(D) In $[Co(NH_3)_6]^{3+}$,$Co$ is in $+3$ oxidation state ($3d^6$ configuration). $NH_3$ is a strong field ligand,causing pairing of electrons,result

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(D) In $[Co(NH_3)_6]^{3+}$,$Co$ is in $+3$ oxidation state ($3d^6$ configuration). $NH_3$ is a strong field ligand,causing pairing of electrons,resulting in $0$ unpaired electrons.In $[CoF_6]^{3-}$,$Co$ is in $+3$ oxidation state ($3d^6$ configuration). $F^-$ is a weak field ligand,not causing pairing,resulting in $4$ unpaired electrons.

Class 12 Chemistry Coordination Compounds Hybridisation and Geometry

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The number of unpaired electrons in $[Co(NH_3)_6]^{3+}$ and $[CoF_6]^{3-}$ are respectively.....

A
$4$ and $4$
B
$0$ and $2$
C
$2$ and $4$
D
$0$ and $4$

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Class 12 Questions

Class 12

Chemistry Questions

Chapter

Coordination Compounds

Subtopic

Hybridisation and Geometry

Match List-$I$ with List-$II$.

List-$I$ List-$II$

$A$. $[MnBr_4]^{2-}$ $I$. $d^2sp^3$ and diamagnetic

$B$. $[FeF_6]^{3-}$ $II$. $sp^3d^2$ and paramagnetic

$C$. $[Co(C_2O_4)_3]^{3-}$ $III$. $sp^3$ and diamagnetic

$D$. $[Ni(CO)_4]$ $IV$. $sp^3$ and paramagnetic

Choose the correct answer from the options given below:

List-$I$	List-$II$
$A$. $[MnBr_4]^{2-}$	$I$. $d^2sp^3$ and diamagnetic
$B$. $[FeF_6]^{3-}$	$II$. $sp^3d^2$ and paramagnetic
$C$. $[Co(C_2O_4)_3]^{3-}$	$III$. $sp^3$ and diamagnetic
$D$. $[Ni(CO)_4]$	$IV$. $sp^3$ and paramagnetic