Q.Iron exhibits +2 and +3 oxidation states. Write their electronic configuration. Which will be more stable ? Why ?

Cr adopts $3d^54s^1$ for half-filled-shell stability; Ti$^{3+}$ has one unpaired electron giving $\mu \approx 1.73$ BM; green chemistry minimises environmental hazard.

(i) Observed configuration of Cr (Z = 24): The expected 'aufbau' configuration would be $[Ar]3d^44s^2$, but chromium actually adopts $[Ar]3d^54s^1$, since a half-filled d-subshell ($3d^5$) with a half-filled s-subshell ($4s^1$) gives extra stability (exchange energy, symmetry).

(ii) Magnetic moment of Ti$^{3+}$ (Z = 22): Ti: $[Ar]3d^24s^2$. Ti$^{3+}$ loses 3 electrons (2 from 4s, 1 from 3d): $[Ar]3d^1$, giving 1 unpaired electron ($n=1$).

$\mu = \sqrt{n(n+2)}\ BM = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73\ BM$

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3d series: $3d^{1-10}4s^{1-2}$. H$_2$SO$_4$: S(=O)$_2$(OH)$_2$. H$_2$S$_2$O$_3$: same but one =O replaced by =S.

General electronic configuration of the 3d (first transition) series: $[Ar]\,3d^{1-10}\,4s^{1-2}$ — the $4s$ orbital fills before $3d$ begins, but once $3d$ starts filling, both subshells are close in energy (with a few well-known exceptions like Cr $= 3d^54s^1$ and Cu $=3d^{10}4s^1$, arising from extra stability of half-filled/fully-filled $d$ subshells).

Structure of sulphuric acid, $H_2SO_4$: a tetrahedral S centre bonded to two terminal oxygens by double bonds ($S=O$) and to two $-OH$ groups.

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