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(A) The hybridization of the given complexes is determined as follows: $(a)$ $[Ni(NH_3)_6]^{+2}$: $Ni^{+2}$ is $3d^8$. $NH_3$ is a weak field ligand,s

Vedclass · Accepted Answer

$a-i, b-ii, c-iii, d-iv$

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(A) The hybridization of the given complexes is determined as follows: $(a)$ $[Ni(NH_3)_6]^{+2}$: $Ni^{+2}$ is $3d^8$. $NH_3$ is a weak field ligand,so it forms an outer orbital complex with $sp^3 d^2$ hybridization. $(b)$ $[PtCl_4]^{-2}$: $Pt^{+2}$ is a $5d^8$ metal ion. $Cl^-$ is a ligand that forces pairing in $5d$ series,resulting in $dsp^2$ hybridization. $(c)$ $[Ni(CO)_4]$: $Ni$ is $3d^8 4s^2$. $CO$ is a strong field ligand,forcing pairing to form $sp^3$ hybridization. $(d)$ $[Co(OX)_3]^{-3}$: $Co^{+3}$ is $3d^6$. $OX^{-2}$ (oxalate) is a strong field ligand,resulting in $d^2 sp^3$ hybridization. Thus,the correct matching is $a-i, b-ii, c-iii, d-iv$.

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

List-$I$	List-$II$
$A. Ni(CO)_4$	$I. sp^3$
$B. [Ni(CN)_4]^{2-}$	$II. sp^3d^2$
$C. [Co(CN)_6]^{3-}$	$III. d^2sp^3$
$D. [CoF_6]^{3-}$	$IV. dsp^2$

Similar Questions

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The complex ion which has no $d$ electrons in the central metal atom is

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

List-$I$ List-$II$

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

$B. [Ni(CN)_4]^{2-}$ $II. sp^3d^2$

$C. [Co(CN)_6]^{3-}$ $III. d^2sp^3$

$D. [CoF_6]^{3-}$ $IV. dsp^2$

Choose the correct answer from the options given below.

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Match List-$I$ with List-$II$ and select the correct answer using the codes given below the lists $:$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)$ $d^2 sp^3$