Question

For a $d$ electron, the orbital angular momentum is:
A: $\sqrt{6}{\displaystyle \frac{h}{2\mathrm{\Pi }}}$
B: $\sqrt{2}{\displaystyle \frac{h}{2\mathrm{\Pi }}}$
C: $\sqrt{6}{\displaystyle \frac{h}{4\mathrm{\Pi }}}$
D: ${\displaystyle \frac{h}{2\mathrm{\Pi }}}$

Answer 1

Hint: Orbital angular momentum is one of the quantum numbers used to specify the position of an electron in an atom. Total orbital angular momentum is the sum of the orbital angular momentum from each of the electrons.
Formula used: Orbital angular momentum$=\sqrt{l\left(l+1\right)}{\displaystyle \frac{h}{2\mathrm{\Pi }}}$
Where $h$ is Planck’s constant and $l$ is the azimuthal quantum number

Complete step by step answer:
The azimuthal quantum number describes the shape of the orbital and determines orbital angular momentum. It is denoted by $l$. Its value is different for different orbitals. For $s$ orbital its value is $0$, for $p$ orbital its value is $1$, for $d$ orbital its value is $2$ and for $f$ orbital its value is $3$. We have to find the orbital angular momentum of $d$ electron in this question. For $d$ electron value of Azimuthal quantum number is $2$ (as stated above). So orbital angular momentum of $d$ electron can be calculated by using the formula stated above that is:
Orbital angular momentum$=\sqrt{l\left(l+1\right)}{\displaystyle \frac{h}{2\mathrm{\Pi }}}$
Azimuthal quantum number$\left(l\right)=2$ (explained above)
Therefore, Orbital angular momentum$=\sqrt{2\left(2+1\right)}{\displaystyle \frac{h}{2\mathrm{\Pi }}}$
                                                                     $=\sqrt{6}{\displaystyle \frac{h}{2\mathrm{\Pi }}}$

So, correct answer is option A that is $\sqrt{6}{\displaystyle \frac{h}{2\mathrm{\Pi }}}$.

Additional Information:-
An electron shell is the set of allowed states that share the same principal quantum number. Electronic configuration of atoms is very important. Many properties of elements can be easily found with the help of electronic configuration. Main exception to electronic configuration is chromium. Atomic number of chromium is $24$ it’s configuration should be $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{6}4s^0$ but it’s configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}4s^1$ because half-filled $d$ subshell is more stable than partially filled $d$ subshell. Due to which one electron from $d$ subshells jumps to $4s$ .
There are four subshells that are, $s,p,d,f$ .
$s$ subshell has zero orbitals that are it can accommodate a maximum of two-electron, $p$ subshell has three orbitals that is it can accommodate a maximum of $6$ electron, $d$ subshell has five orbitals that is it can accommodate a maximum of $10$ electron, $f$ subshell has seven orbitals that is it can accommodate a maximum of $14$ electron.
Note:
Always remember the maximum number of electrons in an orbital can be two that too with opposite spin and the maximum number of electrons in a subshell depends on the number of orbitals it possesses. There are a total of four quantum numbers and no two electrons can have all the four quantum numbers the same. These four quantum numbers are principal quantum number, angular quantum number, magnetic quantum number, and spin quantum number.