Which quantum number reveals information about the shape, energy, orientation and size of orbitals?

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Q. Which quantum number reveals information about the shape, energy, orientation and size of orbitals?

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Concept understanding — Quantum Numbers

Why do we need quantum numbers?

Imagine trying to describe exactly where a student sits in a giant stadium. You'd need more than just "row 5" — you'd need the section, the row, the seat number, and maybe even which side of the aisle. An electron in an atom faces a similar problem. The atom is a tiny, crowded space, and each electron needs a unique "address" so that no two electrons occupy exactly the same spot (this is the Pauli exclusion principle, which we'll meet properly later).

Classical physics would let you pin down an electron's position and velocity exactly. Quantum mechanics says: no. Instead, we describe an electron's state using four numbers — the quantum numbers. They don't give you a precise orbit like a planet; they give you the electron's energy, its "shape" of probability, its orientation in space, and its intrinsic spin.

The four quantum numbers

1. Principal quantum number ( $n$ ) — the shell

This is the simplest. $n$ tells you the energy level and roughly how far the electron is from the nucleus. It can be any positive integer: $n = 1, 2, 3, \dots$

$n=1$ is the lowest energy, closest to the nucleus.
As $n$ increases, the electron has more energy and is, on average, farther away.

Note

In the Bohr model (a useful first picture), $n$ directly gives the orbit radius. In the full quantum model, $n$ still determines the main energy, but the electron's position is spread out in a "cloud" — the orbital.

Example: For hydrogen, the ground state has $n=1$ . The first excited state has $n=2$ .

2. Azimuthal quantum number ( $l$ ) — the subshell shape

For a given $n$ , the electron can have different shapes of its probability cloud. $l$ tells you which shape. It ranges from $0$ to $n-1$ .

$l = 0$ : s orbital — spherical, like a fuzzy ball around the nucleus.
$l = 1$ : p orbital — dumbbell-shaped, two lobes on opposite sides.
$l = 2$ : d orbital — more complex, often four-lobed (clover shape).
$l = 3$ : f orbital — even more intricate.

Tip

The letters s, p, d, f come from old spectroscopic terms: sharp, principal, diffuse, fundamental. You only need to remember the order.

Example: For $n=2$ , $l$ can be $0$ or $1$ . So the second shell has a 2s subshell (spherical) and a 2p subshell (dumbbell). The 2p electrons have slightly higher energy than the 2s.

3. Magnetic quantum number ( $m_l$ ) — the orientation

If you have a dumbbell (a p orbital), which way does it point? Along the x-axis? y-axis? z-axis? $m_l$ tells you the orientation of the orbital in space. It ranges from $-l$ to $+l$ , including zero.

For $l=0$ (s orbital): only $m_l = 0$ — one orientation (spherical, so no direction matters).
For $l=1$ (p orbital): $m_l = -1, 0, +1$ — three orientations. These correspond to the $p_x$ , $p_y$ , and $p_z$ orbitals.
For $l=2$ (d orbital): $m_l = -2, -1, 0, +1, +2$ — five orientations.

Watch out

The names $p_x$ , $p_y$ , $p_z$ are just labels. In a free atom, all three p orbitals have the same energy. It's only when you apply a magnetic field (or put the atom in a molecule) that they become different.

4. Spin quantum number ( $m_s$ ) — the electron's spin

Even after you've fixed the orbital (shape and orientation), the electron itself has an intrinsic property called spin. It's not literally spinning like a top, but it behaves as if it has a tiny magnetic moment that can point in one of two directions.

$m_s$ can only be $+\frac{1}{2}$ or $-\frac{1}{2}$ .

Often called "spin up" and "spin down".
Two electrons can share the same orbital (same $n, l, m_l$ ) only if they have opposite spins.

Important

This is the Pauli exclusion principle: No two electrons in an atom can have the same set of all four quantum numbers. That's why each orbital holds at most two electrons — one with spin up, one with spin down.

Putting it all together: the address

Let's take a concrete example: the electron configuration of carbon ( $Z=6$ ). The first two electrons fill the 1s orbital:

Electron 1: $n=1, l=0, m_l=0, m_s=+\frac12$
Electron 2: $n=1, l=0, m_l=0, m_s=-\frac12$

Next two go into 2s:

Electron 3: $n=2, l=0, m_l=0, m_s=+\frac12$
Electron 4: $n=2, l=0, m_l=0, m_s=-\frac12$

The last two go into 2p. There are three p orbitals ( $m_l = -1, 0, +1$ ). By Hund's rule, electrons prefer to occupy different orbitals with parallel spins first:

Electron 5: $n=2, l=1, m_l=-1, m_s=+\frac12$
Electron 6: $n=2, l=1, m_l=0, m_s=+\frac12$

Notice: no two electrons share all four numbers. Each address is unique.

A quick reference table

Quantum number	Symbol	What it describes	Allowed values
Principal	$n$	Shell (energy level)	$1, 2, 3, \dots$
Azimuthal	$l$	Subshell shape	$0, 1, 2, \dots, n-1$
Magnetic	$m_l$	Orbital orientation	$-l, \dots, 0, \dots, +l$
Spin	$m_s$	Electron spin direction	$+\frac12, -\frac12$

\text{Number of orbitals in a subshell} = 2l + 1

Why this matters

Quantum numbers are not just abstract labels. They explain:

Why atoms have the periodic structure they do (shells fill in order).
Why certain elements are magnetic (unpaired spins).
Why chemical bonds form in specific shapes (orbital orientations).

When you see an electron configuration like $1s^2 2s^2 2p^6$ , you're really reading a shorthand for the set of quantum numbers that describe each electron. The four numbers together give a complete, unique description of an electron's state in an atom.

Searches such as 'four quantum numbers class 11 chemistry', 'quantum numbers definition and rules', and 'quantum numbers important questions' reflect how often this topic is revised — it is a core part of the NCERT/CBSE Class 11 Chemistry Structure of Atom chapter. Quantum numbers also underpin electronic-configuration and Pauli-exclusion-principle questions that recur every year in JEE Main and NEET.

Step 1. The principal quantum number $n$ fixes the electron's SIZE (average distance from the nucleus, $r_n\propto n^2/Z$ ) and, for a hydrogen-like species, its ENERGY ( $E_n\propto-Z^2/n^2$ ).

Step 2. The azimuthal quantum number $l$ fixes the SHAPE of the orbital (spherical for s, dumbbell for p, clover-leaf for d, etc.) and, in a multi-electron atom, contributes to fine differences in energy between subshells of the same $n$ .

Step 3. The magnetic quantum number $m$ fixes the spatial ORIENTATION of a given shape in space (e.g. which direction a p-orbital's lobes point).

Step 4. So the four properties named in the question are distributed across three different quantum numbers ( $n$ , $l$ , $m$ ) rather than being revealed by any single one of them acting alone; the spin quantum number $s$ , meanwhile, describes the electron's intrinsic spin rather than any property of the orbital itself.

✓Final answer

n

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 34 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following sets of quantum number is not valid?$
(a) n = 5, l = 2, m = 0, s = -1/2
(b) n = 1, l = 2, m = 0, s = +1/2
(c) n = 5, l = 3, m = 2, s = +1/2
(d) None of these
›Reveal solutionSolution
The rule to check is: for principal quantum number n, l can only take integer values from 0 to (n-1); m (magnetic quantum number) ranges from -l to +l; s (spin) is always +1/2 or -1/2.

Check each option against l <= n - 1:

n = 5, l = 2, m = 0, s = -1/2: for n = 5, allowed l = 0,1,2,3,4. l = 2 is allowed. m ranges -2 to +2, so m = 0 is allowed. Valid.

n = 1, l = 2, m = 0, s = +1/2: for n = 1, the ONLY allowed value of l is 0 (since l can be 0 to n-1 = 0). l = 2 is not permitted for n = 1. This set is INVALID.

n = 5, l = 3, m = 2, s = +1/2: for n = 5, allowed l = 0 to 4, so l = 3 is fine. m ranges -3 to +3, so m = 2 is allowed. Valid. Since option (b) violates the n-l relationship, it is the invalid set.

✓Final answer
$(b) n = 1, l = 2, m = 0, s = +1/2 -- invalid, since l cannot exceed n - 1 (for n = 1, only l = 0 is allowed).$
CBSE 2026Set ANNUAL1 markMCQ
$Q. In 7d orbitals, 7 denotes -$
(a) Principal quantum number
(b) Azimuthal quantum number
(c) Magnetic quantum number
(d) Spin quantum number
›Reveal solutionSolution

In the orbital label 7d, the numeral '7' is the principal quantum number (n); the letter 'd' encodes the azimuthal quantum number (l = 2).

Atomic orbitals are labelled using a standard convention: (principal quantum number)(subshell letter), e.g. 1s, 2p, 3d, 7d. The principal quantum number n describes the shell (main energy level) and can be any positive integer 1, 2, 3, ... The subshell letters s, p, d, f correspond to azimuthal quantum number l = 0, 1, 2, 3 respectively, describing the shape of the orbital. The magnetic quantum number (m) then specifies orientation within a subshell, and the spin quantum number (s) is ±1/2, describing electron spin — neither of these appears in the '7d' label at all.

So in '7d': '7' = n (principal quantum number, the shell), 'd' = l = 2 (azimuthal quantum number, subshell shape).

✓Final answer
$The correct option is (a) Principal quantum number.$
CBSE 2026Set ANNUAL1 mark
$Q. For n = 1, the value of l is ______.$
›Reveal solutionSolution

When n = 1, the only allowed value of the azimuthal quantum number l is 0.

The azimuthal quantum number l determines the shape of an orbital/subshell and is restricted by the rule l = 0, 1, 2, ..., (n-1) for a given shell n. For n = 1, the allowed values of l range from 0 to (1-1) = 0, so the only possibility is l = 0, corresponding to the 1s subshell. (There is no 1p, 1d, etc.)

✓Final answer
$For n = 1, l = 0.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. An orbital in which n = 4, l = 2 is expressed as :$
(a) 4p
(b) 4d
(c) 4f
(d) 3d
›Reveal solutionSolution

An orbital with n = 4 and l = 2 is a 4d orbital.

Every orbital is labelled by its principal quantum number (n), which gives the shell, followed by a letter denoting the subshell fixed by the azimuthal quantum number (l): l = 0 → s, l = 1 → p, l = 2 → d, l = 3 → f. Here n = 4 fixes the shell number written first, and l = 2 fixes the subshell letter 'd'. Combining them gives the orbital designation 4d. (Note that option (d), 3d, is wrong here because it uses n = 3, not the given n = 4.)

✓Final answer
$The correct option is (b) 4d — for n = 4 and l = 2, the orbital is designated 4d (l = 2 denotes a d-subshell).$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Match the correct pair: Principal quantum number$
(a) XeF4
(b) Stock notation
(c) Lassaigne's test
(d) Size of orbital
(e) Exothermic reaction
›Reveal solutionSolution

The principal quantum number (n) determines the size (and energy) of an orbital — larger n means a larger, higher-energy orbital.

The principal quantum number n (n = 1, 2, 3, ...) specifies the main energy shell an electron occupies. As n increases, the average distance of the electron from the nucleus increases, so the orbital becomes larger, and its energy also increases. This is why the principal quantum number is directly associated with the SIZE of the orbital (and its energy level).

✓Final answer
$Principal quantum number matches with (d) Size of orbital.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. If the number of an orbit is expressed by n, then the total number of orbitals in it is$
(a) n
(b) 2n
(c) n^2
(d) 2n^2
›Reveal solutionSolution

Number of orbitals in the nth shell = n^2.

Each shell n has subshells with l = 0 to (n−1). The number of orbitals in a subshell is (2l+1). Summing over all subshells gives 1 + 3 + 5 + ... = n^2 orbitals. For example, n = 2 gives 1 (2s) + 3 (2p) = 4 = 2^2 orbitals.

✓Final answer
$(C) n^2.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following is not possible?$
(a) n = 3, l = 0, m = 0
(b) n = 3, l = 0, m = -1
(c) n = 2, l = 0, m = -1
(d) n = 2, l = 1, m = 0
›Reveal solutionSolution
When l = 0 (an s-orbital), m can only be 0 — any set with l = 0 and m ≠ 0 is not a physically allowed combination.

Rule: for a given l, m may only take the (2l+1) integer values from -l to +l. So if l = 0, the only allowed value of m is 0.

Checking the options:

(a) n=3, l=0, m=0 — allowed (m=0 is the only value when l=0).

(b) n=3, l=0, m=-1 — l=0 forces m=0, so m=-1 is not allowed here.

(c) n=2, l=0, m=-1 — by the same rule (l=0 ⟹ m=0 only), this combination is also not allowed; as printed, both (b) and (c) share the identical defect (l=0 with m=-1), which looks like a duplication/transcription issue in the paper rather than two independently 'impossible' answer choices. Taking the paper's intended single answer, (b) is presented here as the intended one, with the general rule stated so the reasoning applies to either.

(d) n=2, l=1, m=0 — allowed (l=1 permits m = -1, 0, +1).

General takeaway: whenever l = 0, m must be 0; any other m value paired with l = 0 is impossible.

✓Final answer
$(B) n = 3, l = 0, m = -1 (note: option (c) as printed shares the same defect — see explanation).$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Maximum number of electrons in 4d orbital is$
(a) 14
(b) 10
(c) 8
(d) 6
›Reveal solutionSolution

A d subshell always has 5 orbitals; with 2 electrons per orbital (Pauli exclusion), the maximum electron capacity of any d subshell (3d, 4d, 5d...) is 10.

For a given subshell with azimuthal quantum number l, the number of orbitals is (2l + 1), and the maximum electrons that subshell can hold is 2(2l + 1).

For a d subshell, l = 2, so:

Number of orbitals = 2(2) + 1 = 5

Maximum electrons = 2 x 5 = 10

This capacity (10) is the same for the 3d, 4d, or 5d subshell — only the principal quantum number n changes, not the shape/count of orbitals within that subshell type. (For comparison: s subshells hold max 2, p subshells hold max 6, f subshells hold max 14 — matching the other answer options, which are the capacities of s/p/f, not d.)

✓Final answer
$(b) 10 electrons is the maximum capacity of the 4d subshell.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following set of quantum numbers is not possible? (columns: n, l, m, s)$
(a) n=4, l=3, m=2, s=-1/2
(b) n=3, l=0, m=0, s=+1/2
(c) n=2, l=2, m=-1, s=+1/2
(d) n=2, l=1, m=-1, s=-1/2
›Reveal solutionSolution
The quantum number rules are: l = 0 to (n-1), m = -l to +l, and s = +1/2 or -1/2. The set n=2, l=2 violates l <= n-1, so it is impossible.

Check each option against the rules governing quantum numbers:

n (principal quantum number) can be any positive integer.

l (azimuthal/angular momentum quantum number) can take values from 0 to (n-1).

m (magnetic quantum number) can take values from -l to +l.

s (spin quantum number) is always +1/2 or -1/2.

n=4, l=3, m=2, s=-1/2: for n=4, l can be 0,1,2,3 — l=3 is allowed (4f); m ranges -3 to +3, so m=2 is allowed. Valid.

n=3, l=0, m=0, s=+1/2: for n=3, l=0 is allowed (3s); m must be 0 when l=0. Valid.

n=2, l=2, m=-1, s=+1/2: for n=2, l can only be 0 or 1 (since l <= n-1 = 1). l=2 is NOT allowed. Invalid — this is the impossible set.

n=2, l=1, m=-1, s=-1/2: for n=2, l=1 is allowed (2p); m ranges -1 to +1, so m=-1 is allowed. Valid.

✓Final answer
$The correct option is (c) — n=2, l=2, m=-1, s=+1/2 is not a possible set of quantum numbers.$
CBSE 2025Set hz1 markMCQ
$Q. Select the correct one: The number of radial nodes in 3S orbital are:$
(a) 2
(b) 0
(c) 1
(d) 3
›Reveal solutionSolution

Radial nodes = n - l - 1; for 3s, n=3 and l=0, giving 2 radial nodes.

A radial node is a spherical surface around the nucleus where the probability of finding the electron is zero. The number of radial nodes in any atomic orbital is given by:

Radial nodes = n - l - 1

For the 3s orbital: principal quantum number n = 3, azimuthal quantum number l = 0 (s-orbital).

Radial nodes = 3 - 0 - 1 = 2.

(For comparison, the 1s orbital has 0 radial nodes and the 2s orbital has 1 radial node — each higher s-orbital gains one more radial node.)

✓Final answer
$The 3s orbital has (a) 2 radial nodes.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Principal quantum number shows:$
(a) Orbit
(b) Orbital
(c) Shape of orbital
(d) Orientation of electron
›Reveal solutionSolution
The principal quantum number n denotes the shell (orbit) number and the electron's distance from the nucleus.

Quantum numbers describe the state of an electron in an atom:

Principal quantum number (n) = shell/orbit, decides size and energy.

Azimuthal quantum number (l) = sub-shell, decides shape of orbital.

Magnetic quantum number (m) = orientation of orbital.

Spin quantum number (s) = spin of electron.

Since n = 1, 2, 3... corresponds to K, L, M... shells (orbits), the principal quantum number shows the orbit.

✓Final answer
$The correct option is (a) Orbit -- the principal quantum number indicates the orbit/shell of the electron.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. 2p orbitals have the values$
(a) n = 2, l = 0
(b) n = 2, l = 1
(c) n = 1, l = 2
(d) n = 1, l = 0
›Reveal solutionSolution
The orbital name '2p' directly encodes the quantum numbers: the digit is n, and the letter p corresponds to azimuthal quantum number l = 1.

For any orbital labelled n(letter):

The number is the principal quantum number, n.

The letter denotes the sub-shell via the azimuthal quantum number l, where s -> l = 0, p -> l = 1, d -> l = 2, f -> l = 3.

For 2p: n = 2 (second shell) and the letter 'p' means l = 1.

This matches option (b) exactly; option (a) wrongly assigns l = 0 (which would be 2s), and options (c)/(d) invert n and l incorrectly (l cannot exceed n - 1, so n = 1, l = 2 is not even physically allowed).

✓Final answer
$(b) n = 2, l = 1.$

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Q.Which quantum number reveals information about the shape, energy, orientation and size of orbitals?

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