Out of the following pairs of electrons, identify the pairs of electrons present in degenerate orbitals : (Not

NCERT Exemplar · Q18

Q. Out of the following pairs of electrons, identify the pairs of electrons present in degenerate orbitals : (Note: more than one of the given options may be correct.)

(i)

(a) n = 3, l = 2, m_l = -2, m_s = -1/2

(b) n = 3, l = 2, m_l = -1, m_s = -1/2

(ii)

(a) n = 3, l = 1, m_l = 1, m_s = +1/2

(b) n = 3, l = 2, m_l = 1, m_s = +1/2

(iii)

(a) n = 4, l = 1, m_l = 1, m_s = +1/2

(b) n = 3, l = 2, m_l = 1, m_s = +1/2

(iv)

(a) n = 3, l = 2, m_l = +2, m_s = -1/2

(b) n = 3, l = 2, m_l = +2, m_s = +1/2

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Concept understanding — Energy Level Quantization

Energy Level Quantization: From Intuition to Precision

Imagine you're climbing a smooth ramp. You can stop at any height — 1 metre, 1.5 metres, 2.1 metres — anywhere you like. That's how we intuitively think about energy in everyday life: continuous, like a slide.

Now imagine a staircase. You can stand on step 1, step 2, or step 3 — but you cannot stand halfway between step 2 and step 3. There's no such place. The steps are discrete, not continuous.

Energy level quantization is the idea that in the microscopic world of atoms and molecules, energy behaves like a staircase, not a ramp. Electrons in an atom cannot have just any energy — they can only occupy specific, allowed energy levels. Everything else is forbidden.

Why does this happen? The core intuition

In the classical world, an electron orbiting a nucleus would continuously radiate energy, spiral inward, and crash — atoms would be unstable. But atoms are stable. Nature solved this problem by imposing a rule: the electron's angular momentum (and therefore its energy) can only take certain discrete values.

Think of a guitar string. It can only vibrate at specific frequencies — its fundamental and harmonics. You can't pluck it to produce a frequency halfway between two harmonics. The string's vibration is quantized by its boundaries. Similarly, an electron bound to a nucleus is confined in space, and that confinement forces its energy to be quantized.

Note

Quantization is not a mysterious extra rule — it emerges naturally whenever a wave (like an electron's matter wave) is confined. Confinement creates standing waves, and standing waves only exist at specific frequencies.

The precise statement

For a bound system (like an electron in an atom), the total energy $E$ of the system can only take certain discrete values:

E = E_1, E_2, E_3, \dots

where each $E_n$ is a specific, fixed number. The integer $n$ (1, 2, 3, …) is called the principal quantum number. The lowest energy level ( $n=1$ ) is the ground state; higher levels ( $n>1$ ) are excited states.

For the hydrogen atom, the allowed energies are given by:

E_n = -\frac{13.6 \text{ eV}}{n^2}

So:

$n=1$ : $E_1 = -13.6$ eV (ground state)
$n=2$ : $E_2 = -3.4$ eV
$n=3$ : $E_3 = -1.51$ eV
and so on, approaching 0 eV as $n \to \infty$ (the ionization limit)

Important

The negative sign means the electron is bound to the nucleus. Zero energy corresponds to the electron being free (ionized). The more negative the energy, the more tightly bound the electron.

How do we know this is real?

The most direct evidence comes from atomic spectra. When an electron jumps from a higher energy level to a lower one, it emits a photon of light with energy exactly equal to the difference:

\Delta E = E_{\text{higher}} - E_{\text{lower}} = hf

where $h$ is Planck's constant and $f$ is the frequency of the emitted light. …

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Degenerate orbitals have the same energy. For a hydrogen-like atom, energy depends only on $n$ ; for multi-electron atoms, it depends on $n + l$ . Here, we compare the $(n, l)$ pairs of each option — only electrons with identical $(n, l)$ values occupy degenerate orbitals. The correct pairs are (A) and (D).

The idea of degenerate orbitals is central to understanding atomic structure. Two orbitals are degenerate when they have exactly the same energy. In a hydrogen atom, energy depends only on the principal quantum number $n$ , so all orbitals with the same $n$ are degenerate. But in multi-electron atoms — which is what we deal with in most chemistry problems — the energy of an orbital depends on both $n$ and the azimuthal quantum number $l$ , because of electron-electron repulsion and shielding. The rule is: orbitals with the same $(n, l)$ pair have the same energy; orbitals with different $(n, l)$ pairs have different energies.

The magnetic quantum number $m_l$ and the spin quantum number $m_s$ do not affect the orbital energy in the absence of an external magnetic field. So when we ask whether two electrons are in degenerate orbitals, we only need to check whether their $(n, l)$ values match.

Let’s go through each option.

Option (A):

Electron (a): $n = 3$ , $l = 2$

Electron (b): $n = 3$ , $l = 2$

Both have the same $(n, l) = (3, 2)$ . They differ in $m_l$ and $m_s$ , but that doesn’t change the energy. So these two electrons are in degenerate orbitals — in fact, they are in different $d$ orbitals of the same $3d$ subshell.
Option (B):

Electron (a): $n = 3$ , $l = 1$

Electron (b): $n = 3$ , $l = 2$

Here $l$ differs: one is a $3p$ orbital, the other is a $3d$ orbital. In a multi-electron atom, $3p$ and $3d$ have different energies (the $3p$ is lower). So these are not degenerate.
Option (C):

Electron (a): $n = 4$ , $l = 1$

Electron (b): $n = 3$ , $l = 2$

Both $n$ and $l$ differ. The first is a $4p$ orbital, the second is a $3d$ orbital. Their energies are different — in fact, in multi-electron atoms, $4p$ is higher than $3d$ . Not degenerate.
Option (D): …

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Q.Out of the following pairs of electrons, identify the pairs of electrons present in degenerate orbitals : (Note: more than one of the given options may be correct.)

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