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

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(D) $1$. $[Ni(CO)_4]$: $Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing. Hybridisation is $sp^3$ $(I)$.$2$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) $1$. $[Ni(CO)_4]$: $Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing. Hybridisation is $sp^3$ $(I)$.$2$. $[Cu(NH_3)_4]^{2+}$: $Cu^{2+}$ is $3d^9$. The unpaired electron is in $4p_z$,and the hybridisation is $dsp^2$ $(II)$.$3$. $[Fe(NH_3)_6]^{2+}$: $Fe^{2+}$ is $3d^6$. $NH_3$ is a strong field ligand,causing pairing. Hybridisation is $d^2sp^3$ $(IV)$.$4$. $[Fe(H_2O)_6]^{2+}$: $Fe^{2+}$ is $3d^6$. $H_2O$ is a weak field ligand,no pairing. Hybridisation is $sp^3d^2$ $(III)$.Thus,the correct match is $A-I, B-II, C-IV, D-III$.

List-$I$ (Complexes)	List-$II$ (Hybridisation)
$A$. $[Ni(CO)_4]$	$I$. $sp^3$
$B$. $[Cu(NH_3)_4]^{2+}$	$II$. $dsp^2$
$C$. $[Fe(NH_3)_6]^{2+}$	$III$. $sp^3d^2$
$D$. $[Fe(H_2O)_6]^{2+}$	$IV$. $d^2sp^3$

List-$I$ $(Compounds)$	List-$II$ $(Hybridization)$
$(a)$ $[Ni(NH_3)_6]^{+2}$	$(i)$ $sp^3 d^2$
$(b)$ $[PtCl_4]^{-2}$	$(ii)$ $dsp^2$
$(c)$ $[Ni(CO)_4]$	$(iii)$ $sp^3$
$(d)$ $[Co(OX)_3]^{-3}$	$(iv)$ $d^2 sp^3$

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

List-$I$ (Complexes) List-$II$ (Hybridisation)

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

$B$. $[Cu(NH_3)_4]^{2+}$ $II$. $dsp^2$

$C$. $[Fe(NH_3)_6]^{2+}$ $III$. $sp^3d^2$

$D$. $[Fe(H_2O)_6]^{2+}$ $IV$. $d^2sp^3$

Similar Questions

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The hybridization involved in the complex $[Ni(CN)_4]^{2-}$ is (Atomic number of $Ni = 28$):

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Match List-$I$ with List-$II$: List-$I$ (Complexes) List-$II$ (Hybridisation) $A$. $[Ni(CO)_4]$ $I$. $sp^3$ $B$. $[Cu(NH_3)_4]^{2+}$ $II$. $dsp^2$ $C$. $[Fe(NH_3)_6]^{2+}$ $III$. $sp^3d^2$ $D$. $[Fe(H_2O)_6]^{2+}$ $IV$. $d^2sp^3$