Question

A magnetic moment of 1.73 BM will be shown by one among the following :-

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

Answer 1

Answer: (B)

Magnetic moment 1.73 BM
$\mu =\sqrt{n(n+2)}B.M$
n = no. of unpaired e-
$\mu =1.73$
$1.73=\sqrt{n(n+2)}B.M$
$n=1$
$\ast [CoC{l}_6{]}^{4-}\to C{o}^{+2};d^7$
Cl- (weak field ligand) $t_2{g}^{5}e{g}^2$ unpaired $e^{-}=3$
$\ast [Cu(N{H}_3{)}_4{]}^{2+}C{u}^{+2}-d^9$
$N{H}_3$ Strong field ligand, hybridisation $ds{p}^2$
$\ast$ one $e^{-}$ of $3d$
$\ast [Ni(CN{)}_4{]}^{2-}\to N{i}^{+2}-d^8$ unpaired $e^{-}=0$
$C{N}^{-}$ - Strong field ligand $ds{p}^2$
$\ast TiC{l}_4\to T{i}^{+4}{d}^{\circ }$ unpaired e- = zero.