Group 17 (Halogens) — properties, oxidizing power; HX strengths; interhalogens; pseudohalogens: Question bank

Definition

The four terms every trap on this page leans on — self-contained, no need to leave the page.

Oxidizing power ::: how strongly $X_{2}$ wants to grab electrons and turn into $2 X^{-}$ . We measure it by the standard reduction potential $E^{\circ}$ (in volts) of $X_{2} + 2 e^{-} \to 2 X^{-}$ ; a bigger $E^{\circ}$ means the molecule pulls electrons harder, i.e. a stronger oxidizer. See Electrochemistry — Standard Reduction Potentials.
Electron gain enthalpy ( $Δ_{e g} H$ ) ::: the energy change when a single isolated gaseous atom grabs one electron, $X (g) + e^{-} \to X^{-} (g)$ . More negative = more energy released = "hungrier" atom. It is about a lone atom in the gas phase, nothing about bonds or water.
Bond dissociation enthalpy (bond strength) ::: the energy needed to snap one mole of a bond apart into gaseous atoms, e.g. $H - X (g) \to H (g) + X (g)$ . A big number means a tough, hard-to-break bond. Contrast with $Δ_{e g} H$ : bond enthalpy is about pulling two atoms apart, $Δ_{e g} H$ is about feeding one atom an electron — do not mix them.
Hydration enthalpy ( $Δ_{h y d} H$ ) ::: the energy released when one mole of gaseous ions is surrounded by water molecules, $X^{-} (g) water X^{-} (a q)$ . Small ions pack water tightly and release a lot; it is measured as the enthalpy change of dissolving the bare ion. See Born–Haber Cycle & Hydration Enthalpy.

Three pictures to hold in mind

Before the traps, anchor the three ideas most items lean on.

(1) Why $F_{2}$ is the strongest oxidizer — the energy cycle. Figure s01: a left-to-right chain of boxes $\frac{1}{2} F_{2} (g) \to F (g) \to F^{-} (g) \to F^{-} (a q)$ ; a yellow arrow (atomization, $\frac{1}{2}$ bond dissociation enthalpy) is short because F–F is weak, a red arrow ( $Δ_{e g} H$ ) is modest, and a long green arrow (hydration enthalpy) dominates because $F^{-}$ is tiny. The caption reminds you that the small atomization cost plus huge hydration payback beat F's modest $Δ_{e g} H$ .

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

(2) The X–X bond-strength profile — F is anomalous. Figure s02: a bar chart of single X–X bond dissociation enthalpies (kJ/mol) for F–F (155), Cl–Cl (242), Br–Br (193), I–I (151); the F–F bar is annotated with a red arrow flagging it as anomalously weak due to lone-pair repulsion between the two small atoms.

(3) The shape of $C l F_{3}$ — 5 domains, lone pairs equatorial. Figure s03: a central blue Cl with three green F atoms (two axial up/down, one equatorial right) and two yellow arrows for equatorial lone pairs on the left; the resulting F–Cl–F axis is slightly bent below 90°, giving the T-shaped molecule. See VSEPR Theory & Hybridisation.

True or false — justify

The recurring trick here: a true-sounding reason attached to a claim, where the reason is the real error.

F is the most electronegative element, therefore F has the most negative electron gain enthalpy.

False. Electronegativity describes pull on a bonded pair;

Δ_{e g} H

describes adding an electron to an isolated atom. F's tiny 2p shell is so crowded that adding an electron costs extra repulsion, so Cl has the more negative

Δ_{e g} H

HF is the strongest of the HX acids because F pulls hardest on the H–F bonding pair.

False. In water, acid strength is governed by how easily H–X breaks, i.e. bond dissociation enthalpy. H–F (565 kJ/mol) is the strongest bond plus HF forms strong hydrogen bonds, so HF is the weakest HX acid (

p K_{a} \approx 3.2

, i.e. partially ionised) whereas HCl, HBr, HI all have

p K_{a} < 0

(essentially fully ionised).

F_{2}

is the strongest oxidizer, so F must also have the most negative electron gain enthalpy.

False.

F_{2}

wins on oxidizing power because of its weak F–F bond (easy to atomize) and very high hydration enthalpy of the small

F^{-}

ion — not because of

Δ_{e g} H

, where Cl actually wins.

Because

C l_{2}

can displace

B r^{-}

B r_{2}

can also displace

C l^{-}

False. Displacement runs one way only: the stronger oxidizer displaces the weaker halide. Since

E^{\circ} (C l_{2} / C l^{-}) = + 1.36 > E^{\circ} (B r_{2} / B r^{-}) = + 1.07

, only

C l_{2}

displaces

B r^{-}

; the reverse has

Δ E^{\circ} < 0

(so

Δ G^{\circ} > 0

) and is non-spontaneous.

Interhalogens are less reactive than the parent halogens because they contain two "satisfied" halogens.

False. The

X - X^{'}

bond is weaker than an

X - X

bond (mismatched sizes/EN), so interhalogens are generally more reactive than halogens — except

F_{2}

, which is still king.

X X_{n}^{'}

the smaller, more electronegative halogen sits at the centre.

False. The larger, less electronegative halogen is central because only a big atom has the surface room to hold several small partners (that's why only I/Br reach

n = 5, 7

, e.g.

I F_{7}

Down the group F → I, oxidizing power and atomic radius both increase.

False. Radius increases (new shells added), but oxidizing power decreases — a bigger ion is harder to reduce to

X^{-}

and has a smaller (less exothermic) hydration enthalpy. Linked to Periodic Trends — Electronegativity & Ionization Enthalpy.

The F–F bond, coming from the most electronegative atom, is the strongest single X–X bond.

False. F–F is anomalously weak: the two small F atoms sit so close that their lone pairs repel, weakening the bond. Order is Cl–Cl > Br–Br > F–F > I–I.

Pseudohalogens are called "pseudo" because they contain fake, non-reactive halogen atoms.

False. They contain no halogen atoms at all; they are called pseudohalogens because they mimic halogen chemistry — e.g.

(C N)_{2}

disproportionates in base just like

C l_{2}

, and

C N^{-}

behaves like

X^{-}

Spot the error

Each line has one planted mistake — name it and correct it.

C l_{2} + 2 B r^{-} \to 2 C l^{-} + B r_{2}

is non-spontaneous because

Δ E^{\circ} = 1.07 - 1.36 < 0

Error is using the wrong halves. Using cathode-minus-anode with

C l_{2}

reduced,

Δ E^{\circ} = E^{\circ} (C l_{2}) - E^{\circ} (B r_{2}) = 1.36 - 1.07 = + 0.29 V > 0

, so

Δ G^{\circ} = - n F Δ E^{\circ} < 0

— it is spontaneous.

"HF is a strong acid, only slightly weaker than HCl."

HF is a weak acid (

p K_{a} \approx 3.2

, only partially ionised in water); HCl, HBr, HI are all strong (

p K_{a} < 0

). The short strong H–F bond plus H-bonding keep the proton held.

C l F_{3}

is trigonal planar because it has three bonds."

It has 3 bond pairs and 2 lone pairs = 5 domains (

s p^{3} d

). Lone pairs occupy equatorial positions giving a T-shaped molecule, not trigonal planar. See VSEPR Theory & Hybridisation.

"Thermal stability of HX increases down the group: HF < HCl < HBr < HI."

Reversed. Stability follows bond strength:

H F > H C l > H B r > H I

; HI decomposes most easily.

"Astatine is the most abundant, stable halogen at the bottom, so it's the least reactive metal-like member."

At is radioactive and vanishingly rare, not abundant/stable. The premise is broken; you cannot base trends on a routinely observable At.

"Adding an electron to any halogen always releases energy, and it releases the most for the smallest atom F."

The "smallest = most released" part is wrong. Small size hurts here through electron–electron repulsion, so Cl (not F) releases the most.

Why questions

Answer with the mechanism, not just the fact.

Why does

F_{2}

oxidize more strongly than

C l_{2}

even though Cl has the more negative

Δ_{e g} H

The full reduction cycle also includes atomization (bond dissociation enthalpy) and hydration: F–F is weak (easy to break) and

F^{-}

is tiny (huge hydration enthalpy), and these two terms outweigh F's poorer

Δ_{e g} H

. See Born–Haber Cycle & Hydration Enthalpy.

Why can iodine form

I F_{7}

but chlorine cannot form

C l F_{7}

A large iodine atom has enough surface room to physically pack seven small F atoms around it, while the smaller Cl cannot. (The bonding is better described by 3-centre-4-electron interactions using p-orbitals than by literal

d

-orbital hybrids — the "

s p^{3} d^{3}

" label is only a bookkeeping shorthand.)

Why is HI a stronger acid than HF despite F being far more electronegative?

Acid strength in water tracks how easily H–X breaks and how well the ions are solvated. The long, weak H–I bond (297 kJ/mol) snaps to release

H^{+}

readily; the short, strong H–F bond (565 kJ/mol) plus strong HF hydrogen bonding trap the proton, so HF sits at

p K_{a} \approx 3.2

while HI is

p K_{a} \approx - 10

Why do interhalogens hydrolyse to give a halide plus an oxohalide rather than just decomposing?

The polar

X - X^{'}

bond is attacked by water: the more electronegative

X^{'}

ends up as halide (

X^{' -}

) while the central

X

combines with oxygen/hydroxide to form an oxohalide — mirrors halogen oxoacid chemistry, cf. Oxoacids of Halogens.

Why is electron gain enthalpy the "wrong lens" for ranking HX acidity but relevant elsewhere?

HX acidity is dominated by H–X bond dissociation enthalpy and solvation, not by isolated-atom electron affinity.

Δ_{e g} H

instead matters when we build an anion, e.g. inside the oxidizing-power reduction cycle.

Why are interhalogen compounds diamagnetic?

All electrons in these covalent, even-electron molecules are paired in bonds or lone pairs, so there are no unpaired spins to give paramagnetism.

Edge cases

The degenerate/limiting scenarios the topic quietly assumes you'll notice.

F_{2}

more reactive than every interhalogen?

Yes —

F_{2}

is the exception to "interhalogens are more reactive than halogens." Fluorine's weak F–F bond and extreme oxidizing power keep it more reactive than any

X X_{n}^{'}

What is the shape/hybridisation of a diatomic interhalogen like

C l F

(the

n = 1

limit)?

With only two atoms it is linear by definition — any two points define a straight line, so there is no bond angle to bend and lone pairs on the terminal atoms cannot distort a single axis. Lone-pair geometry only starts mattering from

X X_{3}^{'}

, where three or more bonds compete for space around one centre.

In the pseudohalogen

(C N)_{2}

, what plays the role of the halogen "

X_{2}

The whole

(C N)_{2}

molecule acts as

X_{2}

and

C N^{-}

acts as

X^{-}

; e.g.

(C N)_{2} + 2 O H^{-} \to C N^{-} + O C N^{-} + H_{2} O

mirrors

C l_{2} + 2 O H^{-}

. Other pseudohalide anions that behave like

X^{-}

include thiocyanate

S C N^{-}

, cyanate

O C N^{-}

, and azide

N_{3}^{-}

(which forms HN₃, an acid, and insoluble AgN₃, echoing HCl and AgCl).

For an interhalogen

X X_{n}^{'}

where the two halogens are similar in size (e.g. Cl and F), what limits

n

Similar sizes leave little room advantage, so

n

stays small (

C l F

C l F_{3}

); large-vs-small pairs (I with F) are needed to reach

n = 5, 7

At the boundary of the group, why is astatine chemistry mostly predicted rather than measured?

Every At isotope is radioactive with short half-lives, so bulk samples never accumulate; its "trends" are extrapolated from F→I, not observed directly.

What happens to oxidizing power in the limit of the largest halide,

I^{-}

I_{2} / I^{-}

has the smallest

E^{\circ}

(+0.54 V), so

I_{2}

is the weakest halogen oxidizer and

I^{-}

is the easiest halide to oxidize — the low-oxidizing limit of the trend.