Which of the following complexes is tetrahedr | vedclass

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Question

(C) $1$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $d^8$. $CN^-$ is a strong field ligand,causing pairing. It forms $dsp^2$ hybridization,which is square planar

Vedclass · Accepted Answer

$[Ni(CO)_4]$

Vedclass · Answer

(C) $1$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $d^8$. $CN^-$ is a strong field ligand,causing pairing. It forms $dsp^2$ hybridization,which is square planar and diamagnetic.$2$. $[NiCl_4]^{2-}$: $Ni^{2+}$ is $d^8$. $Cl^-$ is a weak field ligand. It forms $sp^3$ hybridization,which is tetrahedral and paramagnetic.$3$. $[Ni(CO)_4]$: $Ni$ is $d^8s^2$. $CO$ is a strong field ligand,forcing electrons to pair. It forms $sp^3$ hybridization,which is tetrahedral and diamagnetic.$4$. $[Fe(CO)_5]$: $Fe$ is $d^6s^2$. It forms $dsp^3$ hybridization,which is trigonal bipyramidal and diamagnetic.Therefore,the complex that is both tetrahedral and diamagnetic is $[Ni(CO)_4]$.

List-$I$	List-$II$
$A. K_2[Ni(CN)_4]$	$I. sp^3$
$B. [Ni(CO)_4]$	$II. sp^3d^2$
$C. [Co(NH_3)_6]Cl_3$	$III. dsp^2$
$D. Na_3[CoF_6]$	$IV. d^2sp^3$

Which of the following complexes is tetrahedral as well as diamagnetic?

Similar Questions

$STATEMENT-1$: $[Fe(H_2O)_5NO]SO_4$ is paramagnetic.
$STATEMENT-2$: The $Fe$ in $[Fe(H_2O)_5NO]SO_4$ has three unpaired electrons.

Match List-$I$ with List-$II$:

List-$I$ List-$II$

$A. K_2[Ni(CN)_4]$ $I. sp^3$

$B. [Ni(CO)_4]$ $II. sp^3d^2$

$C. [Co(NH_3)_6]Cl_3$ $III. dsp^2$

$D. Na_3[CoF_6]$ $IV. d^2sp^3$

Choose the correct answer from the options given below:

The type of hybridization involved in the metal ion of the $[Ni(H_2O)_6]^{2+}$ complex is:

What is the shape of $Fe(CO)_5$?

The hybridization of $Fe$ in $K_3[Fe(CN)_6]$ is

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