$[\mathrm{Fe(CN)_6}]^{4-}$ has $\mathrm{CN^-}$ ligands, which are STRONG-FIELD ligands. The complex is

1. Find the oxidation state of Fe. Complex charge $-4$, six $\mathrm{CN^-}$ ligands contribute $-6$. So Fe is $+2$. Configuration: Fe$^{2+}$ = [Ar] $3d^6$. 2. Strong-field ligands (large $\Delta_o$): $\mathrm{CN^-}$ is at the top of the spectrochemical series. 3. Strong field → pairing energy is LESS than $\Delta_o$, so electrons fill the lower $t_{2g}$ set FIRST and only pair up there, before populating $e_g$. 4. For 6 d-electrons in a strong-field octahedral case: $t_{2g}^6 \, e_g^0$. ALL six electrons in $t_{2g}$, fully paired. Zero unpaired → DIAMAGNETIC. 5. Option A is the HIGH-SPIN configuration ($t_{2g}^4 e_g^2$, 4 unpaired — would apply to a weak-field ligand like $\mathrm{F^-}$). Option C with 5 unpaired is impossible for $d^6$. Option D is internally inconsistent. _Source: NCERT Class 12 Chemistry Part 1, Ch 5, §5.6.2 (CFT — high spin vs low spin), p. 12–13._