Question

The magnetic moment of $\left[Cu\right(N{H}_3{)}_4{]}^{2+}$ was found to be 1.73 B.M. The number of unpaired electrons in the complex is:

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B)

1

Magnetic moment of a complex having n-unpaired electrons is given by: $\sqrt{n(n+2)}$, where $n$ is the number of unpaired electrons.

$\sqrt{n(n+2)}=1.73$

$\therefore$ $n=1$.