The hybridization and magnetic nature of [Mn(CN)6]4- and [Fe(CN)6]3-, respectively are:

🧠 Both Strong-Field d⁵ — Both Inner-Orbital

Both complexes have CN⁻ as ligand and $\mathrm{d^5}$ as the d-count:

• $[\mathrm{Mn(CN)_6}]^{4-}$: $\mathrm{Mn^{2+}}\,(\mathrm{d^5})$.
• $[\mathrm{Fe(CN)_6}]^{3-}$: $\mathrm{Fe^{3+}}\,(\mathrm{d^5})$.

Strong-field $\mathrm{CN^-}$ forces pairing → low-spin $\mathrm{t_{2g}^5\,e_g^0}$ in both. Both use inner-orbital $\mathrm{d^2sp^3}$ hybridisation. Both have 1 unpaired electron → paramagnetic.

🗺️ Verify Each

$[\mathrm{Mn(CN)_6}]^{4-}$. $\mathrm{Mn^{2+}}$: d⁵. Low-spin → 5 electrons in t_{2g} (one pair × 2 + one single) → 1 unpaired. d²sp³ (inner). Paramagnetic.

$[\mathrm{Fe(CN)_6}]^{3-}$. $\mathrm{Fe^{3+}}$: d⁵. Same configuration as above → 1 unpaired, d²sp³, paramagnetic.

So both: $\mathrm{d^2sp^3}$, paramagnetic — option (2).

⚡ The "$\mathrm{CN^-}$ + d⁵" Universal Result

Whenever you see $\mathrm{CN^-}$ with a d⁵ first-row metal ion, expect:

• Low-spin
• Inner-orbital $\mathrm{d^2sp^3}$
• 1 unpaired electron
• μ ≈ 1.73 BM
• Paramagnetic

Memorise this once and it covers many JEE problems.

⚠️ The "Strong Field → Diamagnetic" Reflex

Strong-field doesn't always mean diamagnetic. d⁵ low-spin still has 1 unpaired electron (an odd-electron system can never be diamagnetic). Diamagnetism requires an even-electron system plus full pairing — e.g., d⁶ low-spin ($\mathrm{t_{2g}^6}$). Always count the unpaired electrons; never assume.

$\boxed{\text{Answer: (2) } \mathrm{d^2sp^3} \text{ and paramagnetic}}$