JEE Advance - Chemistry (1996 - No. 3)

The correct answer is Option B: Zero.

Orbital angular momentum of an electron in any given orbital is determined by the azmuthal quantum number (l), which defines the shape of the orbital. The formula to calculate the magnitude of orbital angular momentum is given by: $$L = \sqrt{l(l+1)} \frac{h}{2\pi}$$ where $L$ is the orbital angular momentum, $l$ is the azmuthal quantum number, and $h$ is Planck's constant.

For an electron in a 2s orbital, the principal quantum number, $n$, is 2, and the azmuthal quantum number, $l$, for an s orbital is 0 (since the value of $l$ ranges from 0 to $n-1$).

Putting $l=0$ in the formula gives:

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

Thus, the orbital angular momentum of an electron in a 2s orbital is Zero, as there is no angular momentum associated with s orbitals owing to their spherical symmetry.