The geometries of Ni(CO)_4 and Ni(PPh_3)_2Cl_ | vedclass

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Question

(D) In $Ni(CO)_4$,the oxidation state of $Ni$ is $0$. The configuration is $[Ar] 3d^8 4s^2$. Due to the strong field ligand $CO$,the electrons pair up

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Both tetrahedral

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(D) In $Ni(CO)_4$,the oxidation state of $Ni$ is $0$. The configuration is $[Ar] 3d^8 4s^2$. Due to the strong field ligand $CO$,the electrons pair up,resulting in $sp^3$ hybridization,which gives a tetrahedral geometry.In $Ni(PPh_3)_2Cl_2$,the oxidation state of $Ni$ is $+2$. The configuration is $[Ar] 3d^8$. $PPh_3$ is a bulky ligand,and $Cl^-$ is a weak field ligand. The steric hindrance caused by $PPh_3$ groups forces the complex to adopt a tetrahedral geometry ($sp^3$ hybridization) rather than square planar.

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

The geometries of $Ni(CO)_4$ and $Ni(PPh_3)_2Cl_2$ are:

Similar Questions

Which one is an example of an octahedral complex?

The coordination complex $[Co(OH_2)_6]^{2+}$ has one unpaired electron. Which of the following statements are true?
$(i)$ The complex is octahedral.
$(ii)$ The complex is an outer orbital complex.
$(iii)$ The complex is diamagnetic.

The geometry of the cuprammonium ion $[Cu(NH_3)_4]^{2+}$ is:

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

$LIST-I$ $LIST-II$

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

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

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

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

$5. Ni(CO)_4$

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

Determine the hybridisation,magnetic nature,and spin-only magnetic moment for $[Co(C_2O_4)_3]^{3-}$

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