Check statement (A): Compare $E^\circ(\text{M}^{3+}/\text{M}^{2+})$ for Fe and Mn.

• For iron: $\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}, \quad E^\circ = +0.77\,\text{V}$
• For manganese: $\text{Mn}^{3+} + e^- \rightarrow \text{Mn}^{2+}, \quad E^\circ \approx +1.51\,\text{V}$

Since $1.51 > 0.77$ the reduction potential for iron is not greater than manganese.

So, (A) is false.

Check statement (B): Higher oxidation states of first-row $d$-block elements are stabilized by oxide ion.

This is a standard trend in transition-metal chemistry. Oxygen, being highly electronegative and small, stabilizes metals in high oxidation states through strong $M-O$ bonds.

Examples:

• $\text{V}_2\text{O}_5$ with V in $+5$
• $\text{CrO}_3$ with Cr in $+6$
• $\text{Mn}_2\text{O}_7$ with Mn in $+7$

Hence, (B) is true.

Check statement (C): Aqueous solution of $\text{Cr}^{2+}$ can liberate hydrogen from dilute acid.

$\text{Cr}^{2+}$ is a strong reducing agent.

Compare standard potentials: $\text{Cr}^{3+} + e^- \rightarrow \text{Cr}^{2+}, \quad E^\circ = -0.41\,\text{V}$ $2\text{H}^+ + 2e^- \rightarrow \text{H}_2, \quad E^\circ = 0.00\,\text{V}$

Since oxidation of $\text{Cr}^{2+}$ to $\text{Cr}^{3+}$ is favorable, it can reduce $\text{H}^+$ to $\text{H}_2$: $2\text{Cr}^{2+} + 2\text{H}^+ \rightarrow 2\text{Cr}^{3+} + \text{H}_2$

Therefore, (C) is true.

Check statement (D): Magnetic moment of $\text{V}^{2+}$ is observed between $4.4 - 5.2$ BM.

Electronic configuration of V: $\text{V}: [\text{Ar}]3d^34s^2$ $\text{V}^{2+}: [\text{Ar}]3d^3$

Number of unpaired electrons, $n = 3$.

Spin-only magnetic moment: $\mu = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\,\text{BM}$

The observed value is around $3.8 - 4.0$ BM, not $4.4 - 5.2$ BM.

So, (D) is false.

Conclusion:

Correct statements are: $\boxed{(B), (C)\text{ only}}$

Hence the correct option is: $\boxed{\text{A}}$