Question

Magnetic moment of $\left[Cu\right(N{H}_3{)}_4{]}^{2+}$ ion is?

• A. 1.414
• B. 1.73
• C. 2.23
• D. 2.38

Answer 1

The correct option is B 1.73

The given complex is $\left[Cu\right(N{H}_3{)}_4{]}^{2+}$

The electronic configuration of Cu is [Ar] $3d^{10}4s^1$

for Cu${}^{2+}$ $\to$ [Ar]$3d^{9}4s^0$

[Ar] $\upharpoonleft \downharpoonright$ $\upharpoonleft \downharpoonright$ $\upharpoonleft \downharpoonright$ $\upharpoonleft \downharpoonright$ $\upharpoonleft$

3d 4s

3d 4s 4p

So, [Cu(NH${}_3{)}_4{]}^{2+}$ $\to$ [Ar] $\upharpoonleft \downharpoonright$ $\upharpoonleft \downharpoonright$ $\upharpoonleft \downharpoonright$ $\upharpoonleft \downharpoonright$ xx xx xx xx $\upharpoonleft$

$N{H}_3$ $N{H}_2$ $N{H}_3$ $N{H}_3$

dsp${}^2$ hybridisation

No. of unpaired $e^{-}$ s =1

Magnetic moment= $\sqrt{n(n+2)}$ BM

Where, n= no.of unpaired $e^{-}$ s

So, magnetic moment = $\sqrt{1(1+2)}=\sqrt{3}$ = 1.73 BM