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(C) $1$. For $[Co(CO)_4]^-$,the oxidation state of $Co$ is $-1$. The electronic configuration is $3d^8 4s^2$. Since $CO$ is a strong field ligand,it c

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$[Co(CO)_4]^-$ and $[Ni(CO)_4]$

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(C) $1$. For $[Co(CO)_4]^-$,the oxidation state of $Co$ is $-1$. The electronic configuration is $3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes pairing of electrons. The geometry is $sp^3$ (tetrahedral) and it is diamagnetic due to the pairing of all electrons.$2$. For $[Ni(CO)_4]$,the oxidation state of $Ni$ is $0$. The electronic configuration is $3d^8 4s^2$. $CO$ causes pairing of electrons,resulting in $sp^3$ hybridization (tetrahedral) and it is diamagnetic.$3$. Other options like $[NiCl_4]^{2-}$ are tetrahedral but paramagnetic,and $[Ni(CN)_4]^{2-}$ is square planar and diamagnetic.

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

Which of the following pairs of complex compounds are both tetrahedral and diamagnetic?

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