Identify the correct set for the [Co(NH_3)_6] | vedclass

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Question

(A) The oxidation state of $Co$ in $[Co(NH_3)_6]^{3+}$ is $+3$ because $NH_3$ is a neutral ligand.The electronic configuration of $Co^{3+}$ is $[Ar] 3

Vedclass · Accepted Answer

$d^2sp^3$,inner orbital complex,zero

Vedclass · Answer

(A) The oxidation state of $Co$ in $[Co(NH_3)_6]^{3+}$ is $+3$ because $NH_3$ is a neutral ligand.The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.Since $NH_3$ is a strong-field ligand,it causes pairing of the $3d$ electrons.This results in two vacant $3d$ orbitals,which participate in $d^2sp^3$ hybridization.Because the inner $3d$ orbitals are used,it is an inner-orbital complex.All $3d$ electrons are paired,so the number of unpaired electrons is $0$.

Column $I$	Column $II$
$(A)$ $sp^3$	$(i)$ $[Co(NH_3)_6]^{3+}$
$(B)$ $dsp^2$	$(ii)$ $[Ni(CO)_4]$
$(C)$ $sp^3d^2$	$(iii)$ $[Pt(NH_3)_2Cl_2]$
$(D)$ $d^2sp^3$	$(iv)$ $[CoF_6]^{3-}$
	$(v)$ $[Fe(CO)_5]$

Identify the correct set for the $[Co(NH_3)_6]^{3+}$ ion regarding the hybridization of $Co^{3+}$,type of complex,and the number of unpaired electrons in the complex ion,respectively.

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