Which of the following complexes has a square | vedclass

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(D) $1$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons. It undergoes $dsp^2$ hybridization,resu

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(D) $1$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons. It undergoes $dsp^2$ hybridization,resulting in a square planar geometry.$2$. $[Cu(NH_3)_4]^{2+}$: $Cu^{2+}$ is $3d^9$. The unpaired electron is in the $4p_z$ orbital,and the four ligands form bonds using $dsp^2$ hybrid orbitals,resulting in a square planar geometry.$3$. $[PtCl_4]^{2-}$: $Pt^{2+}$ is a $5d^8$ ion. $Pt$ is a $5d$ series element,so it always forms square planar complexes with $dsp^2$ hybridization regardless of the ligand strength.Since all three complexes exhibit $dsp^2$ hybridization and square planar geometry,the correct answer is $D$.

$LIST-I$	$LIST-II$
$P. dsp^2$	$1. [FeF_6]^{4-}$
$Q. sp^3$	$2. [Ti(H_2O)_3Cl_3]$
$R. sp^3d^2$	$3. [Cr(NH_3)_6]^{3+}$
$S. d^2sp^3$	$4. [FeCl_4]^{2-}$
	$5. Ni(CO)_4$
	$6. [Ni(CN)_4]^{2-}$

Which of the following complexes has a square planar geometry and $dsp^2$ hybridization?

Similar Questions

The complex ion $[Co(C_2O_4)_3]^{3-}$ has which of the following hybridizations?

Chromium hexacarbonyl,$Cr(CO)_6$,is an octahedral compound involving which type of hybridization?

Match each set of hybrid orbitals from $LIST-I$ with complex(es) given in $LIST-II$. The correct option is

$LIST-I$ $LIST-II$

$P. dsp^2$ $1. [FeF_6]^{4-}$

$Q. sp^3$ $2. [Ti(H_2O)_3Cl_3]$

$R. sp^3d^2$ $3. [Cr(NH_3)_6]^{3+}$

$S. d^2sp^3$ $4. [FeCl_4]^{2-}$

$5. Ni(CO)_4$

$6. [Ni(CN)_4]^{2-}$

Which of the following pairs differ in geometry but resemble in magnetic behavior?

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

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