Match each set of hybrid orbitals from LIST - A with complex(es) given in LIST - B .tg border-collapse:collapse;border-spacing:0;border:none;border-color:#999; .tg tdfont-family:Arial, sans-serif;font-size:14px;padding:10px 5px;border-style:solid;border-width:0px;overflow:hidden;word-break:normal;border-color:#999;color:#444;background-color:#F7FDFA; .tg thfont-family:Arial, sans-serif;font-size:14px;font-weight:normal;padding:10px 5px;border-style:solid;border-width:0px;overflow:hidden;word-break:normal;border-color:#999;color:#fff;background-color:#26ADE4; .tg .tg-yttmfont-weight:bold;font-size:22px;text-align:left .tg .tg-pjk9font-size:22px;text-align:left List - A List - B P. dsp 2 1. [FeF 6 ] 4- Q. sp 3 2. [Ti(H 2 O) 3 Cl 3 ] R. sp 3 d 2 3. [Cr(NH 3 ) 6 ] 3+ S. d 2 sp 3 4. [FeCl 4 ] 2- 5. Ni(CO) 4 6. [Ni(CN) 4 ] 2- The correct option is

Coordination Compounds PYQ — JEE Advanced 2018 · Shift 2 · Q14 with Solution

Match each set of hybrid orbitals from LIST - A with complex(es) given in LIST - B

	List - A		List - B
P.	dsp²	1.	[FeF₆]^4-
Q.	sp³	2.	[Ti(H₂O)₃Cl₃]
R.	sp³d²	3.	[Cr(NH₃)₆]³⁺
S.	d²sp³	4.	[FeCl₄]^2-
		5.	Ni(CO)₄
		6.	[Ni(CN)₄]^2-

The correct option is

A
$P - 5;Q - 4,6;R - 2,3;S - 1$
B
$P - 5,6;Q - 4;R - 3;S - 1,2$
C
$P - 6;Q - 4,5;R - 1;S - 2,3$
D
$P - 4,6;Q - 5,6;R - 1,2;S - 3$

View written solutionFree

Correct answer: C

Identify the geometry and hybridisation of each complex in List-B

We classify each complex using metal oxidation state, $d$ -electron count, ligand strength, and geometry.

(1) $[FeF_6]^{4-}$

Oxidation state of Fe: $x+6(-1)=-4 \Rightarrow x=+2$
So $Fe^{2+}$ is $3d^6$ .
$F^-$ is a weak field ligand, so no pairing occurs in $3d$ orbitals.
Octahedral complex with outer orbital hybridisation: $sp^3d^2$

So, (1) corresponds to $sp^3d^2$ .

(2) $[Ti(H_2O)_3Cl_3]$

$H_2O$ is neutral and $Cl^-$ is $-1$ .
Since the complex is neutral: $x+3(0)+3(-1)=0 \Rightarrow x=+3$
So $Ti^{3+}$ is $3d^1$ .
Coordination number $=6$ , hence octahedral.
Since vacant $3d$ orbitals are available, it can use inner orbital hybridisation: $d^2sp^3$

So, (2) corresponds to $d^2sp^3$ .

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

$NH_3$ is neutral, so Cr oxidation state is $+3$ .
$Cr^{3+}$ is $3d^3$ .
Coordination number $=6$ , octahedral.
Two vacant $3d$ orbitals are available, so inner orbital complex: $d^2sp^3$

So, (3) corresponds to $d^2sp^3$ .

(4) $[FeCl_4]^{2-}$

Oxidation state of Fe: $x+4(-1)=-2 \Rightarrow x=+2$
$Fe^{2+}$ is $3d^6$ .
Coordination number $=4$ .
$Cl^-$ is weak field, so tetrahedral complex forms.
Tetrahedral hybridisation: $sp^3$

So, (4) corresponds to $sp^3$ .

(5) $Ni(CO)_4$

$CO$ is neutral, so Ni oxidation state is $0$ .
$Ni$ atom: $3d^8 4s^2$ .
In the complex, due to pairing, effective configuration becomes $3d^{10}$ .
Coordination number $=4$ , tetrahedral.
Hybridisation: $sp^3$

So, (5) corresponds to $sp^3$ .

(6) $[Ni(CN)_4]^{2-}$

Oxidation state of Ni: $x+4(-1)=-2 \Rightarrow x=+2$
$Ni^{2+}$ is $3d^8$ .
$CN^-$ is strong field ligand, causing pairing.
Coordination number $=4$ and strong field gives square planar geometry.
Square planar hybridisation: $dsp^2$

So, (6) corresponds to $dsp^2$ .

Now match with List-A

$P: dsp^2 \rightarrow 6$
$Q: sp^3 \rightarrow 4,5$
$R: sp^3d^2 \rightarrow 1$
$S: d^2sp^3 \rightarrow 2,3$

So the correct matching is: $P-6,\quad Q-4,5,\quad R-1,\quad S-2,3$

Compare with options

This corresponds to Option C.

Comparison with stored answer

Stored correct answer = C.

Our derived answer also = C.

More from Coordination Compounds

Match each set of hybrid orbitals from LIST - A with complex(es) given in LIST - B

	List - A		List - B
P.	dsp²	1.	[FeF₆]^4-
Q.	sp³	2.	[Ti(H₂O)₃Cl₃]
R.	sp³d²	3.	[Cr(NH₃)₆]³⁺
S.	d²sp³	4.	[FeCl₄]^2-
		5.	Ni(CO)₄
		6.	[Ni(CN)₄]^2-

The correct option is

A
$P - 5;Q - 4,6;R - 2,3;S - 1$
B
$P - 5,6;Q - 4;R - 3;S - 1,2$
C
$P - 6;Q - 4,5;R - 1;S - 2,3$
D
$P - 4,6;Q - 5,6;R - 1,2;S - 3$

View written solutionFree

Correct answer: C

Identify the geometry and hybridisation of each complex in List-B

We classify each complex using metal oxidation state, $d$ -electron count, ligand strength, and geometry.

(1) $[FeF_6]^{4-}$

Oxidation state of Fe: $x+6(-1)=-4 \Rightarrow x=+2$
So $Fe^{2+}$ is $3d^6$ .
$F^-$ is a weak field ligand, so no pairing occurs in $3d$ orbitals.
Octahedral complex with outer orbital hybridisation: $sp^3d^2$

So, (1) corresponds to $sp^3d^2$ .

(2) $[Ti(H_2O)_3Cl_3]$

$H_2O$ is neutral and $Cl^-$ is $-1$ .
Since the complex is neutral: $x+3(0)+3(-1)=0 \Rightarrow x=+3$
So $Ti^{3+}$ is $3d^1$ .
Coordination number $=6$ , hence octahedral.
Since vacant $3d$ orbitals are available, it can use inner orbital hybridisation: $d^2sp^3$

So, (2) corresponds to $d^2sp^3$ .

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

$NH_3$ is neutral, so Cr oxidation state is $+3$ .
$Cr^{3+}$ is $3d^3$ .
Coordination number $=6$ , octahedral.
Two vacant $3d$ orbitals are available, so inner orbital complex: $d^2sp^3$

So, (3) corresponds to $d^2sp^3$ .

(4) $[FeCl_4]^{2-}$

Oxidation state of Fe: $x+4(-1)=-2 \Rightarrow x=+2$
$Fe^{2+}$ is $3d^6$ .
Coordination number $=4$ .
$Cl^-$ is weak field, so tetrahedral complex forms.
Tetrahedral hybridisation: $sp^3$

So, (4) corresponds to $sp^3$ .

(5) $Ni(CO)_4$

$CO$ is neutral, so Ni oxidation state is $0$ .
$Ni$ atom: $3d^8 4s^2$ .
In the complex, due to pairing, effective configuration becomes $3d^{10}$ .
Coordination number $=4$ , tetrahedral.
Hybridisation: $sp^3$

So, (5) corresponds to $sp^3$ .

(6) $[Ni(CN)_4]^{2-}$

Oxidation state of Ni: $x+4(-1)=-2 \Rightarrow x=+2$
$Ni^{2+}$ is $3d^8$ .
$CN^-$ is strong field ligand, causing pairing.
Coordination number $=4$ and strong field gives square planar geometry.
Square planar hybridisation: $dsp^2$

So, (6) corresponds to $dsp^2$ .

Now match with List-A

$P: dsp^2 \rightarrow 6$
$Q: sp^3 \rightarrow 4,5$
$R: sp^3d^2 \rightarrow 1$
$S: d^2sp^3 \rightarrow 2,3$

So the correct matching is: $P-6,\quad Q-4,5,\quad R-1,\quad S-2,3$

Compare with options

This corresponds to Option C.

Comparison with stored answer

Stored correct answer = C.

Our derived answer also = C.

Coordination Compounds question

Coordination Compounds question

(1) $[FeF_6]^{4-}$

(2) $[Ti(H_2O)_3Cl_3]$

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

(4) $[FeCl_4]^{2-}$

(5) $Ni(CO)_4$

(6) $[Ni(CN)_4]^{2-}$

More from Coordination Compounds

Coordination Compounds question

Coordination Compounds question

(1) $[FeF_6]^{4-}$

(2) $[Ti(H_2O)_3Cl_3]$

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

(4) $[FeCl_4]^{2-}$

(5) $Ni(CO)_4$

(6) $[Ni(CN)_4]^{2-}$

More from Coordination Compounds

(1) [FeF6]4−[FeF_6]^{4-}[FeF6​]4−

(2) [Ti(H2O)3Cl3][Ti(H_2O)_3Cl_3][Ti(H2​O)3​Cl3​]

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

(4) [FeCl4]2−[FeCl_4]^{2-}[FeCl4​]2−

(5) Ni(CO)4Ni(CO)_4Ni(CO)4​

(6) [Ni(CN)4]2−[Ni(CN)_4]^{2-}[Ni(CN)4​]2−

More from Coordination Compounds

(1) [FeF6]4−[FeF_6]^{4-}[FeF6​]4−

(2) [Ti(H2O)3Cl3][Ti(H_2O)_3Cl_3][Ti(H2​O)3​Cl3​]

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

(4) [FeCl4]2−[FeCl_4]^{2-}[FeCl4​]2−

(5) Ni(CO)4Ni(CO)_4Ni(CO)4​

(6) [Ni(CN)4]2−[Ni(CN)_4]^{2-}[Ni(CN)4​]2−

More from Coordination Compounds

(1) $[FeF_6]^{4-}$

(2) $[Ti(H_2O)_3Cl_3]$

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

(4) $[FeCl_4]^{2-}$

(5) $Ni(CO)_4$

(6) $[Ni(CN)_4]^{2-}$

(1) $[FeF_6]^{4-}$

(2) $[Ti(H_2O)_3Cl_3]$

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

(4) $[FeCl_4]^{2-}$

(5) $Ni(CO)_4$

(6) $[Ni(CN)_4]^{2-}$