For a $d - $electron, the orbital angular momentum is:
A.$\sqrt 6 \dfrac{h}{{2\pi }}$
B.$\sqrt 2 \dfrac{h}{{2\pi }}$
C.$\dfrac{h}{{2\pi }}$
D.$2\dfrac{h}{{2\pi }}$

Hint: Angular momentum: It is defined for rotation motion. If a body rotates at its fixed axis then angular momentum is generated. It is represented as $L$and angular momentum is calculated as follows: $L = mvr$ and in quantum it is calculated as $\sqrt {l(l + 1)} \dfrac{h}{{2\pi }}$.

Complete step by step answer:
To completely identify where an electron is present, First we have to know about quantum numbers.
Quantum numbers: To completely describe the position of an electron in an atom, quantum numbers are needed. They are further classified into four categories:
Principal quantum number
Azimuthal or angular momentum quantum number
Magnetic quantum number
Spin quantum number
Principal quantum number: It is represented by $n$. This quantum number describes the electron shell, or energy level of an atom. The value of $n$ ranges from $1$ to the shell containing the outermost electron.
Azimuthal or angular momentum quantum number: This quantum number describes the subshell and gives the magnitude of the angular momentum. It is represented by $l$. For $s - $orbital $l = 0$, for $p - $orbital $l = 1$, for $d - $orbital $l = 2$ and so on. The value of $l$ ranges from $0$ to $n - 1$. That’s why the number of orbits for $n = 1$is only $1$ i.e. $s - $orbit. This quantum number decides the shape of an atomic orbital.
Magnetic quantum number: This quantum number describes the energy levels which are available within a subshell. It is represented by $m$. Its value varies from $ - l$ to $l$ including zero.
Spin quantum number: This quantum number describes the spin of an electron within that orbit and gives the projection of the spin angular momentum. It is represented by $s$. Its values are either $ + \dfrac{1}{2}$or $ - \dfrac{1}{2}$
Angular momentum: It is defined for rotation motion. If a body rotates at its fixed axis then angular momentum is generated. It is represented as $L$and angular momentum is calculated as follows: $L = mvr$ and in quantum it is calculated as $\sqrt {l(l + 1)} \dfrac{h}{{2\pi }}$
For $d - $orbital electrons, $l = 2$. So angular momentum will be $\sqrt {2 \times 3} \dfrac{h}{{2\pi }} = \sqrt 6 \dfrac{h}{{2\pi }}$.

So, the correct answer is Option A.

Note:
Angular momentum and angular momentum quantum number are different things. Angular momentum is used to find the momentum of an electron in the orbitals and angular momentum quantum number is used to find the shape of the atomic orbitals.