Total number of unpaired electrons present in | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) $1$. In $[Ni(CO)_{4}]$,$Ni$ is in $0$ oxidation state $(3d^{8} 4s^{2})$. $CO$ is a strong field ligand,causing pairing of electrons. Hybridisation

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) $1$. In $[Ni(CO)_{4}]$,$Ni$ is in $0$ oxidation state $(3d^{8} 4s^{2})$. $CO$ is a strong field ligand,causing pairing of electrons. Hybridisation is $sp^{3}$,unpaired electrons $= 0$.$2$. In $[NiCl_{4}]^{2-}$,$Ni^{2+}$ is $3d^{8}$. $Cl^{-}$ is a weak field ligand,no pairing occurs. Hybridisation is $sp^{3}$,unpaired electrons $= 2$.$3$. In $[PtCl_{2}(NH_{3})_{2}]$,$Pt^{2+}$ is $5d^{8}$. $Pt$ is a $5d$ series metal,so pairing occurs even with weak ligands. Hybridisation is $dsp^{2}$,unpaired electrons $= 0$.$4$. In $[Ni(CN)_{4}]^{2-}$,$Ni^{2+}$ is $3d^{8}$. $CN^{-}$ is a strong field ligand,causing pairing. Hybridisation is $dsp^{2}$,unpaired electrons $= 0$.$5$. In $[Pt(CN)_{4}]^{2-}$,$Pt^{2+}$ is $5d^{8}$. $CN^{-}$ is a strong field ligand. Hybridisation is $dsp^{2}$,unpaired electrons $= 0$.Total unpaired electrons $= 0 + 2 + 0 + 0 + 0 = 2$.

Total number of unpaired electrons present in the central metal atoms/ions of $[Ni(CO)_{4}]$,$[NiCl_{4}]^{2-}$,$[PtCl_{2}(NH_{3})_{2}]$,$[Ni(CN)_{4}]^{2-}$ and $[Pt(CN)_{4}]^{2-}$ is . . . . . . .

Similar Questions

Which of the following complexes is capable of showing optical isomerism?

Which of the following are paramagnetic?
$A$. $[NiCl_4]^{2-}$
$B$. $Ni(CO)_4$
$C$. $[Ni(CN)_4]^{2-}$
$D$. $[Ni(H_2O)_6]^{2+}$
$E$. $Ni(PPh_3)_4$
Choose the correct answer from the options given below:

$[Co(NH_3)_4Cl_2]^+$ exhibits

The magnetic moment of $Fe^{2+}$ is $B.M$.

Which of the following is true about the complex $[PtCl_2(H_2O)(NH_3)]$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module