Which ion has tetrahedral geometry:

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Question

(C) $1$. In $[Fe(CO)_5]$,the geometry is trigonal bipyramidal ($sp^3d$ hybridization).$2$. In $[Co(NH_3)_6]^{2+}$,the geometry is octahedral ($d^2sp^3

Vedclass · Accepted Answer

$[NiCl_4]^{2-}$

Vedclass · Answer

(C) $1$. In $[Fe(CO)_5]$,the geometry is trigonal bipyramidal ($sp^3d$ hybridization).$2$. In $[Co(NH_3)_6]^{2+}$,the geometry is octahedral ($d^2sp^3$ or $sp^3d^2$ hybridization).$3$. In $[NiCl_4]^{2-}$,$Ni^{2+}$ has a $d^8$ configuration. $Cl^-$ is a weak field ligand,so it does not cause pairing. The hybridization is $sp^3$,resulting in a tetrahedral geometry.$4$. In $[Ni(CN)_4]^{2-}$,$Ni^{2+}$ has a $d^8$ configuration. $CN^-$ is a strong field ligand,causing pairing. The hybridization is $dsp^2$,resulting in a square planar geometry.Therefore,the correct option is $C$.

Hybridisation	Geometry	Complex structure
$(A) \ sp^3$	$(i) \ \text{Square planar}$	$(p) \ [Fe(CN)_6]^{3-}$
$(B) \ d^2sp^3$	$(ii) \ \text{Tetrahedral}$	$(q) \ [ZnCl_4]^{2-}$
$(C) \ dsp^2$	$(iii) \ \text{Octahedral}$	$(r) \ [Ni(NH_3)_4]^{2+}$
-	$(iv) \ \text{Linear}$	$(s) \ [Ag(CN)_2]^-$

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