Match List I with List II, (Hybridisation & Magnetic characters) [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 & Magnetic characters)

(A) ${[{MnBr}_{4}]}^{2 -}$ (I) $d^{2} s p^{3}$ & diamagnetic

(B) ${[{FeF}_{6}]}^{3 -}$ (II) $s p^{3} d^{2}$ & paramagnetic

(C) ${[Co {(C_{2} O_{4})}_{3}]}^{3 -}$ (III) $s p^{3}$ & diamagnetic

(D) $[Ni(CO)_{4}]$ (IV) $s p^{3}$ & paramagnetic

Choose the correct answer from the options given below:

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

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

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

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

Ans.
(2)

(A) ${[{MnBr}_{4}]}^{2 -}$

Oxidation number of Mn = +2

${Mn}^{2 +} = {}_{18}{[Ar]}$ $3 d^{5}$

For coordination number 4 as in given complex, possible shapes are square planar (with ${dsp}^{2}$ hybridization) or tetrahedral (with ${sp}^{3}$ hybridization). For ${dsp}^{2}$ hybridization, electrons in 3d orbital need to pair up to make one d orbital vacant. But since bromide is a weak field ligand, it cannot cause pairing.

So this is ${sp}^{3}$ hybridized and paramagnetic.

(B) ${[{FeF}_{6}]}^{3 -}$

Oxidation number of Fe = +3

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

As it has unpaired electrons, it is paramagnetic.

(C) ${[Co {(C_{2} O_{4})}_{3}]}^{3 -}$

Oxidation number of Co = +3

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

With $C o^{3 +}$ , in octahedral complexes, $C_{2} O_{4}^{2 -}$ and $H_{2} O$ act as strong field ligands.

As all electrons are paired, it is diamagnetic.

(D) $[Ni {(CO)}_{4}]$

Oxidation number of Ni = 0

${}_{28}{Ni} = {}_{18}{[Ar]}$ $3 d^{8} 4 s^{2}$

CO is a strong field ligand. It shifts 4s electrons to 3d orbital to give $3 d^{10}$ configuration.

4s orbital becomes vacant. Ni then undergoes ${sp}^{3}$ hybridization and lone pairs from CO are accommodated in ${sp}^{3}$ orbitals.

As all electrons are paired, it is diamagnetic. As hybridization is ${sp}^{3}$ , it is tetrahedral.

Want Us to Email you About Special Offers & Updates?