Q.Define IE1 and IE2. Why IE2 > IE1 for the given atom? Discuss any four factors that affect IE of an element.

Across a period, increasing effective nuclear charge on a constant shell shrinks atoms and pulls electrons more strongly.

1. Atomic radius: Decreases while moving from left to right in a period, because electrons are added to the same principal shell (same $n$) while the nuclear charge (number of protons) also increases. Since electrons in the same shell shield each other poorly, the effective nuclear charge felt by the outer electrons increases, pulling them closer to the nucleus and shrinking the atom.
2. Ionisation energy: Generally increases from left to right in a period, because of the increasing effective nuclear charge and decreasing atomic radius — the outer electron is held more tightly and more energy is required to remove it (some irregularities occur at half-filled/fully-filled stable configurations).
3. Electronegativity: Increases from left to right across a period, because of increasing nuclear charge and decreasing atomic size, so each atom attracts the shared pair of bonding electrons more strongly. OR (give reasons): (a) $H_2O$ is liquid while $H_2S$ is gas: Oxygen is much smaller and more electronegative than sulphur, so the O-H bond is highly polar, allowing $H_2O$ molecules to form strong intermolecular hydrogen bonds, which raises its boiling point so that it is a liquid at room temperature. Sulphur is larger and less electronegative, so the S-H bond in $H_2S$ is only weakly polar, and $H_2S$ cannot form hydrogen bonds; only weak van der Waals forces act between $H_2S$ molecules, giving it a much lower boiling point (so it is a gas at room temperature). (b) Phosphorus forms $PCl_5$ while Nitrogen does not form $NCl_5$: Phosphorus (period 3) has vacant 3d orbitals available in its valence shell, which allows it to expand its octet and use $sp^3d$ hybridisation to form five bonds, giving $PCl_5$. Nitrogen (period 2) has no d-orbitals available (only 2s and 2p), so it cannot expand its octet beyond eight electrons and has a maximum covalency of 4; hence $NCl_5$ does not exist. …

Down group 17: IE decreases; F only $-1$, others $-1$ to $+7$; electron affinity order Cl > F > Br > I.

Trends in the halogen family (group 17).

1. Ionisation Energy. It decreases down the group (F > Cl > Br > I). As we go down, atomic size increases and the outermost electrons are farther from and less strongly held by the nucleus, so less energy is needed to remove an electron. (Halogens have the highest ionisation energies in their periods because of high nuclear charge and small size.)
2. Oxidation States. Fluorine, being the most electronegative element, shows only the $-1$ oxidation state. The other halogens (Cl, Br, I) show $-1$ as well as several positive oxidation states ($+1, +3, +5, +7$) because they have vacant $d$-orbitals that allow expansion of the octet when combined with more electronegative elements (like O or F). …