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

This page is the "no surprises" drill for the Group 17 topic note. The parent note gave you the rules. Here we hit every kind of question those rules can generate — every sign of a cell potential, every degenerate "same element" trap, every shape family, and the exam twists that catch people out.

Before any example, we settle the one number we lean on again and again.

The scenario matrix

Every question this topic throws sits in one of these cells. The examples below are labelled with the cell they fill, and together they touch all of them.

#	Case class	What makes it tricky	Filled by
C1	Displacement, positive $Δ E^{\circ}$	reaction goes — say why	Ex 1
C2	Displacement, negative $Δ E^{\circ}$	reaction fails — beginners force it	Ex 2
C3	Multi-step / chain reasoning	who displaces whom, all four ions	Ex 3
C4	Degenerate / zero input	same halogen, $Δ E^{\circ} = 0$	Ex 4
C5	Acid strength ordering (HX)	bond enthalpy beats electronegativity	Ex 5
C6	Shape / geometry ( $X X_{n}^{'}$ )	count lone pairs, place them	Ex 6, Ex 7
C7	Limiting / anomaly case	why $F$ breaks the pattern	Ex 8
C8	Real-world word problem	translate a lab story into $E^{\circ}$	Ex 9
C9	Exam twist / pseudohalogen	analogy reasoning, no textbook number	Ex 10

Examples

C1 — displacement that WORKS (positive $Δ E^{\circ}$ )

Worked example

Q1. Chlorine gas is bubbled through a solution of potassium bromide (KBr). Does anything happen? Write the reaction and its $Δ E^{\circ}$ .

Forecast: Cl is above Br in the group — decide first whether the higher halogen or the lower one wins, before touching numbers.

Identify the trial reaction: $C l_{2} + 2 B r^{-} \to 2 C l^{-} + B r_{2}$ . Why this step? We always write the candidate as "incoming molecule steals the sitting anion's electrons."
Assign roles: oxidizer = $C l_{2}$ ( $A$ ), the ion being tested = $B r^{-}$ ( $B$ ).
Apply the rule: $Δ E^{\circ} = E^{\circ} (C l_{2}) - E^{\circ} (B r_{2}) = 1.36 - 1.07 = + 0.29 V$ . Why this step? A positive $Δ E^{\circ}$ is our green light for spontaneity.

Answer: $Δ E^{\circ} = + 0.29 V > 0$ → yes, $C l_{2}$ displaces $B r^{-}$ ; the solution turns orange-brown as $B r_{2}$ forms.

Verify: Sanity check against the mnemonic "F Cl Br I" — a higher halogen should always displace a lower halide, and it did. Units: volts subtracted from volts give volts. ✓

C2 — displacement that FAILS (negative $Δ E^{\circ}$ )

Worked example

Q2. Iodine crystals are stirred into a chloride solution (NaCl). Will $I_{2}$ turn $C l^{-}$ into $C l_{2}$ ?

Forecast: I is below Cl. Guess the sign of $Δ E^{\circ}$ before computing.

Trial: $I_{2} + 2 C l^{-} \to 2 I^{-} + C l_{2}$ .
$Δ E^{\circ} = E^{\circ} (I_{2}) - E^{\circ} (C l_{2}) = 0.54 - 1.36 = - 0.82 V$ . Why this step? We test the forward reaction the question proposes, not the reverse.
$- 0.82 < 0$ ⇒ the forward reaction is non-spontaneous. Why this step? Negative cell potential = energy would have to be supplied; nature won't do it for free.

Answer: No. A weaker oxidizer ( $I_{2}$ ) cannot displace a stronger halide's parent. In fact the reverse ( $C l_{2} + 2 I^{-} \to 2 C l^{-} + I_{2}$ ) is the one that runs, with $Δ E^{\circ} = + 0.82 V$ .

Verify: The reverse gives exactly $+ 0.82 V$ (same magnitude, flipped sign) — internally consistent. ✓

C3 — chain reasoning over all four ions

C4 — the degenerate / zero case

C5 — acid strength ordering

Worked example

Q5. Rank $H F, H C l, H B r, H I$ by acid strength in water, and justify with a number.

Forecast: Electronegativity says F should win — but the note warned this is a trap. Predict who actually wins.

Recall the deciding factor is H–X bond dissociation enthalpy (kJ mol⁻¹): $H - F = 565, H - C l = 431, H - B r = 364, H - I = 297$ . Why this step? Acid strength = ease of $H - X$ snapping to release $H^{+}$ ; a weaker bond snaps more easily. See Periodic Trends — Electronegativity & Ionization Enthalpy for why the bond weakens down the group.
Lowest bond enthalpy ⇒ easiest to ionize ⇒ strongest acid: order the enthalpies ascending → $297 < 364 < 431 < 565$ .
Map back to acids: $H I > H B r > H C l > H F$ .

Answer: $H I > H B r > H C l > H F$ . HF is the weakest (a weak acid), despite F being most electronegative.

Verify: Bond enthalpy of $H I$ (297) is lowest → predicted strongest acid → HI. The strength order is exactly the reverse of the bond-enthalpy order. ✓

C6 — geometry: T-shaped interhalogen

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

Worked example

Q6. Predict the shape of $C l F_{3}$ using VSEPR Theory & Hybridisation.

Forecast: 3 fluorines — you might guess flat triangle. Count the electrons before trusting that.

Central Cl valence electrons = 7. Three $C l - F$ bonds use 3 electrons, leaving $7 - 3 = 4$ electrons = 2 lone pairs. Why this step? Lone pairs occupy space and bend the shape; we must count them.
Total electron domains = 3 bonding pairs + 2 lone pairs = 5 ⇒ hybridisation $s p^{3} d$ , base geometry trigonal bipyramidal.
Place both lone pairs in the roomy equatorial positions (look at the figure — the two mint lobes sit in the wide belt). Why this step? Equatorial lone pairs suffer only two 90° neighbours instead of three, minimising repulsion.

Answer: Bent T-shape (a slightly-squeezed "T"), bond angles a touch under $9 0^{\circ}$ .

Verify: Domain count $3 + 2 = 5$ matches $s p^{3} d$ ; with 2 lone pairs the AX₃E₂ class is textbook T-shaped. ✓

C7 — geometry: bigger central atom, more partners

Worked example

Q7. Predict the shapes of $I F_{5}$ and $I F_{7}$ , and explain why iodine (not fluorine) is central.

Forecast: Guess how many lone pairs each has before counting.

$I F_{5}$ : I has 7 electrons; 5 $I - F$ bonds use 5, leaving 2 = 1 lone pair. Domains = $5 + 1 = 6$ ⇒ $s p^{3} d^{2}$ , octahedral base.
One lone pair on an octahedron pushes the five bonds down → square pyramidal (see left panel).
$I F_{7}$ : 7 bonds use all 7 electrons → 0 lone pairs. Domains = 7 ⇒ $s p^{3} d^{3}$ ⇒ pentagonal bipyramidal (right panel). Why this step? No lone pair means the base geometry is the final shape.
Why I central: the large iodine atom has the surface area to seat 5–7 small F atoms; a small F atom simply cannot host that many neighbours.

Answer: $I F_{5}$ = square pyramidal; $I F_{7}$ = pentagonal bipyramidal.

Verify: $I F_{5}$ : $5 bp + 1 lp = 6$ domains ✓; $I F_{7}$ : $7 bp + 0 lp = 7$ domains ✓. Lone-pair counts consistent with 7 central valence electrons.

C8 — the limiting / anomaly case

Worked example

Q8. Both these facts are "F breaks the trend." Explain each with the right energy term. (a) $F_{2}$ is the strongest oxidizer even though Cl has the more negative electron gain enthalpy. (b) The $F - F$ bond enthalpy (155) is lower than $C l - C l$ (242) and even $B r - B r$ (193) kJ mol⁻¹.

Forecast: Which single small-size fact drives both anomalies?

(a) Break the reduction into a Born–Haber Cycle & Hydration Enthalpy chain: bond breaking $\frac{1}{2} Δ_{d i ss} H$ , then $Δ_{e g} H$ , then hydration $Δ_{h y d} H$ . For F: low $Δ_{d i ss} H$ (weak F–F) + very large (negative) $Δ_{h y d} H (F^{-})$ (tiny ion, tight water shell). Why this step? These two terms overpower F's slightly poorer $Δ_{e g} H$ , so $F_{2}$ still wins overall.
(b) The F atom is so small that the three lone pairs on each F sit close together, and the two atoms' lone pairs repel strongly, weakening the shared bond.

Answer: Both anomalies trace to F's tiny size — small ion → huge hydration (makes it strongest oxidizer); small atom → crowded lone-pair repulsion (makes $F - F$ weak).

Verify: $F - F$ (155) is indeed less than both $C l - C l$ (242) and $B r - B r$ (193) — the ordering $155 < 193 < 242$ confirms F is the odd one out, not I. ✓

C9 — real-world word problem

Worked example

Q9. A student wants to make bromine water in the lab. They only have KBr solution and a cylinder of chlorine, plus a cylinder of iodine vapour. Which gas should they bubble in, and why can't the other one work?

Forecast: One of these gases is a stronger oxidizer than $B r_{2}$ , one is weaker. Pick before computing.

Test chlorine: $C l_{2} + 2 B r^{-} \to 2 C l^{-} + B r_{2}$ , $Δ E^{\circ} = 1.36 - 1.07 = + 0.29 V > 0$ → works.
Test iodine: $I_{2} + 2 B r^{-} \to 2 I^{-} + B r_{2}$ , $Δ E^{\circ} = 0.54 - 1.07 = - 0.53 V < 0$ → fails. Why this step? Only a halogen above Br can liberate $B r_{2}$ from bromide.

Answer: Bubble in chlorine ( $+ 0.29 V$ , spontaneous). Iodine cannot ( $- 0.53 V$ ) — it is a weaker oxidizer than bromine.

Verify: Both signs agree with position: Cl above Br → positive; I below Br → negative. Magnitudes are the plain differences of the tabulated $E^{\circ}$ . ✓

C10 — exam twist: pseudohalogen analogy

Worked example

Q10. Cyanogen $(C N)_{2}$ is called a pseudohalogen. Predict its reaction with hot NaOH and name the halogen reaction it mimics. There is no $E^{\circ}$ to look up — reason by analogy.

Forecast: What does $C l_{2}$ do with alkali? Copy that skeleton.

Recall the halogen benchmark: $C l_{2} + 2 O H^{-} \to C l^{-} + O C l^{-} + H_{2} O$ (disproportionation — one Cl goes down to $- 1$ , the other up to $+ 1$ ). Why this step? A pseudohalogen is defined by mimicking $X_{2}$ chemistry, so we transplant the pattern.
Replace $C l$ with the pseudohalide unit $C N$ / its oxo-partner $O C N$ : $(C N)_{2} + 2 O H^{-} \to C N^{-} + O C N^{-} + H_{2} O$

Answer: $(C N)_{2}$ disproportionates in alkali to cyanide $C N^{-}$ and cyanate $O C N^{-}$ — mirroring $C l_{2}$ 's alkaline disproportionation.

Verify: Atom balance both sides: left $2 C, 2 N, 2 O (from 2 O H), 2 H$ ; right $C N^{-} (1 C, 1 N) + O C N^{-} (1 O, 1 C, 1 N) + H_{2} O (2 H, 1 O)$ = $2 C, 2 N, 2 O, 2 H$ . Balanced. ✓

Recall

Recall Cover-the-answers self-test

$Δ E^{\circ}$ for $C l_{2} + 2 B r^{-}$ ? ::: $+ 0.29$ V, spontaneous Does $I_{2}$ oxidize $C l^{-}$ ? ::: No, $Δ E^{\circ} = - 0.82$ V Which ions does $B r_{2}$ oxidize in an F,Cl,Br,I mix? ::: only $I^{-}$ $Δ E^{\circ}$ for a halogen reacting with its own halide? ::: exactly $0$ V Strongest HX acid and why? ::: HI — lowest H–X bond enthalpy (297 kJ/mol) Shape of $C l F_{3}$ / $I F_{5}$ / $I F_{7}$ ? ::: T-shaped / square pyramidal / pentagonal bipyramidal $(C N)_{2}$ with alkali gives? ::: $C N^{-} + O C N^{-}$ (disproportionation like $C l_{2}$ )

Connections

Parent topic (Hinglish)
Electrochemistry — Standard Reduction Potentials
Periodic Trends — Electronegativity & Ionization Enthalpy
Born–Haber Cycle & Hydration Enthalpy
VSEPR Theory & Hybridisation
Oxoacids of Halogens
Group 16 (Oxygen family)