If Hund's rule is violated,then select the CO | vedclass

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

Question

(B) The atomic number of $Ni$ is $28$. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.If $Hund's$ rule is violated,the $8$ electrons in the

Vedclass · Accepted Answer

$d^2sp^3$,diamagnetic

Vedclass · Answer

(B) The atomic number of $Ni$ is $28$. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.If $Hund's$ rule is violated,the $8$ electrons in the $3d$ orbitals would pair up in the lower energy orbitals first.Specifically,the $3d$ orbitals would be filled as: $(d_{xy})^2 (d_{yz})^2 (d_{zx})^2 (d_{x^2-y^2})^2 (d_{z^2})^0$.This leaves two inner $d$-orbitals ($d_{x^2-y^2}$ and $d_{z^2}$) vacant.These two vacant $d$-orbitals,along with one $4s$ and three $4p$ orbitals,undergo $d^2sp^3$ hybridization.Since all electrons are paired in this configuration,the complex becomes diamagnetic.Therefore,the correct statement is $d^2sp^3$ and diamagnetic.

If $Hund's$ rule is violated,then select the $CORRECT$ statement regarding $[Ni(NH_3)_6]^{2+}$:

Similar Questions

Dimethyl glyoxime forms a square planar complex with $Ni^{2+}$. This complex should be

The hybridization observed in the $[PtCl_6]^{2-}$ ion is .......

The hybridisation of metals involved in the following complexes,$[Mn(CN)_6]^{3-}$ and $[CoF_6]^{3-}$,are respectively:

The $[Co(H_2O)_6]^{2+}$ ion has three unpaired electrons. The hybridization of $Co$ in $[Co(H_2O)_6]^{2+}$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module