$[C r (N H_{3})_{6}]^{3 +}$ is paramagnetic while $[N i (C N)_{4}]^{2 -}$ is diamagnetic. Explain, Why?

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Solution

Hexaammine chromium (III) ion, $[C r (N H_{3})_{6}]^{3 +}$ ion
Oxidation state of Cr = +3
$24^{C r = [A r] 3 d^{5} 4 s^{1}}$
$C r^{3} = [A r] 3 d^{3}$
$C r^{3 +}$ provides six empty orbitals to accommodate six electron pairs from six $N H_{3}$ molecules. It involves $d^{2} s p^{3}$ hybridization and hence, octahedral
It is paramagnetic due to the presence of three unpaired electrons.
.
Tetracyanonickelate (II) ion, $[N i (C N)_{4}]^{2 -}$
$N i = [A r] 3 d^{8} 4 s^{2}, N i^{2 +} = [A r] 3 d^{8}$

$C N^{-} b e i n g s t r o n g f i e l d l i g a n d c a u s e s p a i r i n g o f 3 d e l e c t r o n s t h u s, o n e o f t h e 3 d - o r b i t a l b e c o m e s e m p l t y f o r C N^{-} i o n .$

Unpaired electron is not present so it is diamagnetic.

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