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

Last updated date: 24th Jun 2024
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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$= \sqrt {l\left( {l + 1} \right)} \dfrac{h}{{2\Pi }}$
Where $h$ is Planck’s constant and $l$ is the azimuthal quantum number

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$= \sqrt {l\left( {l + 1} \right)} \dfrac{h}{{2\Pi }}$
Azimuthal quantum number$\left( l \right) = 2$ (explained above)
Therefore, Orbital angular momentum$= \sqrt {2\left( {2 + 1} \right)} \dfrac{h}{{2\Pi }}$
$= \sqrt 6 \dfrac{h}{{2\Pi }}$

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

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 $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^6}4{s^0}$ but it’s configuration is $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^5}4{s^1}$ 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.