Match List - I with List - II : .tg border-collapse:collapse;border-spacing:0; .tg tdborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal; .tg thborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal; .tg .tg-0pkyborder-color:inherit;text-align:left;vertical-align:top List - I List -II (A) [PtCl_4]^2 - (I) sp^3d (B) BrF_5 (II) d^2sp^3 (C) PCl_5 (III) dsp^2 (D) [Co(NH_3)_6]^3 + (IV) sp^3d^2 Choose the most appropriate answer from the options given below :

(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

(A)-(II), (B)-(IV), (C)-(I), (D)-(III)

(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

(A)-(II), (B)-(I), (C)-(IV), (D)-(III)

Coordination Compounds PYQ — JEE Main 2022 · 24 Jun · Shift 1 · Q3 with Solution

Match List - I with List - II :

	List - I		List -II
(A)	${[PtC{l_4}]^{2 - }}$	(I)	$s{p^3}d$
(B)	$Br{F_5}$	(II)	${d^2}s{p^3}$
(C)	$PC{l_5}$	(III)	$ds{p^2}$
(D)	${[Co{(N{H_3})_6}]^{3 + }}$	(IV)	$s{p^3}{d^2}$

Choose the most appropriate answer from the options given below :

A
(A)-(II), (B)-(IV), (C)-(I), (D)-(III)
B
(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
C
(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
D
(A)-(II), (B)-(I), (C)-(IV), (D)-(III)

View written solutionFree

Correct answer: B

Determine the hybridisation/geometry of each species

(A) $[PtCl_4]^{2-}$

Oxidation state of Pt: $x+4(-1)=-2 \Rightarrow x=+2$
So, Pt is $Pt^{2+}$ , i.e. $5d^8$ .
For $d^8$ metal ion like $Pt^{2+}$ , the complex $[PtCl_4]^{2-}$ is square planar.
Square planar complexes correspond to: $dsp^2$
Hence, $(A) \to (III)$

(B) $BrF_5$

Central atom Br has 7 valence electrons.
It forms 5 bonds with F and has 1 lone pair.
Total electron pairs around Br = 6.
Electronic geometry = octahedral, molecular shape = square pyramidal.
Hybridisation: $sp^3d^2$
Hence, $(B) \to (IV)$

(C) $PCl_5$

Central atom P forms 5 bonds and has no lone pair.
Geometry = trigonal bipyramidal.
Hybridisation: $sp^3d$
Hence, $(C) \to (I)$

(D) $[Co(NH_3)_6]^{3+}$

Oxidation state of Co: $x+6(0)=+3 \Rightarrow x=+3$
So, Co is $Co^{3+}$ .
Atomic number of Co = 27, so: $Co: [Ar]3d^74s^2$ $Co^{3+}: [Ar]3d^6$
$NH_3$ with $Co^{3+}$ gives a low-spin octahedral complex.
Inner orbital octahedral hybridisation: $d^2sp^3$
Hence, $(D) \to (II)$

Final matching

(A)-(III), \quad (B)-(IV), \quad (C)-(I), \quad (D)-(II)

Compare with options

This corresponds to Option B.

Comparison with stored correct answer

Stored correct answer = B
Derived answer = B
They agree.

More from Coordination Compounds

Match List - I with List - II :

	List - I		List -II
(A)	${[PtC{l_4}]^{2 - }}$	(I)	$s{p^3}d$
(B)	$Br{F_5}$	(II)	${d^2}s{p^3}$
(C)	$PC{l_5}$	(III)	$ds{p^2}$
(D)	${[Co{(N{H_3})_6}]^{3 + }}$	(IV)	$s{p^3}{d^2}$

Choose the most appropriate answer from the options given below :

A
(A)-(II), (B)-(IV), (C)-(I), (D)-(III)
B
(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
C
(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
D
(A)-(II), (B)-(I), (C)-(IV), (D)-(III)

View written solutionFree

Correct answer: B

Determine the hybridisation/geometry of each species

(A) $[PtCl_4]^{2-}$

Oxidation state of Pt: $x+4(-1)=-2 \Rightarrow x=+2$
So, Pt is $Pt^{2+}$ , i.e. $5d^8$ .
For $d^8$ metal ion like $Pt^{2+}$ , the complex $[PtCl_4]^{2-}$ is square planar.
Square planar complexes correspond to: $dsp^2$
Hence, $(A) \to (III)$

(B) $BrF_5$

Central atom Br has 7 valence electrons.
It forms 5 bonds with F and has 1 lone pair.
Total electron pairs around Br = 6.
Electronic geometry = octahedral, molecular shape = square pyramidal.
Hybridisation: $sp^3d^2$
Hence, $(B) \to (IV)$

(C) $PCl_5$

Central atom P forms 5 bonds and has no lone pair.
Geometry = trigonal bipyramidal.
Hybridisation: $sp^3d$
Hence, $(C) \to (I)$

(D) $[Co(NH_3)_6]^{3+}$

Oxidation state of Co: $x+6(0)=+3 \Rightarrow x=+3$
So, Co is $Co^{3+}$ .
Atomic number of Co = 27, so: $Co: [Ar]3d^74s^2$ $Co^{3+}: [Ar]3d^6$
$NH_3$ with $Co^{3+}$ gives a low-spin octahedral complex.
Inner orbital octahedral hybridisation: $d^2sp^3$
Hence, $(D) \to (II)$

Final matching

(A)-(III), \quad (B)-(IV), \quad (C)-(I), \quad (D)-(II)

Compare with options

This corresponds to Option B.

Comparison with stored correct answer

Stored correct answer = B
Derived answer = B
They agree.

Coordination Compounds question

Coordination Compounds question

(A) $[PtCl_4]^{2-}$

(B) $BrF_5$

(C) $PCl_5$

(D) $[Co(NH_3)_6]^{3+}$

More from Coordination Compounds

Coordination Compounds question

Coordination Compounds question

(A) $[PtCl_4]^{2-}$

(B) $BrF_5$

(C) $PCl_5$

(D) $[Co(NH_3)_6]^{3+}$

More from Coordination Compounds

(A) [PtCl4]2−[PtCl_4]^{2-}[PtCl4​]2−

(B) BrF5BrF_5BrF5​

(C) PCl5PCl_5PCl5​

(D) [Co(NH3)6]3+[Co(NH_3)_6]^{3+}[Co(NH3​)6​]3+

More from Coordination Compounds

(A) [PtCl4]2−[PtCl_4]^{2-}[PtCl4​]2−

(B) BrF5BrF_5BrF5​

(C) PCl5PCl_5PCl5​

(D) [Co(NH3)6]3+[Co(NH_3)_6]^{3+}[Co(NH3​)6​]3+

More from Coordination Compounds

(A) $[PtCl_4]^{2-}$

(B) $BrF_5$

(C) $PCl_5$

(D) $[Co(NH_3)_6]^{3+}$

(A) $[PtCl_4]^{2-}$

(B) $BrF_5$

(C) $PCl_5$

(D) $[Co(NH_3)_6]^{3+}$