Atomic Orbitals and Quantum Numbers — NEET Chemistry MCQs with Solutions
Free NEET Chemistry Atomic Orbitals and Quantum Numbers MCQs with step-by-step solutions (56 questions). Part of Atomic Structure. Practise online on Prepizo — no login needed.
▶ Practise Atomic Orbitals and Quantum Numbers online (free)
Questions with solutions
Q1 — Atomic Orbitals and Quantum Numbers · easy
For electrons in '2s' and '2p' orbitals, the orbital angular momentum values, respectively, are
A. $\sqrt{2}\,\dfrac{h}{2\pi}$ and $0$
B. $\dfrac{h}{2\pi}$ and $\sqrt{2}\,\dfrac{h}{2\pi}$
C. $0$ and $\sqrt{6}\,\dfrac{h}{2\pi}$
D. $0$ and $\sqrt{2}\,\dfrac{h}{2\pi}$ ✓ Correct
Solution: Orbital angular momentum $= \sqrt{l(l+1)}\,\dfrac{h}{2\pi}$. For 2s, $l = 0$, so it is 0. For 2p, $l = 1$, so it is $\sqrt{2}\,\dfrac{h}{2\pi}$.
Q2 — Atomic Orbitals and Quantum Numbers · medium
Which one of the following statements about an electron occupying the 1s orbital in a hydrogen atom is incorrect? (Bohr's radius is represented by $a_0$)
A. The electron can be found at a distance $2a_0$ from the nucleus
B. The 1s orbital is spherically symmetrical
C. The probability density of finding the electron is maximum at the nucleus
D. The total energy of the electron is maximum when it is at a distance $a_0$ from the nucleus ✓ Correct
Solution: The total energy of an electron in a stationary state (1s) is constant — it does not depend on its distance from the nucleus. The other three statements are true: the 1s orbital is spherical, its probability density $\psi^2$ is maximum at the nucleus, and the electron has a non-zero probability of being found at any finite distance, including $2a_0$.
Q3 — Atomic Orbitals and Quantum Numbers · easy
Consider the ground state of a chromium atom (Z = 24). How many electrons have azimuthal quantum number $l = 1$ and $l = 2$, respectively?
A. 16 and 5
B. 16 and 4
C. 12 and 4
D. 12 and 5 ✓ Correct
Solution: Cr: [Ar] 3d⁵ 4s¹, i.e. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹. Electrons with $l = 1$ (p): 2p⁶ + 3p⁶ = 12. Electrons with $l = 2$ (d): 3d⁵ = 5.
Q4 — Atomic Orbitals and Quantum Numbers · medium
Correct statements for an element with atomic number 9 are:
A. There can be 5 electrons for which $m_s = +\tfrac{1}{2}$ and 4 electrons for which $m_s = -\tfrac{1}{2}$.
B. There is only one electron in the p_z orbital.
C. The last electron goes to an orbital with $n = 2$ and $l = 1$.
D. The sum of angular nodes of all the atomic orbitals is 4.
Choose the correct answer from the options given below:
A. A and C only ✓ Correct
B. A and B only
C. A, C and D only
D. C and D only
Solution: Fluorine (Z = 9): 1s² 2s² 2p⁵. A: of the 9 electrons, 5 can have $m_s = +\tfrac{1}{2}$ and 4 have $m_s = -\tfrac{1}{2}$ — true. C: the last electron enters 2p ($n = 2$, $l = 1$) — true. B: the singly occupied 2p orbital is not necessarily p_z — false. D: angular nodes of 1s, 2s, 2pₓ, 2pᵧ, 2p_z = 0 + 0 + 1 + 1 + 1 = 3, not 4 — false.
Q5 — Atomic Orbitals and Quantum Numbers · medium
For the hydrogen atom, the orbital(s) with lowest energy is/are:
A. 4s
B. 3pₓ
C. $3d_{x^2-y^2}$
D. $3d_{z^2}$
E. 4p_z
Choose the correct answer from the options given below:
A. B only
B. A only
C. A and E only
D. B, C and D only ✓ Correct
Solution: For hydrogen (a one-electron atom), the energy depends only on $n$. The orbitals with $n = 3$ (3pₓ, $3d_{x^2-y^2}$, $3d_{z^2}$) all have the same, lowest energy of the list; 4s and 4p_z ($n = 4$) lie higher.
Q6 — Atomic Orbitals and Quantum Numbers · easy
In a multielectron atom, which of the following orbitals described by three quantum numbers will have the same energy in the absence of electric and magnetic fields?
A. $n = 1, l = 0, m_l = 0$
B. $n = 2, l = 0, m_l = 0$
C. $n = 2, l = 1, m_l = 1$
D. $n = 3, l = 2, m_l = 1$
E. $n = 3, l = 2, m_l = 0$
Choose the correct answer from the options given below:
A. A and B only
B. D and E only ✓ Correct
C. B and C only
D. C and D only
Solution: In a multielectron atom, orbitals with the same $n$ and the same $l$ are degenerate. D and E are both 3d orbitals ($n = 3$, $l = 2$) differing only in $m_l$, so they have the same energy.
Q7 — Atomic Orbitals and Quantum Numbers · medium
Which of the following is/are not correct with respect to the energy of atomic orbitals of a hydrogen atom?
A. 1s < 2p < 3d < 4s
B. 1s < 2s = 2p < 3s = 3p
C. 1s < 2s < 2p < 3s < 3p
D. 1s < 2s < 4s < 3d
Choose the correct answer from the options given below:
A. A and C only
B. A and B only
C. B and D only
D. C and D only ✓ Correct
Solution: For hydrogen, orbital energy depends only on $n$. A (1s < 2p < 3d < 4s, i.e. $n$ = 1 < 2 < 3 < 4) and B (2s = 2p, 3s = 3p) are correct. C is wrong because 2s = 2p (not 2s < 2p), and D is wrong because 3d ($n = 3$) lies below 4s ($n = 4$).
Q8 — Atomic Orbitals and Quantum Numbers · medium
Compare the energies of the following sets of quantum numbers for a multielectron system.
A. $n = 4, l = 1$
B. $n = 4, l = 2$
C. $n = 3, l = 1$
D. $n = 3, l = 2$
E. $n = 4, l = 0$
Choose the correct answer from the options given below:
A. E > C > A > D > B
B. C < E < D < A < B ✓ Correct
C. B > A > C > E > D
D. E < C < D < A < B
Solution: Apply the $(n+l)$ rule: C = 3p (4), E = 4s (4), D = 3d (5), A = 4p (5), B = 4d (6). For equal $(n+l)$, the lower $n$ is lower in energy: 3p < 4s and 3d < 4p. Hence C < E < D < A < B.
Q9 — Atomic Orbitals and Quantum Numbers · easy
The number of radial node(s) for a 3p orbital is
A. 3
B. 2
C. 1 ✓ Correct
D. 4
Solution: Radial nodes $= n - l - 1 = 3 - 1 - 1 = 1$.
Q10 — Atomic Orbitals and Quantum Numbers · easy
The four quantum numbers for the electron in the outermost orbital of potassium (atomic number 19) are
A. $n = 4,\ l = 2,\ m = -1,\ s = +\tfrac{1}{2}$
B. $n = 3,\ l = 0,\ m = 1,\ s = +\tfrac{1}{2}$
C. $n = 2,\ l = 0,\ m = 0,\ s = +\tfrac{1}{2}$
D. $n = 4,\ l = 0,\ m = 0,\ s = +\tfrac{1}{2}$ ✓ Correct
Solution: K (Z = 19): [Ar] 4s¹. The outermost electron is in 4s, so $n = 4$, $l = 0$, $m = 0$, $s = \pm\tfrac{1}{2}$.
Q11 — Atomic Orbitals and Quantum Numbers · medium
The wave function $\psi$ of 2s is given by $\psi_{2s} = \dfrac{1}{2\sqrt{2\pi}}\left(\dfrac{1}{a_0}\right)^{1/2}\left(2 - \dfrac{r}{a_0}\right)e^{-r/2a_0}$. At $r = r_0$, a radial node is formed. Thus $r_0$ in terms of $a_0$ is
A. $r_0 = \dfrac{a_0}{2}$
B. $r_0 = 2a_0$ ✓ Correct
C. $r_0 = 4a_0$
D. $r_0 = a_0$
Solution: At a radial node $\psi = 0$. The exponential term is never zero, so $2 - \dfrac{r_0}{a_0} = 0$, giving $r_0 = 2a_0$.
Q12 — Atomic Orbitals and Quantum Numbers · easy
The maximum number of electrons that can be accommodated in a shell with $n = 4$ is
A. 72
B. 16
C. 32 ✓ Correct
D. 50
Solution: Maximum electrons in a shell $= 2n^2 = 2 \times 4^2 = 32$.
Q13 — Atomic Orbitals and Quantum Numbers · medium
Arrange the following orbitals in decreasing order of energy (multielectron atom).
A. $n = 3, l = 0, m = 0$
B. $n = 4, l = 0, m = 0$
C. $n = 3, l = 1, m = 0$
D. $n = 3, l = 2, m = 1$
A. D > B > C > A ✓ Correct
B. B > D > C > A
C. A > C > B > D
D. D > B > A > C
Solution: A = 3s ($n+l = 3$), B = 4s (4), C = 3p (4), D = 3d (5). Between 4s and 3p (equal $n+l$), the one with lower $n$ (3p) is lower. Decreasing order: 3d > 4s > 3p > 3s, i.e. D > B > C > A.
Q14 — Atomic Orbitals and Quantum Numbers · easy
Which of the following sets of quantum numbers is not allowed?
A. $n = 3,\ l = 2,\ m_l = 0,\ s = +\tfrac{1}{2}$
B. $n = 3,\ l = 2,\ m_l = -2,\ s = +\tfrac{1}{2}$
C. $n = 3,\ l = 3,\ m_l = -3,\ s = -\tfrac{1}{2}$ ✓ Correct
D. $n = 3,\ l = 0,\ m_l = 0,\ s = +\tfrac{1}{2}$
Solution: $l$ can take values from 0 to $n - 1$. For $n = 3$, $l$ can be 0, 1 or 2; $l = 3$ is not allowed.
Q15 — Atomic Orbitals and Quantum Numbers · medium
The correct decreasing order of energy for the orbitals having the following sets of quantum numbers:
A. $n = 3, l = 0, m = 0$
B. $n = 4, l = 0, m = 0$
C. $n = 3, l = 1, m = 0$
D. $n = 3, l = 2, m = 1$
A. D > B > C > A ✓ Correct
B. B > D > C > A
C. C > B > D > A
D. B > C > D > A
Solution: A = 3s, B = 4s, C = 3p, D = 3d. By the $(n+l)$ rule: 3d (5) > 4s (4) > 3p (4, lower $n$ so lower than 4s) > 3s (3). So D > B > C > A.
Q16 — Atomic Orbitals and Quantum Numbers · easy
Identify the incorrect statement from the following.
A. A circular path around the nucleus in which an electron moves is proposed as Bohr's orbit
B. An orbital is the one-electron wave function ($\psi$) in an atom
C. The existence of Bohr's orbits is supported by the hydrogen spectrum
D. An atomic orbital is characterised by the quantum numbers $n$ and $l$ only ✓ Correct
Solution: An atomic orbital is characterised by three quantum numbers: $n$, $l$ and $m_l$ — not by $n$ and $l$ only. The other statements are correct.
Q17 — Atomic Orbitals and Quantum Numbers · medium
Given below are the quantum numbers for 4 electrons.
A. $n = 3, l = 2, m_l = 1, m_s = +\tfrac{1}{2}$
B. $n = 4, l = 1, m_l = 0, m_s = +\tfrac{1}{2}$
C. $n = 4, l = 2, m_l = -2, m_s = -\tfrac{1}{2}$
D. $n = 3, l = 1, m_l = -1, m_s = +\tfrac{1}{2}$
The correct order of increasing energy is
A. D < B < A < C
B. D < A < B < C ✓ Correct
C. B < D < A < C
D. B < D < C < A
Solution: A = 3d ($n+l = 5$), B = 4p (5), C = 4d (6), D = 3p (4). For equal $(n+l)$, lower $n$ is lower: 3d < 4p. Increasing order: 3p < 3d < 4p < 4d, i.e. D < A < B < C.
Q18 — Atomic Orbitals and Quantum Numbers · medium
Consider the following pairs of electrons:
(A) i. $n = 3, l = 1, m_l = 1, m_s = +\tfrac{1}{2}$ ii. $n = 3, l = 2, m_l = 1, m_s = +\tfrac{1}{2}$
(B) i. $n = 3, l = 2, m_l = -2, m_s = -\tfrac{1}{2}$ ii. $n = 3, l = 2, m_l = -1, m_s = -\tfrac{1}{2}$
(C) i. $n = 4, l = 2, m_l = 2, m_s = +\tfrac{1}{2}$ ii. $n = 3, l = 2, m_l = 2, m_s = +\tfrac{1}{2}$
The pair(s) of electrons present in degenerate orbitals is/are
A. only (A)
B. only (B) ✓ Correct
C. only (C)
D. (B) and (C)
Solution: Degenerate orbitals have the same $n$ and the same $l$. Only pair (B) satisfies this (both are 3d orbitals). In (A) the orbitals are 3p and 3d; in (C) they are 4d and 3d.
Q19 — Atomic Orbitals and Quantum Numbers · easy
The number of radial and angular nodes in a 4d orbital are, respectively,
A. 1 and 2 ✓ Correct
B. 3 and 2
C. 1 and 0
D. 2 and 1
Solution: Radial nodes $= n - l - 1 = 4 - 2 - 1 = 1$; angular nodes $= l = 2$.
Q20 — Atomic Orbitals and Quantum Numbers · medium
Consider the following statements:
(A) The principal quantum number $n$ is a positive integer with values $n = 1, 2, 3, \ldots$
(B) The azimuthal quantum number $l$ for a given $n$ can have values $l = 0, 1, 2, \ldots, n$.
(C) The magnetic orbital quantum number $m_l$ for a particular $l$ has $(2l + 1)$ values.
(D) $\pm\tfrac{1}{2}$ are the two possible orientations of electron spin.
(E) For $l = 5$, there will be a total of 9 orbitals.
Which of the above statements are correct?
A. (A), (B) and (C)
B. (A), (C), (D) and (E)
C. (A), (C) and (D) ✓ Correct
D. (A), (B), (C) and (D)
Solution: B is wrong: $l$ goes from 0 to $n - 1$, not up to $n$. E is wrong: for $l = 5$ there are $2l + 1 = 11$ orbitals, not 9. A, C and D are correct.
Q21 — Atomic Orbitals and Quantum Numbers · easy
Which of the following statements are correct?
(A) The electronic configuration of Cr is [Ar] 3d⁵ 4s¹.
(B) The magnetic quantum number may have a negative value.
(C) In the ground state of an atom, the orbitals are filled in order of their increasing energies.
(D) The total number of nodes is given by $n - 2$.
Choose the most appropriate answer from the options given below:
A. (A), (C) and (D) only
B. (A) and (B) only
C. (A) and (C) only
D. (A), (B) and (C) only ✓ Correct
Solution: A is correct (half-filled d gives extra stability), B is correct ($m_l$ ranges from $-l$ to $+l$), C is the Aufbau principle. D is wrong: total nodes $= n - 1$, not $n - 2$.
Q22 — Atomic Orbitals and Quantum Numbers · easy
A certain orbital has no angular nodes and two radial nodes. The orbital is
A. 3p
B. 2p
C. 3s ✓ Correct
D. 2s
Solution: Angular nodes $= l = 0$, so it is an s orbital. Radial nodes $= n - l - 1 = 2$ gives $n = 3$. Hence 3s.
Q23 — Atomic Orbitals and Quantum Numbers · easy
The orbital having two radial as well as two angular nodes is
A. 4f
B. 4d
C. 5d ✓ Correct
D. 3p
Solution: Two angular nodes means $l = 2$ (a d orbital). Two radial nodes means $n - l - 1 = 2$, so $n = 5$. The orbital is 5d.
Q24 — Atomic Orbitals and Quantum Numbers · medium
The number of subshells associated with $n = 4$ and $m_l = -2$ quantum numbers is
A. 8
B. 4
C. 16
D. 2 ✓ Correct
Solution: For $n = 4$, $l$ can be 0, 1, 2, 3. The value $m_l = -2$ is possible only when $l \geq 2$, i.e. for the 4d and 4f subshells — 2 subshells.
Q25 — Atomic Orbitals and Quantum Numbers · medium
The correct statement about probability density (except at infinite distance from the nucleus) is
A. it can be zero for the 1s orbital
B. it can be negative for the 2p orbital
C. it can be zero for the 3p orbital ✓ Correct
D. it can never be zero for the 2s orbital
Solution: Probability density $\psi^2$ is never negative. The 1s orbital has no nodes, so $\psi^2$ is never zero at finite $r$. The 2s orbital has one radial node where $\psi^2 = 0$, so option 4 is false. The 3p orbital has one radial node and one nodal plane, so its probability density can be zero.
Q26 — Atomic Orbitals and Quantum Numbers · easy
The number of orbitals associated with the quantum numbers $n = 5$, $m_s = +\tfrac{1}{2}$ is
A. 25 ✓ Correct
B. 11
C. 15
D. 50
Solution: The number of orbitals in a shell is $n^2 = 5^2 = 25$. The spin quantum number describes the electron, not the orbital, so $m_s = +\tfrac{1}{2}$ does not reduce the count; each of the 25 orbitals can hold one electron with $m_s = +\tfrac{1}{2}$.
Q27 — Atomic Orbitals and Quantum Numbers · easy
The size of the isoelectronic species Cl⁻, Ar and Ca²⁺ is affected by
A. the principal quantum number of the valence shell
B. the nuclear charge ✓ Correct
C. electron–electron interaction in the outer orbitals
D. the azimuthal quantum number of the valence shell
Solution: Isoelectronic species have the same number of electrons (18) and the same valence-shell configuration, so their size differences arise from the different nuclear charges: greater nuclear charge pulls the electrons in more strongly (Ca²⁺ < Ar < Cl⁻).
Q28 — Atomic Orbitals and Quantum Numbers · medium
The quantum numbers of four electrons are given below:
I. $n = 4, l = 2, m_l = -2, m_s = -\tfrac{1}{2}$
II. $n = 3, l = 2, m_l = -2, m_s = +\tfrac{1}{2}$
III. $n = 4, l = 1, m_l = 0, m_s = +\tfrac{1}{2}$
IV. $n = 3, l = 1, m_l = 1, m_s = -\tfrac{1}{2}$
The correct order of their increasing energies will be
A. IV < II < III < I ✓ Correct
B. IV < III < II < I
C. I < III < II < IV
D. I < II < III < IV
Solution: I = 4d ($n+l = 6$), II = 3d (5), III = 4p (5), IV = 3p (4). For equal $(n+l)$, lower $n$ is lower in energy: 3d < 4p. Increasing order: 3p < 3d < 4p < 4d, i.e. IV < II < III < I.
Q29 — Atomic Orbitals and Quantum Numbers · easy
The total number of orbitals associated with the principal quantum number 5 is
A. 20
B. 25 ✓ Correct
C. 10
D. 5
Solution: The number of orbitals in a shell is $n^2 = 5^2 = 25$.
Q30 — Atomic Orbitals and Quantum Numbers · easy
If the principal quantum number $n = 6$, the correct sequence of filling of electrons will be
A. $ns \rightarrow np \rightarrow (n-1)d \rightarrow (n-2)f$
B. $ns \rightarrow (n-2)f \rightarrow (n-1)d \rightarrow np$ ✓ Correct
C. $ns \rightarrow (n-1)d \rightarrow (n-2)f \rightarrow np$
D. $ns \rightarrow (n-2)f \rightarrow np \rightarrow (n-1)d$
Solution: For $n = 6$ the Aufbau order is 6s → 4f → 5d → 6p, i.e. $ns \rightarrow (n-2)f \rightarrow (n-1)d \rightarrow np$, as given by the increasing $(n+l)$ values 6, 7, 7, 7 (with lower $n$ first among equals).