 2S + 1. For 4G, written 4G in the paper:\[ \begin{aligned} L &= 4\ (\text{G}) \\ 2S + 1 &= 4 \Rightarrow S = 3/2 \end{aligned} \]The other pairs misread the letter (2 is D, 3 is F, 5 is H) or the multiplicity (S = 1/2 would be a doublet).

Q: The ratio of the energy of the electron in the ground state of hydrogen atom to that of the electron in the first excited state of Be3+ is :

The ratio is 1 : 4.For a one-electron atom or ion of nuclear charge Z, the energy of level n is\[ E_{n} = -13.6\,\frac{Z^{2}}{n^{2}}\ \text{eV} \]Hydrogen ground state: Z = 1, n = 1. Be3+ is one-electron beryllium, Z = 4, and its first excited state is n = 2:\[ \begin{aligned} E_{H} &= -13.6\ \text{eV} \\ E_{Be^{3+}} &= -13.6 \times \tfrac{16}{4} = -54.4\ \text{eV} \end{aligned} \]The ratio is 13.6 : 54.4 = 1 : 4. 1 : 16 forgets the n2 = 4 in the denominator (it is the ground-state ratio), and 1 : 8 and 2 : 9 come from using Z or n instead of their squares.

Q: The ionization energy of hydrogen atom is 13.6 eV; the ionization energy for the ground state of Li2+ is approximately :

The ionization energy of Li2+ is about 122.4 eV.Li2+ has one electron, so it is hydrogen-like, and the ground-state energy of a hydrogen-like ion scales with the square of the nuclear charge:\[ \begin{aligned} IE &= 13.6\,Z^{2}\ \text{eV} \\ &= 13.6 \times 3^{2} = 122.4\ \text{eV} \end{aligned} \]27.2 eV and 40.8 eV scale with Z1 (for Z = 2 and Z = 3), and 54.4 eV is the value for He+ (Z = 2, 13.6 × 4). Only the Z2 scaling with Z = 3 gives the value for lithium.

Q: Calculate the wavelength of an electron accelerated through a potential of 16 V :(Given : (2 me)½ = 5.4 × 10−25 kg.m/S.V½ )

The wavelength is 0.307 nm.An electron accelerated through V volts has kinetic energy eV, so its momentum is \(p = \sqrt{2m_{e}eV}\), and the de Broglie wavelength is \(\lambda = h/p\). Using the given factor \(\sqrt{2m_{e}e} = 5.4\times10^{-25}\) in SI units and h = 6.626 × 10−34 J s:\[ \begin{aligned} p &= 5.4\times10^{-25} \times \sqrt{16} \\ &= 2.16\times10^{-24}\ \text{kg m s}^{-1} \\ \lambda &= \frac{6.626\times10^{-34}}{2.16\times10^{-24}} \\ &= 3.07\times10^{-10}\ \text{m} = 0.307\ \text{nm} \end{aligned} \]A check: the familiar shortcut \(\lambda \approx 1.226/\sqrt{V}\) nm gives 1.226/4 = 0.307 nm. The other options are the same digits shifted by a factor of 10, or 0.123 nm, which corresponds to 100 V.

Q: Number of unpaired electrons in Fe++ is :

Fe2+ has 4 unpaired electrons.Iron (Z = 26) is [Ar] 3d6 4s2. Forming Fe2+ removes the two 4s electrons, leaving 3d6. By Hund's rule the six d electrons fill the five d orbitals singly first, then one pairs up: one orbital holds a pair and four hold one electron each, so 4 are unpaired (in the free ion or a high-spin complex).5 unpaired electrons belongs to Fe3+ (3d5). 2 and 3 would need d2, d3 or a different count.

Q: The allowed term symbol for microstate (1− 0−) is :

The microstate belongs to the term 3P.The notation lists the ml values of the two electrons with their spins: one electron in ml = 1 and one in ml = 0, both with spin down. So ML = 1 + 0 = 1 and MS = −1/2 − 1/2 = −1.MS = −1 is possible only in a triplet (S = 1), so singlet terms 1P and 1G are ruled out. The question is set on a p2 configuration (ml = +1, 0, −1), whose terms are 3P, 1D and 1S; the only triplet is 3P, and ML = 1 lies within its range of −1 to +1. A 3F term arises only from d2, which is not the configuration intended here.

Q: The following sketch represents the probability density distribution for :

The sketch represents the radial probability for a 3s orbital, option A.A radial distribution curve of an orbital with quantum numbers n and l has n − l maxima. The sketch has three maxima, so n − l = 3, and for the lobes to rise in height with the largest one farthest out the orbital is an s orbital. With l = 0 this gives n = 3, a 3s orbital, whose last maximum lies far out (here near 15 a0) and is the tallest.A 3d orbital (B) has n − l = 1, so a single peak; a 2p orbital (C) also has one peak; and 1p (D) does not exist because l must be less than n.

Q: The most probable distance of 1s electron in hydrogen like atom with atomic number 'z' is given by :

The most probable distance is a0/z.For a 1s orbital ψ is proportional to e−zr/a0, so the radial distribution function is P(r) = 4πr2ψ2, proportional to r2e−2zr/a0. Setting its derivative to zero:\[ \begin{aligned} \frac{d}{dr}\left(r^{2}e^{-2zr/a_{0}}\right) &= 0 \\ 2r - \frac{2z r^{2}}{a_{0}} &= 0 \\ r &= \frac{a_{0}}{z} \end{aligned} \]A larger nuclear charge pulls the electron closer, so the distance must decrease as z increases, which rules out z.a0 and z3.a0; z3/a0 does not even have units of length.

Q: The ground state energy of hydrogen like species is proportional to :

The ground-state energy is proportional to −Rz2/n2.The Bohr energy of a one-electron species with nuclear charge z is\[ E_{n} = -\frac{R z^{2}}{n^{2}} \]The energy is negative because the electron is bound, it grows with the square of the nuclear charge (He+ lies four times as deep as H), and it falls off as 1/n2.−R/n2 is correct only for hydrogen itself (z = 1). Rz2/n2 has the wrong sign for a bound state, and Rz/n has the wrong dependence on both z and n.

Q: A particular metal complex has an electronic configuration of np1 nf1. The number of microstates possible in an isolated gaseous state for this metal complex is :

The number of microstates is 84.The two electrons are in different subshells (one p, one f), so they are not equivalent and the Pauli principle places no restriction between them. Each can occupy any of its spin-orbitals independently: a p subshell has 3 × 2 = 6 spin-orbitals and an f subshell 7 × 2 = 14.\[ N = 6 \times 14 = 84 \]14 and 6 are the counts for the f electron or the p electron alone. 20 is their sum, which would be right only for choosing one position in either subshell, not for placing both electrons.

41.

The total orbital angular momentum quantum number L and spin quantum S of the term symbol 4G is :

A.2, 1/2
B.3, 3/2
C.4, 3/2
D.5, 1/2

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

L and S are 4 and 3/2.

In a term symbol ^2S+1L, the letter gives L (S = 0, P = 1, D = 2, F = 3, G = 4) and the left superscript gives the multiplicity 2S + 1. For ⁴G, written 4G in the paper:

\[ \begin{aligned} L &= 4\ (\text{G}) \\ 2S + 1 &= 4 \Rightarrow S = 3/2 \end{aligned} \]

The other pairs misread the letter (2 is D, 3 is F, 5 is H) or the multiplicity (S = 1/2 would be a doublet).

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42.

The ratio of the energy of the electron in the ground state of hydrogen atom to that of the electron in the first excited state of Be³⁺ is :

A.1 : 4
B.1 : 8
C.2 : 9
D.1 : 16

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The ratio is 1 : 4.

For a one-electron atom or ion of nuclear charge Z, the energy of level n is

\[ E_{n} = -13.6\,\frac{Z^{2}}{n^{2}}\ \text{eV} \]

Hydrogen ground state: Z = 1, n = 1. Be³⁺ is one-electron beryllium, Z = 4, and its first excited state is n = 2:

\[ \begin{aligned} E_{H} &= -13.6\ \text{eV} \\ E_{Be^{3+}} &= -13.6 \times \tfrac{16}{4} = -54.4\ \text{eV} \end{aligned} \]

The ratio is 13.6 : 54.4 = 1 : 4. 1 : 16 forgets the n² = 4 in the denominator (it is the ground-state ratio), and 1 : 8 and 2 : 9 come from using Z or n instead of their squares.

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43.

The ionization energy of hydrogen atom is 13.6 eV; the ionization energy for the ground state of Li²⁺ is approximately :

A.27.2 eV
B.40.8 eV
C.54.4 eV
D.122.4 eV

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The ionization energy of Li²⁺ is about 122.4 eV.

Li²⁺ has one electron, so it is hydrogen-like, and the ground-state energy of a hydrogen-like ion scales with the square of the nuclear charge:

\[ \begin{aligned} IE &= 13.6\,Z^{2}\ \text{eV} \\ &= 13.6 \times 3^{2} = 122.4\ \text{eV} \end{aligned} \]

27.2 eV and 40.8 eV scale with Z¹ (for Z = 2 and Z = 3), and 54.4 eV is the value for He⁺ (Z = 2, 13.6 × 4). Only the Z² scaling with Z = 3 gives the value for lithium.

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44.

Calculate the wavelength of an electron accelerated through a potential of 16 V :

(Given : (2 me)^½ = 5.4 × 10⁻²⁵ kg.m/S.V^½ )

A.0.307 nm
B.0.123 nm
C.0.0123 nm
D.3.07 nm

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The wavelength is 0.307 nm.

An electron accelerated through V volts has kinetic energy eV, so its momentum is \(p = \sqrt{2m_{e}eV}\), and the de Broglie wavelength is \(\lambda = h/p\). Using the given factor \(\sqrt{2m_{e}e} = 5.4\times10^{-25}\) in SI units and h = 6.626 × 10⁻³⁴ J s:

\[ \begin{aligned} p &= 5.4\times10^{-25} \times \sqrt{16} \\ &= 2.16\times10^{-24}\ \text{kg m s}^{-1} \\ \lambda &= \frac{6.626\times10^{-34}}{2.16\times10^{-24}} \\ &= 3.07\times10^{-10}\ \text{m} = 0.307\ \text{nm} \end{aligned} \]

A check: the familiar shortcut \(\lambda \approx 1.226/\sqrt{V}\) nm gives 1.226/4 = 0.307 nm. The other options are the same digits shifted by a factor of 10, or 0.123 nm, which corresponds to 100 V.

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45.

Number of unpaired electrons in Fe⁺⁺ is :

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

Fe²⁺ has 4 unpaired electrons.

Iron (Z = 26) is [Ar] 3d⁶ 4s². Forming Fe²⁺ removes the two 4s electrons, leaving 3d⁶. By Hund's rule the six d electrons fill the five d orbitals singly first, then one pairs up: one orbital holds a pair and four hold one electron each, so 4 are unpaired (in the free ion or a high-spin complex).

5 unpaired electrons belongs to Fe³⁺ (3d⁵). 2 and 3 would need d², d³ or a different count.

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46.

The allowed term symbol for microstate (1⁻ 0⁻) is :

A.\(^{3}\)F
B.\(^{3}\)P
C.\(^{1}\)P
D.\(^{1}\)G

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The microstate belongs to the term ³P.

The notation lists the m_l values of the two electrons with their spins: one electron in m_l = 1 and one in m_l = 0, both with spin down. So M_L = 1 + 0 = 1 and M_S = −1/2 − 1/2 = −1.

M_S = −1 is possible only in a triplet (S = 1), so singlet terms ¹P and ¹G are ruled out. The question is set on a p² configuration (m_l = +1, 0, −1), whose terms are ³P, ¹D and ¹S; the only triplet is ³P, and M_L = 1 lies within its range of −1 to +1. A ³F term arises only from d², which is not the configuration intended here.

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47.

The following sketch represents the probability density distribution for :

Plot of P against r over a0 from 0 to 25 with three maxima that rise in height, the last and largest near 15

A.3s orbital
B.3d orbital
C.2p orbital
D.1p orbital

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The sketch represents the radial probability for a 3s orbital, option A.

A radial distribution curve of an orbital with quantum numbers n and l has n − l maxima. The sketch has three maxima, so n − l = 3, and for the lobes to rise in height with the largest one farthest out the orbital is an s orbital. With l = 0 this gives n = 3, a 3s orbital, whose last maximum lies far out (here near 15 a₀) and is the tallest.

A 3d orbital (B) has n − l = 1, so a single peak; a 2p orbital (C) also has one peak; and 1p (D) does not exist because l must be less than n.

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48.

The most probable distance of 1s electron in hydrogen like atom with atomic number 'z' is given by :

A.\(\dfrac{z^{3}}{a_{0}}\)
B.\(z.a_{0}\)
C.\(\dfrac{a_{0}}{z}\)
D.\(z^{3}.a_{0}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The most probable distance is a₀/z.

For a 1s orbital ψ is proportional to e^−zr/a₀, so the radial distribution function is P(r) = 4πr²ψ², proportional to r²e^−2zr/a₀. Setting its derivative to zero:

\[ \begin{aligned} \frac{d}{dr}\left(r^{2}e^{-2zr/a_{0}}\right) &= 0 \\ 2r - \frac{2z r^{2}}{a_{0}} &= 0 \\ r &= \frac{a_{0}}{z} \end{aligned} \]

A larger nuclear charge pulls the electron closer, so the distance must decrease as z increases, which rules out z.a₀ and z³.a₀; z³/a₀ does not even have units of length.

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49.

The ground state energy of hydrogen like species is proportional to :

A.\(\dfrac{-R}{n^{2}}\)
B.\(\dfrac{-Rz^{2}}{n^{2}}\)
C.\(\dfrac{Rz^{2}}{n^{2}}\)
D.\(\dfrac{Rz}{n}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The ground-state energy is proportional to −Rz²/n².

The Bohr energy of a one-electron species with nuclear charge z is

\[ E_{n} = -\frac{R z^{2}}{n^{2}} \]

The energy is negative because the electron is bound, it grows with the square of the nuclear charge (He⁺ lies four times as deep as H), and it falls off as 1/n².

−R/n² is correct only for hydrogen itself (z = 1). Rz²/n² has the wrong sign for a bound state, and Rz/n has the wrong dependence on both z and n.

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50.

A particular metal complex has an electronic configuration of np¹ nf¹. The number of microstates possible in an isolated gaseous state for this metal complex is :

A.14
B.6
C.20
D.84

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

The number of microstates is 84.

The two electrons are in different subshells (one p, one f), so they are not equivalent and the Pauli principle places no restriction between them. Each can occupy any of its spin-orbitals independently: a p subshell has 3 × 2 = 6 spin-orbitals and an f subshell 7 × 2 = 14.

\[ N = 6 \times 14 = 84 \]

14 and 6 are the counts for the f electron or the p electron alone. 20 is their sum, which would be right only for choosing one position in either subshell, not for placing both electrons.

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