Q10 / 75 · JEE Main 2026 (21 January Shift 2) · Chemistry

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

  • Correct answer
    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

$\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} \Rightarrow$ CFSE value is zero because of $\mathrm{d}^5$ configuration with Weak Field Ligands in coordination number 6 $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$.

$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} \Rightarrow \mathrm{CFSE}$ value is $-0.6 \Delta_0$ because of $\mathrm{d}^4$ configuration with Weak Field Ligands in coordination number 6.

For : $\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right], \mu=\sqrt{1(1+2)}=\sqrt{3}$ B.M.

For : $\mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right], \mu=\sqrt{0}$ B.M.

Hence, the correct answer is option: 'A'.

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