Import Question JSON

$\text{HINT: Second period ionization energy does not increases in regular manner.}$ $\text{Explanation:}$ $\text{For option (1)}$ $\text{First ionization energy is the energy required to remove an electron from outermost shell.}$ $\text{Hence, correct order is B} < \text{C} < \text{O} < \text{N.}$ $\text{For option (2)}$ $\text{Electron gain enthalpy is the energy required to gain an electron in the outermost shell.}$ $\text{Hence, the correct order is I} < \text{Br} < \text{F} < \text{Cl.}$ $\text{For option (3)}$ $\text{As we move down the group in alkali metal, metallic radius increases Li} < \text{Na} < \text{K} < \text{Rb.}$ $\text{For option (4)}$ $\text{In case of isoelectronic species, as positive charge decreases or negative charge increases the ionic size of the species increases and vice-versa Al}^{3+} < \text{Mg}^{2+} < \text{Na}^+ < \text{F}^-.$$ $\text{The incorrect match is first.}$ $\text{Detailed explanation for option 1:}$ $\text{The first ionization energy generally increases across a period from left to right. However, there are exceptions due to electronic configuration effects:}$ $\text{B (1s}^2\text{2s}^2\text{2p}^1\text{): The 2p electron is easier to remove than 2s electrons}$ $\text{C (1s}^2\text{2s}^2\text{2p}^2\text{): Higher nuclear charge, more difficult to remove electron}$ $\text{N (1s}^2\text{2s}^2\text{2p}^3\text{): Half-filled p orbital provides extra stability}$ $\text{O (1s}^2\text{2s}^2\text{2p}^4\text{): Electron-electron repulsion in paired p orbital}$ $\text{Due to the extra stability of half-filled p orbitals in nitrogen, it has higher first ionization energy than oxygen, despite oxygen having higher nuclear charge.}$ $\text{Therefore, the correct order should be: B} < \text{C} < \text{O} < \text{N}$ $\text{The given order B} < \text{C} < \text{N} < \text{O is incorrect because it places N before O, which contradicts the stability of half-filled orbitals.}$