Question

An aqueous solution of metal ion $M_1$ reacts separately with reagents $Q$ and $R$ in excess to give tetrahedral and square planar complexes, respectively. An aqueous solution of another metal ion $M_2$ always forms tetrahedral complexes with these reagents. The aqueous solution of $M_2$ on reaction with reagent $S$ gives the white precipitate which dissolves in excess of $S$. The reactions are summarized in the given below. Tetrahedral $\underset{excess}{\overset{Q}{\leftarrow }}{M}_1\underset{excess}{\overset{R}{\to }}$ Square planar

Tetrahedral $\underset{excess}{\overset{Q}{\leftarrow }}M2\underset{excess}{\overset{R}{\to }}$ Tetrahedral

$\downarrow S$, Stoichiometric amount

White precipitate $\underset{excess}{\overset{S}{\to }}$ precipitate dissolves

• A. $1.73BM$
• B. $4.90BM$
• C. $3.87BM$
• D. $2.83BM$

Answer 1

The correct option is D $2.83BM$

Spin magnetic moment of $N{i}^{2+}$
Electronic configuration: $3d^8$
$\therefore n=2$
Spin magnetic moment : $\sqrt{n(n+2)}=2.82BM$