Indian Navy 10+2 B.Tech Cadet Inorganic Chemistry — practice questions

Atomic radius **decreases** from left to right across a period. Why: nuclear charge ($Z$) increases by 1 as you move right, pulling electrons closer. The added electron enters the same shell so screening barely increases — the net effect is stronger attraction and smaller radius. Across period 2: Li (152 pm) → Be → B → C → N → O → F (72 pm). Roughly halves. Down a group, atomic radius *increases* (electrons go into higher principal shells, further from nucleus).

First ionization energy generally increases across a period because nuclear charge grows and atomic radius shrinks, making it harder to remove an outer electron. In period 2: Li (520) < Be (899) ≈ B (801) < C (1086) < N (1402) > O (1314) < F (1681) < **Ne (2081)** kJ/mol. Neon has the highest because it has a stable octet ($1s^2 2s^2 2p^6$, full $2p$). Removing an electron requires breaking that closed shell. Note two anomalies: B < Be (removing a $2p$ electron easier than $2s$) and O < N (paired $2p$ electron in O has extra repulsion).

Manganese has the configuration $[\mathrm{Ar}]3d^5 4s^2$. All five 3d electrons and the two 4s electrons can participate in bonding so it can lose anywhere from one to seven electrons giving oxidation states $+2$ through $+7$. **Examples:** $\mathrm{Mn}^{2+}$ in $\mathrm{MnSO_4}$, $\mathrm{Mn}^{4+}$ in $\mathrm{MnO_2}$, $\mathrm{Mn}^{7+}$ in $\mathrm{KMnO_4}$.