The hybridization of [Ni(CO)_4] is: | vedclass

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Question

(A) In the complex $[Ni(CO)_4]$,the central metal atom is Nickel $(Ni)$.The atomic number of $Ni$ is $28$. Its ground state electronic configuration i

Vedclass · Accepted Answer

$sp^3$

Vedclass · Answer

(A) In the complex $[Ni(CO)_4]$,the central metal atom is Nickel $(Ni)$.The atomic number of $Ni$ is $28$. Its ground state electronic configuration is $[Ar] 3d^8 4s^2$.In $[Ni(CO)_4]$,the oxidation state of $Ni$ is $0$ because $CO$ is a neutral ligand.$CO$ is a strong field ligand,which causes the pairing of electrons in the $3d$ orbital. The two electrons from the $4s$ orbital shift into the $3d$ orbital,making it fully filled $(3d^{10})$.This leaves one $4s$ orbital and three $4p$ orbitals vacant,which undergo $sp^3$ hybridization to accommodate the four electron pairs donated by four $CO$ ligands.Therefore,the hybridization of $[Ni(CO)_4]$ is $sp^3$.

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

The hybridization of $[Ni(CO)_4]$ is:

Similar Questions

$A$ complex involving $dsp^2$ hybridization has:

Match the following hybridization in Column $I$ with the corresponding coordination complexes in Column $II$.
$A. sp^3$ $(i). [Co(NH_3)_6]^{3+}$
$B. dsp^2$ $(ii). [Ni(CO)_4]$
$C. sp^3d^2$ $(iii). [Pt(NH_3)_2Cl_2]$
$D. d^2sp^3$ $(iv). [CoF_6]^{3-}$
$(v). [Fe(CO)_5]$

The complex ion $[Fe(CN)_6]^{4-}$ contains

$[Cu(NH_3)_4]^{2+}$ shows the following hybridization:

Predict the number of unpaired electrons in the square planar $[Pt(CN)_{4}]^{2-}$ ion.

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