Chemistry · Coordination Compounds

JEE Main 2026 — 21 January, Evening Shift — Question 57

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

Statement I: Crystal Field Stabilization Energy (CFSE) of [Cr(H2O)6]2+\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} is greater than that of [Mn(H2O)6]2+\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}.

Statement II: Potassium ferricyanide has a greater spin-only magnetic moment than sodium Ferrocyanide.

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

  1. Option A:

    Both Statement I and Statement II are true

  2. Option B:

    Both Statement I and Statement II are false

  3. Option C:

    Statement I is true but Statement II is false

  4. Option D:

    Statement I is false but Statement II is true

    Correct

Answer: D

Step-by-step solution

[Mn(H2O)6]2+⇒\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} \Rightarrow CFSE value is zero because of d5\mathrm{d}^{5} configuraiton with WFL in coordination number 6 [Cr(H2O)6]2+\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}. [Cr(H2O)6]2+⇒\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} \Rightarrow CFSE value is −0.6Δ0-0.6 \Delta_{0} because of d4\mathrm{d}^{4} configuraiton with WFL in coordination number 6.

For : K3[Fe(CN)6],μ=1(1+2)=3\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right], \mu=\sqrt{1(1+2)}=\sqrt{3} B.M. For : Na4[Fe(CN)6],μ=0\mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right], \mu=\sqrt{0} B.M.

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Subject
Chemistry
Chapter
Coordination Compounds
Topic
Properties and importance of Coordination Complexes