How many hybrid orbitals of cobalt ion are in | vedclass

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

Question

(A) The complex is $[Co(NH_3)_6]^{3+}$.In this complex,the central metal ion is $Co^{3+}$.The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.Sin

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) The complex is $[Co(NH_3)_6]^{3+}$.In this complex,the central metal ion is $Co^{3+}$.The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.Since $NH_3$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbitals.This results in two empty $3d$ orbitals,one $4s$ orbital,and three $4p$ orbitals being available for hybridization.Thus,the hybridization is $d^2sp^3$.This involves a total of $6$ hybrid orbitals to accommodate the $6$ lone pairs donated by the $6$ $NH_3$ ligands.

List-$I$ (Complex/ion)	List-$II$ (Shape/geometry)
$A$. $[Pt(Cl)_2(NH_3)_2]$	$I$. Octahedral
$B$. $[Co(NH_3)_6]Cl_3$	$II$. Trigonal bipyramidal
$C$. $[NiCl_4]^{2-}$	$III$. Square planar
$D$. $[Fe(CO)_5]$	$IV$. Tetrahedral

How many hybrid orbitals of cobalt ion are involved in the formation of the hexaamminecobalt$(III)$ chloride complex?

Similar Questions

The geometry possessed by $[Ni(CO)_4]$ is

In Wilkinson's catalyst,the hybridization of the central metal ion and its shape are respectively:

In the complex $[SbF_5]^{2-}$,$sp^3d$ hybridization is present. The geometry of the complex is

Match List-$I$ with List-$II$:
List-$I$ (Complex/ion) List-$II$ (Shape/geometry)
$A$. $[Pt(Cl)_2(NH_3)_2]$ $I$. Octahedral
$B$. $[Co(NH_3)_6]Cl_3$ $II$. Trigonal bipyramidal
$C$. $[NiCl_4]^{2-}$ $III$. Square planar
$D$. $[Fe(CO)_5]$ $IV$. Tetrahedral

The hexachloridocobaltate $(III)$ ion is a high-spin complex. What is the possible hybridisation state of the cobalt ion in this complex?

Vedclass Test Series

Exam Paper Generator

Online Exam Module