Question bank — Ionization energy — first, second, …; trends and anomalies (e.g. B - Be)

False. $I{E}_2>I{E}_1$ is only guaranteed for the same element (same $Z$, fewer electrons). Across elements, e.g. $I{E}_2$ of Li (75.6 eV, core-breaking) is huge, but $I{E}_1$ of He (24.6 eV) exceeds $I{E}_1$ of Li — you must compare within one species.

True. You are pulling a negative electron away from a net-positive core, which always requires supplying energy against Coulomb attraction — the process is endothermic by definition.

False. $Z$ actually increases down a group. IE falls because a new shell raises $n$ (bigger radius) and inner shells shield strongly, so $Z_{\text{eff}}$ barely grows while distance grows a lot — see Atomic Radius Trends.

False. Electrons in the same shell shield each other poorly (Slater ≈ 0.35 each vs 0.85 for inner). So each added proton wins the tug-of-war, $Z_{\text{eff}}$ rises, and IE generally climbs.

False. Boron is perfectly stable; the dip exists because beryllium is unusually stable (filled $2s^2$). We compare the ease of removal, and B's lone $2p$ electron is higher-energy and shielded by $2s^2$, so it leaves more easily.

False. Shielding is nearly identical for adjacent same-shell atoms. The dip is from electron–electron pairing repulsion in O's doubly-occupied $2p$ orbital plus the exchange stability of N's half-filled $2p^3$.

False. Jumps are gradual and small within a shell; a sudden large jump appears only when you break into a filled inner (noble-gas) core, which reveals how many valence electrons the atom had.

True. A noble gas has a filled valence shell, maximal $Z_{\text{eff}}$ for that period, and no easily-lost electron, so pulling one out costs the most energy of any element in that row.

The conclusion is right but the rule is wrong. F > O does hold, yet the same "more protons" logic falsely predicts B > Be and N > O, both of which fail. Protons are the default cause, but orbital energy and pairing can override it.

Wrong. Na is $[\text{Ne}]3s^1$; $I{E}_1$ removes the lone $3s$ electron, but $I{E}_2$ must break the stable $[\text{Ne}]$ core, so $I{E}_2$ is roughly nine times larger — a core-breaking cliff, not a small step.

Wrong — they are different processes. IE removes an electron ($X\to X^{+}+e^{-}$); Electron Affinity adds one ($X+e^{-}\to X^{-}$). They involve different species and different orbital energies, so their magnitudes are unrelated in general.

Close but overstated. Francium sits below Cs, yet relativistic effects and measurement make Cs's $I{E}_1$ (3.89 eV) among the lowest; the clean "down a group falls" rule bends at the very bottom. Never assert "lowest of all" from the trend alone.

Wrong. It is a hydrogen-like approximation; real multi-electron orbitals aren't hydrogenic and Slater's $S$ is an estimate. For Li it gives ≈ 5.75 eV vs 5.39 eV measured — right ballpark, not exact.

Backwards. Higher orbital energy means the electron is less tightly bound — closer to escaping. That, plus $2s^2$ shielding, is exactly why B's electron leaves more easily than Be's.

In gas the atoms are isolated, so no lattice energy, bonding, or neighbour interactions contaminate the number — you measure the pure atom-to-cation energy.

N's $2p^3$ is half-filled: one electron per orbital, maximal exchange stabilisation and no same-orbital repulsion. O's $2p^4$ has one paired orbital whose electrons repel, so removal relieves that repulsion and is easier — the pairing penalty beats the extra proton.

Each removal leaves the same $Z$ but one fewer electron, so $Z_{\text{eff}}$ on the survivors rises and mutual repulsion drops — Coulomb's law then grips the next electron harder (see Coulomb's Law).

The filled $s^2$ is a stable, low-energy arrangement; the next element must place an electron in the higher, $s$-shielded $p$ orbital, which is easier to remove — so IE dips at the $p^1$ element.

Both come from rising $Z_{\text{eff}}$: a stronger effective pull draws the same-shell electrons inward (smaller radius, see Atomic Radius Trends) and makes any one of them harder to remove (higher IE). One cause, two visible effects.

The jump appears when you finally strip into the noble-gas core. If the cliff comes after removing $g$ electrons, the atom had exactly $g$ valence electrons, which equals its group number for main-group elements.

Both reflect how tightly an atom holds electrons: a strong grip (high $Z_{\text{eff}}$, small radius) makes an atom reluctant to lose electrons (high IE) and eager to attract them in bonds (high Electronegativity).

He ($1s^2$, $Z_{\text{eff}}\approx 1.7$, $n=1$) is far larger than Li ($2s^1$, $Z_{\text{eff}}\approx 1.3$, $n=2$). The smaller $n$ and higher $Z_{\text{eff}}$ both push He's IE up — distance dominates the comparison.

IE rises only gently across a transition series: added $d$ electrons shield the outer $s$ electrons fairly well, so $Z_{\text{eff}}$ on the valence $s$ grows slowly — the trend is flatter than in a main-group period.

No. Since removing the first electron always raises $Z_{\text{eff}}$ on the remainder and lowers repulsion, the same species always shows $I{E}_1<I{E}_2<I{E}_3\dots$; a decrease would violate Coulomb's law for a more-positive core.

With one electron there is nothing to shield, so $S=0$, $Z_{\text{eff}}=Z=1$, $n=1$, giving $IE=13.6\times 1^2/1^2=13.6$ eV — which is the exact experimental value, because hydrogen is genuinely hydrogen-like.

IE crashes from Ne (21.6 eV) to Na (5.14 eV): Na starts a new shell ($n=3$) with a lone, well-shielded $3s^1$ electron. Starting a fresh outer shell is the single largest downward step in the whole IE curve.

The one with smaller $n$. Since $IE\propto Z_{\text{eff}}^2/n^2$, a smaller shell places the electron closer to the nucleus, raising $|E_n|$ and thus IE — distance is squared in the denominator.