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

Open in App

Solution

$[C r (N H_{3})_{6}]^{3 +}$ is paramagnetic due to presence of 3 unpaired electrons.

$C r (Z = 24)$ has electronic configuration $[A r] 3 d^{5} 4 s^{1}$ .

$C r^{3 +}$ has electronic configuration $[A r] 3 d^{3}$ .

It undergoes $d^{2} s p^{3}$ hybridisation to form six hybrid orbitals which are occupied by electron pairs donated by six ammonia ligands.

$[N i (C N)_{4}]^{2 -}$ is diamagnetic as all electrons are paired.

Ni (Z=28) has electronic configuration $[A r] 3 d^{8} 4 s^{2}$ .

$N i^{2 +}$ has electronic configuration $[A r] 3 d^{8}$ .

It undergoes $d s p^{2}$ hybridisation to form 4 hybrid orbitals which are occupied by electron pairs donated by four cyanide ligands.

Suggest Corrections

Similar questions

[C r (N H_{3})_{6}]^{3 +}

[N i (C N)_{4}]^{2 -}

^{'} [N i (C N)_{4}]^{2 -}

[N i C l_{4}]^{2 -}

C u^{+}

C u^{2 +}

(Z = 29)

N i (C O)_{4}, [N i (C N)_{4}]^{2 -}

N i C l_{4}^{2 -}

N i (C O)_{4}

[N i (C N)_{4}]^{2 -}

N i C l_{4}^{2 -}

Join BYJU'S Learning Program