Number of angular nodes for 4d orbital is _____. (i) 4 (ii) 3 (iii) 2 (iv) 1

NCERT Exemplar · Q9

Q. Number of angular nodes for 4d orbital is __________.

(i) 4

(ii) 3

(iii) 2

(iv) 1

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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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Angular nodes depend only on the azimuthal quantum number $l$ , not on the principal quantum number $n$ . For a 4d orbital, $l = 2$ , so the number of angular nodes is $l = 2$ . The correct option is (C).

The idea of angular nodes comes directly from the shape of atomic orbitals. In quantum mechanics, an orbital is described by three quantum numbers: $n$ (principal), $l$ (azimuthal), and $m_l$ (magnetic). The angular nodes are surfaces where the probability of finding an electron is zero, and they arise from the angular part of the wavefunction — specifically from the spherical harmonics $Y_{l}^{m}(\theta, \phi)$ .

The number of angular nodes is always equal to the value of $l$ . This is a fixed rule: an s orbital ( $l=0$ ) has zero angular nodes, a p orbital ( $l=1$ ) has one angular node (a plane), a d orbital ( $l=2$ ) has two angular nodes, and so on. The principal quantum number $n$ only affects the total number of nodes (radial + angular), not the angular ones alone.

Let’s apply this to the 4d orbital.

Identify the azimuthal quantum number $l$ .

For a d orbital, $l = 2$ . The number 4 in "4d" is the principal quantum number $n$ , which tells you the energy level and the number of radial nodes, but it does not change $l$ .
Recall the relation for angular nodes.

Angular nodes = $l$ . This is a direct consequence of the spherical harmonics: the function $Y_{l}^{m}$ has $l$ nodal surfaces (planes or cones) that pass through the nucleus. …

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Q.Number of angular nodes for 4d orbital is __________.

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