Given below are two statements : one is labelled as Statement I and the other is labelled as Statement II : Statements-I : The number of paramagnetic species among [ CoF 6 ] 3- , [ TiF 6 ] 3- , V 2 O 5 and [ Fe ( CN ) 6 ] 3- is 3 . Statement-II : K 4 [ Fe ( CN ) 6 ]< K 3 [ Fe ( CN ) 6 ]< [ Fe ( H 2 O ) 6 ] SO 4 × H 2 O < [ Fe ( H 2 O ) 6 ] Cl 3 is the correct order in terms of number of unpaired electron(s) in the complexes. In the light of the above statements, choose the correct answer from the options given below.

Both statement-I and statement-II are true

Both statement-I and statement-II are false

Statement-I is true but statement-II is false

Statement-I is false but statement-II is true

Chemistry · Coordination Compounds

JEE Main 2026 — 24 January, Morning Shift — Question 54

Given below are two statements : one is labelled as Statement I and the other is labelled as Statement II :

Statements-I : The number of paramagnetic species among $\left[\mathrm{CoF}_{6}\right]^{3-}, \quad\left[\mathrm{TiF}_{6}\right]^{3-}, \quad \mathrm{V}_{2} \mathrm{O}_{5}$ and $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ is 3 .

Statement-II : $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]<\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]< \left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{SO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}<\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3}$ is the correct order in terms of number of unpaired electron(s) in the complexes.

In the light of the above statements, choose the correct answer from the options given below.

Option A:
Both statement-I and statement-II are true
Correct
Option B:
Both statement-I and statement-II are false
Option C:
Statement-I is true but statement-II is false
Option D:
Statement-I is false but statement-II is true

Answer: A

Step-by-step solution

Paramagnetic species : $\left[\mathrm{CoF}_{6}\right]^{3-},\left[\mathrm{TiF}_{6}\right]^{3-},\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$ Diamagnetic species : $\mathrm{V}_{2} \mathrm{O}_{5}$ In $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \Rightarrow$ No. of unpaired electron $=0$ $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \Rightarrow$ No. of unpaired electron $=1$ $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{SO}_{4} \cdot \mathrm{H}_{2} \mathrm{O} \Rightarrow$ No. of unpaired electron $=4$ $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3} \Rightarrow$ No. of unpaired electron $=5$

Answer key and solution verified before publishing.

Practise Coordination Compounds

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2026
Paper: 24 January, Morning Shift
Subject: Chemistry
Chapter: Coordination Compounds
Topic: Properties and importance of Coordination Complexes

Answer: A

Step-by-step solution

Practise Coordination Compounds

More Coordination Compounds questions from this paper