Question

The magnetic moment of a complexion is $2.83BM$. The complex ion is

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

Answer 1

Answer: (A)

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

The spin only magnetic moment is given by the formula ${\mu }_{eff}=\sqrt{n(n+2)}B.M.$

The magnetic moment of 2.83 BM. corresponds to two unpaired electrons.

$V^{+3}$ ion in $\left[V\right(H_{2}O{)}_6{]}^{3+}$ has outer electronic configuration of $3d^2$ and contains two unpaired electrons.

The central metal ion in $\left[Cr\right(H_{2}O{)}_6{]}^{3+},\left[Cu\right(CN{)}_4{]}^{2-}and[MnC{l}_4{]}^{2-}$ contains 3,1 and 5 unpaired electrons respectively.

Hence, the correct option is A.