JEE Chemistry 2025 Question with Solution

Given: We need the orbital angular momentum for electrons in $2s$ and $2p$ orbitals.

Find: The corresponding values of orbital angular momentum and the correct option.

The orbital angular momentum of an electron in an atom is given by

$$L = \sqrt{l(l+1)} \frac{h}{2\pi} = \sqrt{l(l+1)} \hbar$$

where $l$ is the azimuthal quantum number.

For a $2s$ orbital, $l = 0$. Therefore,

$$L_{2s} = \sqrt{0(0+1)} \frac{h}{2\pi} = 0$$

For a $2p$ orbital, $l = 1$. Therefore,

$$L_{2p} = \sqrt{1(1+1)} \frac{h}{2\pi} = \sqrt{2} \frac{h}{2\pi}$$

Thus, the orbital angular momentum values for $2s$ and $2p$ orbitals are $0$ and $\sqrt{2} \frac{h}{2\pi}$ respectively. The correct option is D.

Given: Orbital angular momentum depends only on the azimuthal quantum number $l$.

Find: The values for electrons in $2s$ and $2p$ orbitals.

Recall the subshell values:

• For an $s$ orbital, $l = 0$
• For a $p$ orbital, $l = 1$

Use the formula

$$L = \sqrt{l(l+1)} \frac{h}{2\pi}$$

For $2s$ orbital:

$$L = \sqrt{0(0+1)} \frac{h}{2\pi} = 0$$

For $2p$ orbital:

$$L = \sqrt{1(1+1)} \frac{h}{2\pi} = \sqrt{2} \frac{h}{2\pi}$$

Therefore, the required values are $0$ and $\sqrt{2} \frac{h}{2\pi}$ respectively, so the answer is D.