Question

What is the electronic configuration of chromium in ${[Cr{(N{H_3})_6}]^{3 + }}$?

Answer 1

Hint: Chromium is a d block element. It has a total of twenty-four electrons in its outermost shell. Chromium has the ability to possess three oxidation states being $ + 2, + 3\,and\, + 6$.

Complete answer:
Chromium is a block element. It has twenty-four electrons in its outermost shells
The configuration of chromium generally is $[Ar]3{d^5}4{s^1}$
The electrons in the outermost electrons are all unpaired. The allotment of electrons when to be done, there are two rules that need to be followed. The allotment of electrons is to be such that it is the most stable. Now the configuration of electrons having all paired or all unpaired, is the most stable state than any other state for a tom.
Therefore, here chromium has the ability to lose electrons easily. It can consider the carbon of d block elements.
Now in ${[Cr{(N{H_3})_6}]^{3 + }}$
$N{H_3}$ is a weak ligand that is not able cause the pairing of the electrons in the d orbital? The oxidation state of $N{H_3}$ is zero, therefore calculating the oxidation state of chromium.
Let the oxidation state of chromium be x
$
  x - (0) \times 6 = + 3 \\
  x = + 3 \\
 $
The presence of three unpaired electrons makes the complex paramagnetic
Now the electronic configuration of chromium in $ + 3$ oxidation state is: -
$Cr = [Ar]3{d^3}4{s^0}$
There are three unpaired electrons. the six ligands of $N{H_3}$ occupy the empty spaces in the third and fourth orbital.
$Cr = [Ar]3{d^3}4{s^0}$ is the electronic configuration of chromium in ${[Cr{(N{H_3})_6}]^{3 + }}$.

Note:
Chromium is used to harden steel in the industries. It is also used to manufacture stainless steel, which is steel which doesn't rust. It is also used in producing several alloys.