Which of the following is an incorrect pair?

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(C) $1$. For $[Cu(NH_3)_4]^{2+}$,the geometry is square planar ($dsp^2$ hybridization).$2$. In $[Ni(CO)_4]$,$CO$ is a neutral ligand.$3$. For $[Fe(CN)

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$[Fe(CN)_6]^{3-}$: $sp^3d^2$ hybridization

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(C) $1$. For $[Cu(NH_3)_4]^{2+}$,the geometry is square planar ($dsp^2$ hybridization).$2$. In $[Ni(CO)_4]$,$CO$ is a neutral ligand.$3$. For $[Fe(CN)_6]^{3-}$,$Fe^{3+}$ is $3d^5$. Since $CN^-$ is a strong field ligand,it causes pairing,resulting in $d^2sp^3$ hybridization (inner orbital complex),not $sp^3d^2$.$4$. For $[Co(en)_3]^{3+}$,$Co^{3+}$ has $Z=27$,so $Co^{3+} = 24$ electrons. $en$ donates $2 	imes 3 = 6$ electrons. Total $EAN = 24 + 12 = 36$,which follows the $EAN$ rule (Krypton configuration).

List-$I$ (Hybridisation)	List-$II$ (Compound/ion)
$A. sp^3d$	$I. [PtCl_4]^{2-}$
$B. sp^3d^2$	$II. SF_6$
$C. dsp^2$	$III. BCl_3$
$D. dsp^3$	$IV. PCl_5$
	$V. ClF_3$

Similar Questions

Match the following:
List-$I$ (Hybridisation) List-$II$ (Compound/ion)
$A. sp^3d$ $I. [PtCl_4]^{2-}$
$B. sp^3d^2$ $II. SF_6$
$C. dsp^2$ $III. BCl_3$
$D. dsp^3$ $IV. PCl_5$
$V. ClF_3$

The correct match is:

The hybridization states of the central metal atom in the complexes $[Ni(CO)_4]$,$[Ni(CN)_4]^{2-}$,and $[NiCl_4]^{2-}$ are respectively:

State the limitations of valence bond theory.

Which of the following configurations will show an outer orbital octahedral complex?

According to Werner’s theory,the geometry of the complex is determined by:

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