1. First identify which complexes are homoleptic

A homoleptic complex has only one kind of ligand attached to the metal.

• (A) $[\mathrm{FeO}_4]^{2-}$ : only $\mathrm{O}^{2-}$ ligands $\Rightarrow$ homoleptic
• (B) $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}$ : only $\mathrm{CN}^-$ ligands $\Rightarrow$ homoleptic
• (C) $[\mathrm{Fe}(\mathrm{CN})_5\mathrm{NO}]^{2-}$ : $\mathrm{CN}^-$ and $\mathrm{NO}$ both present $\Rightarrow$ heteroleptic
• (D) $[\mathrm{CoCl}_4]^{2-}$ : only $\mathrm{Cl}^-$ ligands $\Rightarrow$ homoleptic
• (E) $[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_3\mathrm{F}_3]$ : $\mathrm{H_2O}$ and $\mathrm{F}^-$ both present $\Rightarrow$ heteroleptic

So only (A), (B), (D) are homoleptic. Now check whether the central metal has an odd number of $d$ electrons.

2. Option (A): $[\mathrm{FeO}_4]^{2-}$

Let oxidation state of Fe be $x$.

$x + 4(-2) = -2$ $x - 8 = -2$ $x = +6$

Fe: $Z=26$, ground state $[\mathrm{Ar}]3d^64s^2$

For $\mathrm{Fe}^{6+}$, remove $2$ electrons from $4s$ and $4$ from $3d$:

$\mathrm{Fe}^{6+} : 3d^2$

Number of $d$ electrons $=2$ (even), so not required.

3. Option (B): $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}$

Let oxidation state of Fe be $x$.

$x + 6(-1) = -3$ $x - 6 = -3$ $x = +3$

So,

$\mathrm{Fe}^{3+} : 3d^5$

Number of $d$ electrons $=5$ (odd), so required.

4. Option (D): $[\mathrm{CoCl}_4]^{2-}$

Let oxidation state of Co be $x$.

$x + 4(-1) = -2$ $x - 4 = -2$ $x = +2$

Co: $Z=27$, ground state $[\mathrm{Ar}]3d^74s^2$

For $\mathrm{Co}^{2+}$:

$\mathrm{Co}^{2+} : 3d^7$

Number of $d$ electrons $=7$ (odd), so required.

5. Now collect the correct homoleptic complexes with odd number of $d$ electrons

These are:

• (B) $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}$
• (D) $[\mathrm{CoCl}_4]^{2-}$

Thus the correct choice is:

$\boxed{\text{(B) and (D) only}}$

which corresponds to Option C.

6. Comparison with stored correct answer

Stored correct answer = C

Derived answer = C

So they agree.