Match List–I with List–II. (Hybridisation of central metal ion) [2025]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Match List–I with List–II. [2025]

List-I List-II

(Complex) (Hybridisation of central metal ion)

(A) ${[{CoF}_{6}]}^{3 -}$ (I) $d^{2} s p^{3}$

(B) ${[{NiCl}_{4}]}^{2 -}$ (II) $s p^{3}$

(C) ${[Co {({NH}_{3})}_{6}]}^{3 +}$ (III) $s p^{3} d^{2}$

(D) ${[Ni {(CN)}_{4}]}^{2 -}$ (IV) $d s p^{2}$

Choose the correct answer from the options given below:

1 (A)–(III), (B)–(IV), (C)–(I), (D)–(II)

2 (A)–(III), (B)–(II), (C)–(I), (D)–(IV)

3 (A)–(I), (B)–(II), (C)–(III), (D)–(IV)

4 (A)–(I), (B)–(IV), (C)–(III), (D)–(II)

Ans.
(2)

$(A) {[{CoF}_{6}]}^{3 -}$

$Oxidation number of Co=+3$

${Co}^{3 +} = {}_{18}{[Ar]}$ $3 d^{6}$

$F^{-} is a weak field ligand. It cannot pair up d electrons. Thus 4 d orbitals are used for s p^{3} d^{2} hybridization. Six e^{-} pairs from six F^{-} are accommodated in these six s p^{3} d^{2} orbitals.$

$(B) {[{NiCl}_{4}]}^{2 -}$

$Oxidation number of Ni=+2$

${Ni}^{2 +} : {}_{18}{[Ar]}$ $3 d^{8}$

${Cl}^{-} is a weak field ligand; hence does not cause pairing of 3 d^{8} electrons. {Ni}^{2 +} undergoes s p^{3} hybridization and lone pairs from {Cl}^{-} are accommodated in s p^{3} orbitals.$

$(C) {[Co {({NH}_{3})}_{6}]}^{3 +}$

$Oxidation number of Co=+3$

${Co}^{3 +} = {}_{18}{[Ar]}$ $3 d^{6}$

${NH}_{3} is a strong field ligand. It pairs up d electrons. After pairing d electrons, two 3 d orbitals are vacant. {Co}^{3 +} undergoes d^{2} s p^{3} hybridization by using two 3 d orbitals. Six e^{-} pairs from six {NH}_{3} are accommodated in these six d^{2} s p^{3} orbitals.$

$(D) {[Ni {(CN)}_{4}]}^{2 -}$

$Oxidation number of Ni=+2$

${Ni}^{2 +} : {}_{18}{[Ar]}$ $3 d^{8}$

${CN}^{-} is a strong field ligand; hence causes pairing of 3 d^{8} electrons, then causes d s p^{2} hybridization using 3 d, 4 s, and 4 p orbitals.$

Want Us to Email you About Special Offers & Updates?