Question

The magnetic moment of a complex iron is $2.83BM$. The complex ion of the following is?

• A. ${\left[V{\left(H_{2}O\right)}_6\right]}^{3+}$
• B. ${\left[Cr{\left(H_{2}O\right)}_6\right]}^{3+}$
• C. ${\left[Cu{\left(CN\right)}_4\right]}^{2-}$
• D. ${\left[Cu{CI}_4^2\right]}^{2-}$

Answer 1

The correct option is A ${\left[V{\left(H_{2}O\right)}_6\right]}^{3+}$

Magnetic moment is given by

$\mu =\sqrt{n(n+2)}2.83=\sqrt{n(n+2)}(2.83{)}^2=n(n+2)8.00=n^2+2n{n}^2+2n-8=0(n+4\left)\right(n-2)=0n=2n=-4\leftarrow$ not be consider

Now we find out from the option which complex ion has $2$ unpaired electrons.

Option $A\left[V\right(H_{2}O{)}_6{]}^{3+}\Rightarrow V^{3+}$

The electronic configuration of $V^{3+}$ is

$V^{3+}\to \left[Ar\right]4s^{0}3d^2$

So $V^{3+}$ has $2$ unpaired $3d$ electrons

Answer $\left[V\right(H_{2}O{)}_6{]}^{3+}$