Question

Geometry, hybridisation and magnetic moment of the ions ${\left[Ni{\left(CN\right)}_4\right]}^{2-},{\left[MnB{r}_4\right]}^{2-}$ and ${\left[Fe{F}_6\right]}^{4-}$ respectively are:

• A. tetrahedral, square planar, octahedral $:s{p}^3,ds{p}^2,s{p}^3{d}^2:5.9,0,4.9$
• B. tetrahedral, square planar, octahedral $:ds{p}^2,s{p}^3,s{p}^3{d}^2:0,5.9,4.9$
• C. square planar, tetrahedral, octahedral $:ds{p}^2,s{p}^3,d^{2}s{p}^3:5.9,4.9,0$
• D. square planar, tetrahedral, octahedral $:ds{p}^2,s{p}^3,s{p}^3{d}^2:0,5.9,4.9$

Answer 1

The correct option is D square planar, tetrahedral, octahedral $:ds{p}^2,s{p}^3,s{p}^3{d}^2:0,5.9,4.9$

In ${\left[Ni{\left(CN\right)}_4\right]}^{2-},$
$N{i}^{+2},d^{10},ds{p}^2$ hybridization with square planar structure & zero unpaired electron so zero magnetic moment.
In ${\left[MnB{r}_4\right]}^{2-}$,
$M{n}^{+2},d^5,s{p}^3$ hybridization with tetrahedral structure & 5 unpaired electron so 5.89 magnetic moment.
In ${\left[Fe{F}_6\right]}^{4-}$,
$F{e}^{+2},d^6,s{p}^3{d}^2$ hybridization, so octahedral geometry & 4 unpaired electron so 4.89 magnetic moment.
(As magnetic moment $\mu =\sqrt{n(n+2)}$ , where n is no. of unpaired electron).