JEE Main 2023 — Chemistry Inorganic Chemistry

$[\mathrm{Fe}({H}_{2}O{)}_{6}{]}^{3+}$[Fe(H2​O)6​]3+ has magnetic moment value of $5.92\mathrm{BM}$5.92BM because ${H}_{2}O$H2​O is a weak ligand and pairing will not occur. It involves ${\mathrm{sp}}^{3}{d}^{2}$sp3d2 hybridisation with five unpaired electrons (because magnetic moment is equal to $5.92\mathrm{BM}$5.92BM) and outer orbital complex is formed. Magnetic moment can be calculated by the formula ${\mu }_{m}=\sqrt{n(n+2)}$μm​=n(n+2)​ $\mathrm{BM}$BM where, ${\mu }_{m}=$μm​=magnetic moment, $n=$n=number of unpaired electron(s).

$${\mu }_{m}=1.74\mathrm{BM},n=1 {\mu }_{m}=5.92\mathrm{BM},n=5$$μm​=1.74BM,n=1μm​=5.92BM,n=5

In $[\mathrm{Fe}(\mathrm{CN}{)}_{6}{]}^{3–}$[Fe(CN)6​]3–complex ${\mathrm{CN}}^{-}$CN− is stronger ligand than ${H}_{2}O$H2​O according to spectrochemical series. Hence, four electrons will pair itself. It involves ${d}^{2}{\mathrm{sp}}^{3}$d2sp3hybridisation with one unpaired electron and inner orbital complex is formed and has a value of only $1.74\mathrm{BM}$1.74BM.

In Bothe the complexes the oxidation state of iron is +3.