Exercises — Group 17 (Halogens) — properties, oxidizing power; HX strengths; interhalogens; pseudohalogens

The halogens are F, Cl, Br, I, At (and the synthetic Ts, tennessine, if you want the modern list). Valence configuration: $n{s}^{2}n{p}^5$ — that is two electrons in the s-subshell and five in the p-subshell, so $2+5=7$ valence electrons, i.e. one short of the noble-gas eight. Radioactive: At (astatine). Why it matters: "one electron short" is the seed of every trend — the atom's whole chemistry is "grab one electron."

Most negative ${\Delta }_{eg}H$: Cl, not F. Most electronegative: F. They are different atoms. See Periodic Trends — Electronegativity & Ionization Enthalpy. Why: electronegativity measures pull on a bonded pair; ${\Delta }_{eg}H$ measures energy released when a lone, gaseous atom captures an electron. F is so tiny that the incoming electron is squeezed into a crowded $2p$ shell → electron–electron repulsion partly cancels the energy gain.

Proposed reaction: $C{l}_2+2B{r}^{-}\to 2C{l}^{-}+B{r}_2$. Here $C{l}_2$ is reduced (cathode, $+1.36$) and $B{r}^{-}$ is oxidised (anode; we flip its reduction, $+1.07$). $\Delta E^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }=1.36-1.07=+0.29\text{ V}$ Positive ⇒ spontaneous. Yes, chlorine displaces bromide. Why this rule: a positive cell potential means $\Delta G=-nF{E}^{\circ }<0$. See Electrochemistry — Standard Reduction Potentials.

Central Br has 7 valence electrons. Five Br–F bonds consume 5 → $7-5=2$ electrons left = 1 lone pair. Electron domains = 5 bond pairs + 1 lone pair = 6 domains ⇒ hybridisation ==$s{p}^3{d}^2$==, octahedral electron geometry. One position is a lone pair ⇒ molecular shape = square pyramidal. See VSEPR Theory & Hybridisation. Look at figure below for the lone-pair-pushes-atoms-down picture.

Acid strength of $HX$ = ease of breaking $H-X$ to release $H^{+}$. The dominant factor is bond strength (not electronegativity). Weaker bond ⇒ easier ionisation ⇒ stronger acid. Since bond enthalpy falls $F>Cl>Br>I$, acidity rises the opposite way: $HF<HCl<HBr<HI$ Check the gaps: $565\to 297$ kJ mol⁻¹ is a 268 kJ mol⁻¹ drop — a huge energetic advantage for HI. HF is a weak acid; the other three are strong.

Split $\frac{1}{2}{X}_2+e^{-}\to X^{-}(aq)$ into three energy steps: $\frac{1}{2}{X}_2\stackrel{\frac{1}{2}{\Delta }_{diss}H}{\to }X(g)\stackrel{{\Delta }_{eg}H}{\to }{X}^{-}(g)\stackrel{{\Delta }_{hyd}H}{\to }{X}^{-}(aq)$ The more energy released overall, the larger $E^{\circ }$, the stronger the oxidiser. For F, two terms overpower its modest ${\Delta }_{eg}H$:

1. $\frac{1}{2}{\Delta }_{diss}H(F_2)$ is small — the F–F bond is weak (lone-pair repulsion between two tiny atoms), so little energy is spent breaking it.
2. ${\Delta }_{hyd}H(F^{-})$ is very large (magnitude) — the tiny $F^{-}$ ion binds water strongly. Together they win, so $E^{\circ }(F_2/F^{-})=+2.87\text{ V}$, the highest of all. See Born–Haber Cycle & Hydration Enthalpy.

$B{r}_2$ can oxidise a halide only if it is a stronger oxidiser than that halide's parent $X_2$ (i.e. $\Delta E^{\circ }>0$).

vs $C{l}^{-}$: $B{r}_2+2C{l}^{-}\to 2B{r}^{-}+C{l}_2$, $\Delta E^{\circ }=1.07-1.36=-0.29\text{ V}<0$ → no reaction. vs $I^{-}$: $B{r}_2+2I^{-}\to 2B{r}^{-}+I_2$, $\Delta E^{\circ }=1.07-0.54=+0.53\text{ V}>0$ → spontaneous, iodide oxidised. vs $B{r}^{-}$: same species, no change.

Conclusion: only I⁻ is oxidised to I₂; Cl⁻ is untouched. This is the classic "a halogen displaces only the halides below it."

(a) Why $I{F}_7$ but not $Cl{F}_7$: the central atom must be large enough to fit seven small F atoms around it and must offer enough space/orbitals ($s{p}^3{d}^3$, needing accessible d-orbitals). Iodine is large (period 5); chlorine (period 3) is too small — seven F atoms would crowd/repel intolerably and Cl lacks the low-lying d-orbitals I uses. Hence the highest fluoride of Cl is $Cl{F}_5$, while I reaches $I{F}_7$ (pentagonal bipyramidal). (b) Why more reactive: the $X-X^{\prime }$ bond between two different halogens is weaker than a symmetric $X-X$ bond (unequal size/electronegativity → poorer overlap, polar strain). A weaker bond breaks more easily ⇒ interhalogens are more reactive than the parent $X_2$ (except vs $F_2$, which is uniquely aggressive).

Pseudohalogens disproportionate in alkali exactly like $C{l}_2$: $(CN{)}_2+2O{H}^{-}\to C{N}^{-}+OC{N}^{-}+H_{2}O$ Compare $C{l}_2+2O{H}^{-}\to C{l}^{-}+OC{l}^{-}+H_{2}O$ — same pattern. Oxidation state of C:

• In $(CN{)}_2$, treat each CN unit; average C oxidation state $=+3$ (with N $=-3$).
• In $C{N}^{-}$ (reduced product), C $=+2$.
• In $OC{N}^{-}$ (oxidised product), C $=+4$ (O $=-2$, N $=-3$: $x+(-2)+(-3)=-1\Rightarrow x=+4$). So C goes $+3\to +2$ (reduction) and $+3\to +4$ (oxidation) — a genuine disproportionation, confirming the "pseudo-halogen" label.

Thermodynamics of displacement: $F_2+2C{l}^{-}\to 2F^{-}+C{l}_2$, $\Delta E^{\circ }=2.87-1.36=+1.51\text{ V}>0$ → strongly spontaneous, so the displacement direction is correct. But the practical claim is wrong: $F_2$ is such a powerful oxidiser that it oxidises water itself before/alongside any halide: $2F_2+2H_{2}O\to 4HF+O_2$ ($E^{\circ }(O_2/H_{2}O)=+1.23$ V, so $\Delta E^{\circ }=2.87-1.23=+1.64\text{ V}>0$). Hence you cannot get a "clean" $F_2/C{l}^{-}$ displacement in water — the solvent is consumed. This is exactly why fluorine chemistry is done in anhydrous conditions. Lesson: a favourable $\Delta E^{\circ }$ tells you a reaction can go, but with an oxidiser this strong you must check every oxidisable species present — including the solvent.

• Solved L1–L2 unaided → you own the facts.
• Solved through L3 → you can reason with trends.
• Solved L4 → you can synthesise cycles + geometry + potentials.
• Solved L5 → you spotted the solvent-oxidation twist — mastery.