Match List-I with List-II and select the corr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) $1. [Ni(NH_3)_6]^{2+}$: $Ni^{2+}$ is $3d^8$. $NH_3$ is a weak field ligand,so it forms an outer-orbital complex with $sp^3d^2$ hybridization.$2. [

Vedclass · Accepted Answer

$2-3-1-4$

Vedclass · Answer

(A) $1. [Ni(NH_3)_6]^{2+}$: $Ni^{2+}$ is $3d^8$. $NH_3$ is a weak field ligand,so it forms an outer-orbital complex with $sp^3d^2$ hybridization.$2. [PtCl_4]^{2-}$: $Pt^{2+}$ is a $5d^8$ metal ion. $4d$ and $5d$ series metals always form square planar complexes with $dsp^2$ hybridization regardless of the ligand strength.$3. [Ni(CO)_4]$: $Ni$ is $3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing,resulting in $sp^3$ hybridization.$4. [Co(ox)_3]^{3-}$: $Co^{3+}$ is $3d^6$. $ox^{2-}$ is a strong field ligand,causing pairing,resulting in $d^2sp^3$ hybridization.Therefore,the correct matching is $A-2, B-3, C-1, D-4$.

List-$I$ (Compound)	List-$II$ (Hybridization of Central atom)
$A. [Ni(NH_3)_6]^{2+}$	$1. sp^3$
$B. [PtCl_4]^{2-}$	$2. sp^3d^2$
$C. [Ni(CO)_4]$	$3. dsp^2$
$D. [Co(ox)_3]^{3-}$	$4. d^2sp^3$

Column $I$	Column $II$
$(A)$ $sp^3$	$(i)$ $[Co(NH_3)_6]^{3+}$
$(B)$ $dsp^2$	$(ii)$ $[Ni(CO)_4]$
$(C)$ $sp^3d^2$	$(iii)$ $[Pt(NH_3)_2Cl_2]$
$(D)$ $d^2sp^3$	$(iv)$ $[CoF_6]^{3-}$
	$(v)$ $[Fe(CO)_5]$

Similar Questions

Which of the following statements is incorrect for the complex $[Cr(NH_3)_6]Cl_3$?

The hybridization and magnetic behaviour of the cobalt ion in the $[Co(NH_3)_6]^{3+}$ complex,respectively,are:

What is the coordination geometry around the manganese atom in decacarbonyldimanganese$(0)$?

The coordination environment of $Ca^{2+}$ ion in its complex with $EDTA^{4-}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Match List-$I$ with List-$II$ and select the correct answer using the codes given below the lists. List-$I$ (Compound) List-$II$ (Hybridization of Central atom) $A. [Ni(NH_3)_6]^{2+}$ $1. sp^3$ $B. [PtCl_4]^{2-}$ $2. sp^3d^2$ $C. [Ni(CO)_4]$ $3. dsp^2$ $D. [Co(ox)_3]^{3-}$ $4. d^2sp^3$ $A-B-C-D$