1. Find oxidation states and $d$-electron counts

For $[Mn(CN)_6]^{4-}$:

• Let oxidation state of Mn be $x$.
• Since $CN^-$ is a $-1$ ligand, $x+6(-1)=-4$ $x-6=-4\Rightarrow x=+2$
• Mn: $Z=25$, ground state $[Ar]3d^54s^2$
• $Mn^{2+}$: $[Ar]3d^5$

For $[Fe(CN)_6]^{3-}$:

• Let oxidation state of Fe be $x$. $x+6(-1)=-3$ $x-6=-3\Rightarrow x=+3$
• Fe: $Z=26$, ground state $[Ar]3d^64s^2$
• $Fe^{3+}$: $[Ar]3d^5$

So, both complexes contain a $d^5$ metal ion.

2. Nature of ligand $CN^-$

$CN^-$ is a strong-field ligand. Therefore, in octahedral complexes it tends to cause pairing and usually forms inner orbital complexes with hybridization $d^2sp^3$ when possible.

3. First complex: $[Mn(CN)_6]^{4-}$

• Metal ion: $Mn^{2+}$, which is $3d^5$
• In presence of strong-field ligand $CN^-$, electrons pair up as much as possible in the $t_{2g}$ set.
• Octahedral low-spin configuration: $t_{2g}^5e_g^0$
• This leaves one unpaired electron.
• Since two $3d$ orbitals become available for bonding, the hybridization is: $d^2sp^3$
• Magnetic nature: paramagnetic

4. Second complex: $[Fe(CN)_6]^{3-}$

• Metal ion: $Fe^{3+}$, also $3d^5$
• With strong-field ligand $CN^-$, low-spin octahedral configuration: $t_{2g}^5e_g^0$
• This also has one unpaired electron.
• Therefore magnetic nature is paramagnetic.

5. Match with options

The question asks for the hybridization of $[Mn(CN)_6]^{4-}$ and the magnetic nature of $[Fe(CN)_6]^{3-}$, respectively.

• $[Mn(CN)_6]^{4-}$: $d^2sp^3$
• $[Fe(CN)_6]^{3-}$: paramagnetic

So the correct option is:

$\boxed{\text{B: } d^2sp^3 \text{ and paramagnetic}}$

6. Comparison with stored correct answer

Stored correct answer: B

My derived answer: B

They match.