Triple degenerate d- orbitals are exactly hal | vedclass

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

Question

(B) The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$. For $Co^{2+}$,the configuration is $3d^7$.In a tetrahedral field,the $d-$ orbitals split

Vedclass · Accepted Answer

$Co^{2+}$ (tetrahedral field)

Vedclass · Answer

(B) The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$. For $Co^{2+}$,the configuration is $3d^7$.In a tetrahedral field,the $d-$ orbitals split into $e_g$ (lower energy,doubly degenerate) and $t_{2g}$ (higher energy,triply degenerate) sets.The $7$ electrons fill as follows: $e_g^4 t_{2g}^3$.The $t_{2g}$ orbitals are triply degenerate and contain $3$ electrons,meaning they are exactly half-filled ($3/6$ capacity).Thus,$Co^{2+}$ in a tetrahedral field satisfies the condition.

Triple degenerate $d-$ orbitals are exactly half-filled in

Similar Questions

Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:
$(i)$ $[Fe(CN)_6]^{4-}$
$(ii)$ $[FeF_6]^{3-}$
$(iii)$ $[Co(C_2O_4)_3]^{3-}$
$(iv)$ $[CoF_6]^{3-}$

Which of the following pairs of compounds is planar?

The hybridization of the central atom in $NH_3, [PtCl_4]^{2-}, PCl_5$,and $BCl_3$ respectively is:

Nickel combines with a uninegative monodentate ligand $(X^{-})$ to form a paramagnetic complex $[NiX_4]^{2-}$. The hybridisation involved and the number of unpaired electrons present in the complex are respectively:

What is the number of $d$-electrons in the metal ion of the tetrahedral complex $K_2[NiCl_4]$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module