JEE Main 2026 — Jan 24, Shift 1 — Chemistry solution

JEE Main2026Jan 24, Shift 1Chemistry

Given below are two statements

Statement-I : The number of paramagnetic species among $[C o F_{6}]^{3 -}$ , $[T i F_{6}]^{3 -}$ , $V_{2} O_{5}$ and $[F e (C N)_{6}]^{3 -}$ is 3.

Statement-II : $K_{4} [F e (C N)_{6}] < K_{3} [F e (C N)_{6}] < [F e (H_{2} O)_{6}] S O_{4} \cdot H_{2} O < [F e (H_{2} O)_{6}] C l_{3}$ 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.

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

Solution

Statement-I: Counting unpaired electrons:

$[C o F_{6}]^{3 -}$ : $C o^{3 +}$ ( $d^{6}$ ), high spin with weak $F^{-}$ → 4 unpaired → paramagnetic.

$[T i F_{6}]^{3 -}$ : $T i^{3 +}$ ( $d^{1}$ ) → 1 unpaired → paramagnetic.

$V_{2} O_{5}$ : V is in $+ 5$ state ( $d^{0}$ ) → diamagnetic.

$[F e (C N)_{6}]^{3 -}$ : $F e^{3 +}$ ( $d^{5}$ ), low spin with strong $C N^{-}$ → 1 unpaired → paramagnetic.

Paramagnetic count = 3. True.

Statement-II: Unpaired electrons:

$K_{4} [F e (C N)_{6}]$ : $F e^{2 +}$ , low spin → 0

$K_{3} [F e (C N)_{6}]$ : $F e^{3 +}$ , low spin → 1

$[F e (H_{2} O)_{6}] S O_{4} \cdot H_{2} O$ : $F e^{2 +}$ , high spin → 4

$[F e (H_{2} O)_{6}] C l_{3}$ : $F e^{3 +}$ , high spin → 5

Order: $0 < 1 < 4 < 5$ . True.

$∴$ Both statements are correct.

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