Match List-I with List-II :- List-I (Coo | vedclass

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(A) For option $(A)$: $[Cr(CN)_6]^{3-}$,$Cr^{3+}$ is $3d^3$. Since $CN^-$ is a strong field ligand,the electrons remain in $t_{2g}$ orbitals. Number o

Vedclass · Accepted Answer

$A-II, B-IV, C-I, D-III$

Vedclass · Answer

(A) For option $(A)$: $[Cr(CN)_6]^{3-}$,$Cr^{3+}$ is $3d^3$. Since $CN^-$ is a strong field ligand,the electrons remain in $t_{2g}$ orbitals. Number of unpaired electrons $= 3$.For option $(B)$: $[Fe(H_2O)_6]^{2+}$,$Fe^{2+}$ is $3d^6$. Since $H_2O$ is a weak field ligand,the configuration is $t_{2g}^4 e_g^2$. Number of unpaired electrons $= 4$.For option $(C)$: $[Co(NH_3)_6]^{3+}$,$Co^{3+}$ is $3d^6$. Since $NH_3$ is a strong field ligand,all electrons are paired. Number of unpaired electrons $= 0$.For option $(D)$: $[Ni(NH_3)_6]^{2+}$,$Ni^{2+}$ is $3d^8$. The configuration is $t_{2g}^6 e_g^2$. Number of unpaired electrons $= 2$.Thus,the correct matching is $A-II, B-IV, C-I, D-III$.

List-$I$ (Coordination compounds)	List-$II$ (Number of unpaired electrons)
$A$. $[Cr(CN)_6]^{3-}$	$I$. $0$
$B$. $[Fe(H_2O)_6]^{2+}$	$II$. $3$
$C$. $[Co(NH_3)_6]^{3+}$	$III$. $2$
$D$. $[Ni(NH_3)_6]^{2+}$	$IV$. $4$

List-$I$ (Complexes)	List-$II$ (Types)
$a$. $[Co(NH_3)_5NO_2]Cl_2$ and $[Co(NH_3)_5ONO]Cl_2$	$i$. Linkage isomerism
$b$. $[Cr(NH_3)_6][Co(CN)_6]$ and $[Cr(CN)_6][Co(NH_3)_6]$	$ii$. Coordination isomerism
$c$. $[Co(NH_3)_5(SO_4)]Br$ and $[Co(NH_3)_5Br]SO_4$	$iii$. Ionisation isomerism
$d$. $[Cr(H_2O)_6]Cl_3$ and $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$	$iv$. Solvate isomerism

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