Which of the following has dsp^2 hybridizatio | vedclass

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(D) To determine the hybridization,we look at the coordination number and the nature of the central metal ion. $NiCl_4^{2-}$ involves $Ni^{2+}$ with a

Vedclass · Accepted Answer

$PtCl_4^{2-}$

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(D) To determine the hybridization,we look at the coordination number and the nature of the central metal ion. $NiCl_4^{2-}$ involves $Ni^{2+}$ with a weak field ligand $Cl^-$,resulting in $sp^3$ hybridization (tetrahedral). $SCl_4$ has $sp^3d$ hybridization due to the presence of a lone pair on sulfur. $NH_4^+$ has $sp^3$ hybridization. $PtCl_4^{2-}$ involves $Pt^{2+}$ (a $5d$ series metal),which always forms square planar complexes with strong or weak field ligands due to high crystal field splitting energy,resulting in $dsp^2$ hybridization.

List-$I$	List-$II$
$A. Ni(CO)_4$	$I. sp^3$
$B. [Ni(CN)_4]^{2-}$	$II. sp^3d^2$
$C. [Co(CN)_6]^{3-}$	$III. d^2sp^3$
$D. [CoF_6]^{3-}$	$IV. dsp^2$

Which of the following has $dsp^2$ hybridization?

Similar Questions

Match List-$I$ with List-$II$.

List-$I$ List-$II$

$A. Ni(CO)_4$ $I. sp^3$

$B. [Ni(CN)_4]^{2-}$ $II. sp^3d^2$

$C. [Co(CN)_6]^{3-}$ $III. d^2sp^3$

$D. [CoF_6]^{3-}$ $IV. dsp^2$

Choose the correct answer from the options given below.

For $K_3[Co(CN)_6]$,which of the following statements is not correct?

Discuss the $[CoF_6]^{3-}$ complex species based on Valence Bond Theory $(VBT)$.

The hybridisation and magnetic behavior of the complex $[Ni(CO)_4]$ are

The number of unpaired electrons responsible for the paramagnetic nature of the following complex species are respectively $:$
$[Fe(CN)_6]^{3-}, [FeF_6]^{3-}, [CoF_6]^{3-}, [Mn(CN)_6]^{3-}$

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