Use spin-only magnetic moment formula

The spin-only magnetic moment is $\mu = \sqrt{n(n+2)}\, \text{BM}$ where $n$ is the number of unpaired electrons.

Find electronic configurations of the ions

(i) $Cu^+$

Neutral $Cu$: $[Ar]3d^{10}4s^1$ For $Cu^+$, one electron is removed: $Cu^+ : [Ar]3d^{10}$ Number of unpaired electrons: $n=0$ So, $\mu = \sqrt{0(0+2)} = 0$

(ii) $Cu^{2+}$

From $Cu$, remove first $4s$ electron and then one $3d$ electron: $Cu^{2+} : [Ar]3d^9$ Number of unpaired electrons: $n=1$ So, $\mu = \sqrt{1(1+2)} = \sqrt{3}$

(iii) $Cr^{2+}$

Neutral $Cr$: $[Ar]3d^5 4s^1$ Remove two electrons: $Cr^{2+} : [Ar]3d^4$ Number of unpaired electrons in $3d^4$: $n=4$ So, $\mu = \sqrt{4(4+2)} = \sqrt{24}$

(iv) $Cr^{3+}$

Remove three electrons from $Cr$: $Cr^{3+} : [Ar]3d^3$ Number of unpaired electrons: $n=3$ So, $\mu = \sqrt{3(3+2)} = \sqrt{15}$

Compare the magnetic moments

Their values are: $Cr^{2+} : \sqrt{24}$ $Cr^{3+} : \sqrt{15}$ $Cu^{2+} : \sqrt{3}$ $Cu^{+} : 0$

Therefore, decreasing order is: $Cr^{2+} > Cr^{3+} > Cu^{2+} > Cu^+$

Match with the options

This corresponds to Option C.

Comparison with stored correct answer

Stored correct answer = C.

My derived answer also = C. Hence, they agree.