Question

A magnetic moment of $1.73$ BM will be shown by which one among the following compounds?

• A. $\left[Cu\right(N{H}_3{)}_4{]}^{2+}$
• B. $\left[Ni\right(CN{)}_4{]}^{2-}$
• C. $TiC{l}_4$
• D. $[CoC{l}_6{]}^{4-}$

Answer 1

Answer: (A)

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

Electronic configuration of $C{u}^{2+}$ ion in $\left[Cu\right(N{H}_3{)}_4{]}^{2+}$.
$C{u}^{2+}$ ion $=\left[Ar\right]3d^{9}4s^0$.
$\therefore C{u}^{2+}$ ion has one unpaired electron.
Magnetic moment of $\left[Cu\right(N{H}_3{)}_4{]}^{2+}\left(\mu \right)=\sqrt{n(n+2)}BM$
where, $n=$no. of unpaired electrons
$\mu =\sqrt{1(1+2)}=\sqrt{3}=1.73$BM
Whereas $N{i}^{2+}$ in $\left[Ni\right(CN{)}_4{]}^{2-},T{i}^{4+}$ in $TiC{l}_4$ and $C{o}^{2+}$ ion $[COC{l}_6{]}^{4-}$ has $2,0$ and $3$ unpaired electrons respectively.