1. First, identify which complexes are homoleptic

A homoleptic complex has only one kind of ligand.

• (A) $[\mathrm{Fe}(\mathrm{CN})_5\mathrm{NO}]^{2-}$ has $\mathrm{CN}^-$ and $\mathrm{NO}$, so it is heteroleptic.
• (B) $[\mathrm{CoF}_6]^{3-}$ is homoleptic.
• (C) $[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$ is homoleptic.
• (D) $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$ is homoleptic.
• (E) $[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ is homoleptic.

So among the given complexes, we need to check (B), (C), (D), (E) for low spin.

2. Check each homoleptic complex

(B) $[\mathrm{CoF}_6]^{3-}$

• Oxidation state of Co: $x + 6(-1) = -3 \Rightarrow x = +3$
• So metal ion is $\mathrm{Co}^{3+}$.
• Electronic configuration of Co: $\mathrm{Co}: [\mathrm{Ar}]3d^74s^2$ $\mathrm{Co}^{3+}: 3d^6$
• Geometry is octahedral.
• $\mathrm{F}^-$ is a weak field ligand, so it causes small splitting.
• Hence electrons remain unpaired as much as possible: high spin.

So (B) is not low spin.

(C) $[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$

• Oxidation state of Fe: $x + 6(-1) = -4 \Rightarrow x = +2$
• So metal ion is $\mathrm{Fe}^{2+}$.
• Electronic configuration: $\mathrm{Fe}^{2+}: 3d^6$
• $\mathrm{CN}^-$ is a strong field ligand, producing large crystal field splitting.
• Therefore electrons pair in lower $t_{2g}$ orbitals first.
• Octahedral arrangement becomes: $t_{2g}^6 e_g^0$
• This is low spin.

So (C) is low spin.

(D) $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$

• Oxidation state of Co is $+3$.
• Thus $\mathrm{Co}^{3+} = 3d^6$.
• $\mathrm{NH}_3$ is stronger than $\mathrm{H_2O}$ and for $\mathrm{Co}^{3+}$ the crystal field splitting is sufficiently large.
• Hence octahedral $d^6$ becomes: $t_{2g}^6 e_g^0$
• Therefore it is low spin.

So (D) is low spin.

(E) $[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_6]^{2+}$

• Oxidation state of Cr: $x + 6(0) = +2 \Rightarrow x = +2$
• So metal ion is $\mathrm{Cr}^{2+}$.
• Electronic configuration: $\mathrm{Cr}: [\mathrm{Ar}]3d^54s^1$ $\mathrm{Cr}^{2+}: 3d^4$
• $\mathrm{H_2O}$ is a weak field ligand.
• So octahedral complex is high spin.

So (E) is not low spin.

3. Conclusion

Among the homoleptic complexes, the low spin ones are: $\boxed{(C)\ \text{and}\ (D)\ \text{only}}$

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

4. Comparison with stored correct answer

Stored correct answer: A
Derived answer: A

They match.