How many hybrid orbitals of cobalt ion are in | vedclass

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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)	List-$II$ (Hybridisation of central metal ion)
$A$. $[CoF_6]^{3-}$	$I$. $d^2sp^3$
$B$. $[NiCl_4]^{2-}$	$II$. $sp^3$
$C$. $[Co(NH_3)_6]^{3+}$	$III$. $sp^3d^2$
$D$. $[Ni(CN)_4]^{2-}$	$IV$. $dsp^2$

List-$I$	List-$II$
$A. K_2[Ni(CN)_4]$	$I. sp^3$
$B. [Ni(CO)_4]$	$II. sp^3d^2$
$C. [Co(NH_3)_6]Cl_3$	$III. dsp^2$
$D. Na_3[CoF_6]$	$IV. d^2sp^3$

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

Similar Questions

Match List-$I$ with List-$II$.
List-$I$ (Complex) List-$II$ (Hybridisation of central metal ion)
$A$. $[CoF_6]^{3-}$ $I$. $d^2sp^3$
$B$. $[NiCl_4]^{2-}$ $II$. $sp^3$
$C$. $[Co(NH_3)_6]^{3+}$ $III$. $sp^3d^2$
$D$. $[Ni(CN)_4]^{2-}$ $IV$. $dsp^2$
Choose the correct answer from the options given below:

The shape of the $[Fe(CN)_6]^{4-}$ ion is:

What type of hybridisation is present in square planar geometry of complex $[Ni(CN)_4]^{2-}$?

Match List-$I$ with List-$II$:

List-$I$ List-$II$

$A. K_2[Ni(CN)_4]$ $I. sp^3$

$B. [Ni(CO)_4]$ $II. sp^3d^2$

$C. [Co(NH_3)_6]Cl_3$ $III. dsp^2$

$D. Na_3[CoF_6]$ $IV. d^2sp^3$

Choose the correct answer from the options given below:

Which of the following cobalt-$(III)$ complex is paramagnetic and high spin complex?

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