Which of the following complexes formed by Ni | vedclass

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(C) According to the spectrochemical series,$CN^-$ and $CO$ are strong field ligands,while $Cl^-$ is a weak field ligand. $Ni^{2+}$ $(3d^8)$ with four

Vedclass · Accepted Answer

$[Ni(Cl)_4]^{2-}$

Vedclass · Answer

(C) According to the spectrochemical series,$CN^-$ and $CO$ are strong field ligands,while $Cl^-$ is a weak field ligand. $Ni^{2+}$ $(3d^8)$ with four strong field ligands forms a square planar,diamagnetic complex like $[Ni(CN)_4]^{2-}$. $Ni(0)$ $(3d^8 4s^2)$ with four strong field ligands forms a tetrahedral,diamagnetic complex like $[Ni(CO)_4]$. $Ni^{2+}$ $(3d^8)$ with four weak field ligands $(Cl^-)$ forms a tetrahedral,paramagnetic complex $[Ni(Cl)_4]^{2-}$. In $[Ni(Cl)_4]^{2-}$,the $Ni^{2+}$ ion has an electronic configuration of $[Ar] 3d^8$. Due to the weak field ligand,no pairing occurs,resulting in two unpaired electrons in the $d$-orbitals,making it paramagnetic.

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

Which of the following complexes formed by $Nickel$ is tetrahedral and paramagnetic?

Similar Questions

Statement $1$: $[Ti(H_2O)_6]^{4+}$ is coloured while $[Sc(H_2O)_6]^{3+}$ is colourless.
Statement $2$: $d-d$ transition is not possible in $[Sc(H_2O)_6]^{3+}$.

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 statements is not true for the reaction given below?
$[Cu(H_2O)_4]^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+} + 4H_2O$

What is true for $[Fe(CN)_6]^{3-}$ and $[FeF_6]^{3-}$?

The shape of $[Cu(NH_3)_4]^{2+}$ is square planar. The $Cu^{2+}$ ion in this complex is:

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