The orbital angular momentum (L) of an electron in an atom is given by the formula:

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

where $l$ is the azimuthal quantum number or orbital quantum number, and $h$ is the Planck constant. The azimuthal quantum number $l$ determines the shape of the orbital and for a given electron shell $n$, $l$ can have values from 0 to $n-1$.

For a d-electron, the value of $l$ is 2, since the orbitals are designated as follows:

• s-orbital: $l = 0$
• p-orbital: $l = 1$
• d-orbital: $l = 2$
• f-orbital: $l = 3$

Substituting $l = 2$ into the formula for orbital angular momentum, we get:

$$L = \sqrt{2(2 + 1)}\left(\frac{h}{2\pi}\right) = \sqrt{6}\left(\frac{h}{2\pi}\right)$$

This shows that the correct option for the orbital angular momentum of a d-electron is Option A:

$$\sqrt{6}\left(\frac{h}{2\pi}\right)$$