Explain why [Co(NH_{3})_{6}]^{3+} is an inner | vedclass

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

Question

(N/A) $[Co(NH_{3})_{6}]^{3+}$$[Ni(NH_{3})_{6}]^{2+}$Oxidation state of cobalt $= +3$Oxidation state of nickel $= +2$Electronic configuration of cobalt

Vedclass · Accepted Answer

Vedclass · Answer

(N/A) $[Co(NH_{3})_{6}]^{3+}$$[Ni(NH_{3})_{6}]^{2+}$Oxidation state of cobalt $= +3$Oxidation state of nickel $= +2$Electronic configuration of cobalt ion $(Co^{3+})$ $= 3d^{6}$Electronic configuration of nickel ion $(Ni^{2+})$ $= 3d^{8}$$NH_{3}$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals. This makes two $3d$ orbitals available for $d^{2}sp^{3}$ hybridization. Hence,it is an inner orbital complex.Even with $NH_{3}$ as a ligand,the $3d^{8}$ configuration leaves only one $3d$ orbital empty after pairing. Thus,it cannot undergo $d^{2}sp^{3}$ hybridization and instead undergoes $sp^{3}d^{2}$ hybridization using $4d$ orbitals. Hence,it is an outer orbital complex.

List-$I$ (Molecules/ion)	List-$II$ (Hybridisation of central atom)
$A. PF_5$	$I. dsp^2$
$B. SF_6$	$II. sp^3d$
$C. Ni(CO)_4$	$III. sp^3d^2$
$D. [PtCl_4]^{2-}$	$IV. sp^3$

Explain why $[Co(NH_{3})_{6}]^{3+}$ is an inner orbital complex whereas $[Ni(NH_{3})_{6}]^{2+}$ is an outer orbital complex.

Similar Questions

What will be the type of hybridization if the shapes are square planar,octahedral,and linear,respectively?

Match the List-$I$ with List-$II$
List-$I$ (Molecules/ion) List-$II$ (Hybridisation of central atom)
$A. PF_5$ $I. dsp^2$
$B. SF_6$ $II. sp^3d$
$C. Ni(CO)_4$ $III. sp^3d^2$
$D. [PtCl_4]^{2-}$ $IV. sp^3$
Choose the correct answer from the options given below $:$

Which of the following statements is not true?

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

The shape of the manganate ion $(MnO_4^{2-})$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$[Co(NH_{3})_{6}]^{3+}$	$[Ni(NH_{3})_{6}]^{2+}$
Oxidation state of cobalt $= +3$	Oxidation state of nickel $= +2$
Electronic configuration of cobalt ion $(Co^{3+})$ $= 3d^{6}$	Electronic configuration of nickel ion $(Ni^{2+})$ $= 3d^{8}$
$NH_{3}$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals. This makes two $3d$ orbitals available for $d^{2}sp^{3}$ hybridization. Hence,it is an inner orbital complex.	Even with $NH_{3}$ as a ligand,the $3d^{8}$ configuration leaves only one $3d$ orbital empty after pairing. Thus,it cannot undergo $d^{2}sp^{3}$ hybridization and instead undergoes $sp^{3}d^{2}$ hybridization using $4d$ orbitals. Hence,it is an outer orbital complex.