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Structure of Atom
Quantum Numbers
13 questions from 11 papers, each with its answer.
- questions
- 13
- papers
- 11
- years asked, 2017–2026
- 1 of 10
- of Structure of Atom
- 20.97%
Appears in
Concepts tested
- Allowed values of quantum numbers 2
- Capacity of a subshell 2
- Orbital angular momentum 2
- Electrons in an orbital
- number of m values for a given l
- Number of orbitals in a shell
- One orbital per set of n, l, m
- Orbital from quantum numbers
- Quantum numbers of a valence electron
- Significance of quantum numbers
All Questions
2023 1 question
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The relation between $n_m$ ($n_m$ = the number of permissible values of magnetic quantum number $(m)$) for a given value of azimuthal quantum number $(l)$, is
$n_m=2l+1$, hence $l=\frac{n_m-1}{2}$.
$n_m=2l+1$ As there are $(2l+1)$ number of permissible values of magnetic quantum number. Hence, $l=\frac{n_m-1}{2}$
2015 AIPMT-I 1 question
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The angular momentum of electron in 'd' orbital is equal to:Orbital angular momentum $= \sqrt{l(l+1)}\,\hbar$
For a d orbital, l = 2: $\sqrt{2(2+1)}\,\hbar = \sqrt{6}\,\hbar$
2014 AIPMT 1 question
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What is the maximum number of orbitals that can be identified with the following quantum numbers? n = 3, l = 1, $m_l$ = 0n = 3, l = 1, $m_l$ = 0 specifies a single orbital, the $3p_z$ orbital.
2013 NEET 1 question
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What is the maximum number of electrons that can be associated with the following set of quantum numbers? n = 3, l = 1 and m = −1n = 3, l = 1 and m = −1 represent one 3p orbital, which can hold a maximum of 2 electrons.
2012 AIPMT-MAINS 1 question
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The orbital angular momentum of a p-electron is given as:Orbital angular momentum $= \sqrt{l(l+1)}\,\frac{h}{2\pi}$
For a p-electron l = 1: $\sqrt{1(1+1)}\,\frac{h}{2\pi} = \sqrt2\times\frac{h}{2\pi}$
$= \frac{h}{\sqrt2\pi}$
2012 AIPMT-PRE 2 questions
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Maximum number of electrons in a subshell with l = 3 and n = 4 is:n = 4, l = 3 is the 4f subshell.
Number of orbitals $= 2l + 1 = 7$
Each orbital holds 2 electrons, so maximum electrons $= 2(2l + 1) = 14$ -
The correct set of four quantum numbers for the valence electron of rubidium atom (Z = 37) isRb (Z = 37): $[Kr]5s^1$
The valence electron is in the 5s orbital: n = 5, l = 0 (s orbital), m = 0, s = $+\frac{1}{2}$
2011 AIPMT-PRE 1 question
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The total number of atomic orbitals in fourth energy level of an atom isThe number of orbitals in a shell with principal quantum number n is $n^2$.
For n = 4 the subshells are 4s (1 orbital), 4p (3 orbitals), 4d (5 orbitals) and 4f (7 orbitals).
Total orbitals = 1 + 3 + 5 + 7 = 16
Equivalently, $n^2 = 4^2 = 16$
So the fourth energy level has 16 atomic orbitals (which can hold 32 electrons).
2009 AIPMT 2 questions
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Maximum number of electrons in a subshell of an atom is determined by the followingA subshell with azimuthal quantum number l has $(2l + 1)$ orbitals, one for each value of $m_l$ from −l to +l.
Each orbital can hold a maximum of 2 electrons with opposite spins (Pauli exclusion principle).
Maximum number of electrons in a subshell = $2(2l + 1) = 4l + 2$
Check: s (l = 0) holds 2, p (l = 1) holds 6, d (l = 2) holds 10, f (l = 3) holds 14.
($2n^2$ is the maximum number of electrons in a shell, not a subshell.) -
Which of the following is not permissible arrangement of electrons in an atom?Rules: l can take values 0 to (n − 1); m can take values from −l to +l; s is $+\frac{1}{2}$ or $-\frac{1}{2}$.
n = 3, l = 2, m = –2: l = 2 is allowed for n = 3, and m = –2 lies within −2 to +2. Permissible.
n = 4, l = 0, m = 0: allowed (4s orbital). Permissible.
n = 5, l = 3, m = 0: l = 3 is allowed for n = 5, and m = 0 lies within −3 to +3. Permissible.
n = 3, l = 2, m = –3: for l = 2, m can only be −2, −1, 0, +1, +2; the value −3 is not possible.
Hence the arrangement n = 3, l = 2, m = –3, s = $-\frac{1}{2}$ is not permissible.
Note: the source prints the first and the fourth options identically; the first has been restored to m = –2, as in the original paper.
2007 AIPMT-PRE 1 question
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Consider the following sets of quantum numbers:
(i) $n = 3$, $l = 0$, $m = 0$, $s = +\dfrac{1}{2}$
(ii) $n = 2$, $l = 2$, $m = 1$, $s = +\dfrac{1}{2}$
(iii) $n = 4$, $l = 3$, $m = -2$, $s = -\dfrac{1}{2}$
(iv) $n = 1$, $l = 0$, $m = -1$, $s = -\dfrac{1}{2}$
(v) $n = 3$, $l = 2$, $m = 3$, $s = +\dfrac{1}{2}$
Which of the following sets of quantum numbers is not possible?Rules: for a given $n$, $l$ varies from 0 to $(n - 1)$; for a given $l$, $m$ varies from $-l$ to $+l$; and $s = \pm\dfrac{1}{2}$.
(i) $n = 3$, $l = 0$, $m = 0$: allowed (3s orbital).
(ii) $n = 2$, $l = 2$: not possible, because for $n = 2$, $l$ can be only 0 or 1.
(iii) $n = 4$, $l = 3$, $m = -2$: allowed (4f orbital).
(iv) $n = 1$, $l = 0$, $m = -1$: not possible, because for $l = 0$, $m$ can be only 0.
(v) $n = 3$, $l = 2$, $m = 3$: not possible, because for $l = 2$, $m$ varies only from $-2$ to $+2$.
So the sets that are not possible are (ii), (iv) and (v).
2006 AIPMT-PRE 1 question
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The orientation of an atomic orbital is governed by :-The principal quantum number ($n$) determines the size and, to a large extent, the energy of the orbital.
The azimuthal quantum number ($l$) determines the shape of the orbital (s, p, d, f).
The magnetic quantum number ($m_l$) determines the orientation of the orbital in space relative to the others; for a given $l$ it takes the $(2l + 1)$ values from $-l$ to $+l$.
For example, for $l = 1$, $m_l = -1, 0, +1$ correspond to the three orientations $p_x$, $p_y$ and $p_z$.
The spin quantum number ($m_s$) describes the spin of the electron, not the orbital.
So the orientation of an atomic orbital is governed by the magnetic quantum number.
2001 AIPMT 1 question
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The following quantum numbers are possible for how many orbitals: n = 3, l = 2, m = +2The three quantum numbers n, l and $m_l$ together specify exactly one orbital.
n = 3, l = 2 is the 3d subshell, which has five orbitals with $m_l$ = –2, –1, 0, +1, +2.
Only one of these has $m_l = +2$.
So the given quantum numbers are possible for 1 orbital (which can hold 2 electrons with $m_s = \pm\dfrac{1}{2}$).