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. Hybridisati

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

Out of the following two coordination entities,which is chiral (optically active)?
$(a)$ $cis-[CrCl_{2}(ox)_{2}]^{3-}$
$(b)$ $trans-[CrCl_{2}(ox)_{2}]^{3-}$

Which of the following would exhibit ionisation isomerism?

Which of the following does not exhibit geometrical isomerism?

The complex that shows Facial - Meridional isomerism is

The magnitude of magnetic moment (spin only) of $[NiCl_4]^{2-}$ will be ........ $B.M.$

Vedclass Test Series

Exam Paper Generator

Online Exam Module