JEE Advanced 2018 · Paper 2 · Chemistry Q15 · Dronacharya Exams

My attemptsTake a paper

← JEE Advanced 2018 · Paper 2

ChemistryList match+3 / -1Question 33 of 54

Match each set of hybrid orbitals from LIST-I with complex(es) given in LIST-II.

The correct option is

LIST-I

P
$dsp^2$
Q
$sp^3$
R
$sp^3d^2$
S
$d^2sp^3$

LIST-II

1
$[\mathrm{FeF_6}]^{4-}$
2
$[\mathrm{Ti(H_2O)_3Cl_3}]$
3
$[\mathrm{Cr(NH_3)_6}]^{3+}$
4
$[\mathrm{FeCl_4}]^{2-}$
5
$\mathrm{Ni(CO)_4}$
6
$[\mathrm{Ni(CN)_4}]^{2-}$

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

Answer and worked solution

Official answer: C

$\ce{[FeF6]^{4-}}$ : Fe²⁺ with weak-field F⁻ is high spin, an outer-orbital $sp^3d^2$ complex. That is R.
$\ce{[Ti(H2O)3Cl3]}$ : Ti³⁺ is $d^1$ , so two inner $d$ orbitals are free: $d^2sp^3$ . That is S.
$\ce{[Cr(NH3)6]^{3+}}$ : Cr³⁺ is $d^3$ : $d^2sp^3$ . That is S.
$\ce{[FeCl4]^{2-}}$ : tetrahedral, $sp^3$ . That is Q.
$\ce{Ni(CO)4}$ : Ni(0) with $3d^{10}$ , tetrahedral $sp^3$ . That is Q.
$\ce{[Ni(CN)4]^{2-}}$ : Ni²⁺ with strong-field CN⁻ is square planar, $dsp^2$ . That is P.

P → 6; Q → 4, 5; R → 1; S → 2, 3.

(C)

Topics:Valence bond theory Hybridisation

← Previous question Next question →