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

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(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

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(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.

$A$. $[CoF_6]^{3-}$	$M$. $sp^3d^2$,paramagnetic
$B$. $[Mn(CN)_6]^{3-}$	$N$. $sp^3d^2$,diamagnetic
$C$. $[Co(C_2O_4)_3]^{3-}$	$O$. $d^2sp^3$,paramagnetic
$D$. $[Zn(H_2O)_6]^{2+}$	$P$. $d^2sp^3$,diamagnetic

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

Match the correct option.
$A$. $[CoF_6]^{3-}$ $M$. $sp^3d^2$,paramagnetic
$B$. $[Mn(CN)_6]^{3-}$ $N$. $sp^3d^2$,diamagnetic
$C$. $[Co(C_2O_4)_3]^{3-}$ $O$. $d^2sp^3$,paramagnetic
$D$. $[Zn(H_2O)_6]^{2+}$ $P$. $d^2sp^3$,diamagnetic

Among $[Co(CN)_4]^{4-}, [Co(CO)_3(NO)], XeF_4, [PCl_4]^{+}, [PdCl_4]^{2-}, [ICl_4]^{-}, [Cu(CN)_4]^{3-}$ and $P_4$,the total number of species with tetrahedral geometry is:

$d^{2}sp^{3}$ hybridisation of the atomic orbitals gives

The type of hybridisation and magnetic property of the complex $[MnCl_6]^{3-}$,respectively,are :

In which of the following configurations will there be the possibility of both paramagnetism and diamagnetism,depending on the nature of the ligands?

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$[Co(NH_{3})_{6}]^{3+}$	$[Ni(NH_{3})_{6}]^{2+}$
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.