Question

Number of electrons in chromium for which ${\text{n}} + {\text{l}} = 4$ is:
A.$4$
B.$5$
C.$7$
D.$10$

Answer 1

Hint: To answer this question, you must recall the electronic configuration of chromium atoms. Also you must be familiar with the values of the angular quantum number of various orbitals.

Complete step by step answer:
The principal quantum number $\left( {\text{n}} \right)$ is a positive integer. It determines the size and energy of the orbitals. We know that the size of energy shells increases with increasing ${\text{n}}$. Thus we can say that the orbital size also increases with ${\text{n}}$.
Azimuthal quantum number $\left( {\text{l}} \right)$is an integer having all values between $0$and ${\text{n}} - 1$. It is known as a subsidiary quantum number and is used to represent a subshell. The azimuthal quantum number is also used to define the shape of an orbital.
We know that, each value of $l$ is designated with letters as, $s\left( {l = 0} \right),p\left( {l = 1} \right),d\left( {l = 2} \right),f\left( {l = 3} \right)$ and so on.
The electronic configuration of chromium is,
${\text{Cr}}:{\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^6}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{s}}^{\text{1}}}{\text{3}}{{\text{d}}^{\text{5}}}$
Calculating the value of ${\text{n}} + {\text{l}}$ for each orbital,
$1s = 1$,$2s = 2$,$2p = 3$,$3s = 3$,$3p = 4$,$$,$$.
We can clearly see that the ${\text{n}} + {\text{l}}$ value is four for two of the orbitals, i.e. for 4s and 3p, which have one and six electrons respectively.
Thus, the total number of electrons for which ${\text{n}} + {\text{l}} = 4$ are 7.

Hence, the correct option is option C.

Note:
It is essential to note that chromium has an exceptional electronic configuration.
Considering the general trend, the electronic configuration of chromium atom is expected to be,
${\text{Cr}}:{\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^6}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{s}}^2}{\text{3}}{{\text{d}}^4}$
But, one electron from the $4s$ orbital shifts to the $3d$ orbital. This completes five electrons in the $3d$ orbital making it half filled. Half filled orbitals have greater stability due to symmetry and greater exchange energy. Thus the electronic configuration of chromium atom is
${\text{Cr}}:{\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^6}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{s}}^{\text{1}}}{\text{3}}{{\text{d}}^{\text{5}}}$.