JEE Mains Chemistry · The p-Block Elements

Groups 17 and 18: Halogens and Noble Gases

Fluorine, chlorine, bromine and iodine (ns² np⁵), their redox chemistry, oxoacids and interhalogen compounds, and the xenon fluorides of the noble gases, whose shapes follow from counting lone pairs.

All The p-Block Elements notes JEE Mains Chemistry strategy

Why this matters

Thirty-five PYQs, twenty-nine of them multiple choice, and four from 2026: the largest page of the chapter. Nine test halogen properties such as bond enthalpy, electron gain enthalpy and the boiling points of HX; thirteen are halogen redox, from disproportionation to the silver-halide and chlorine tests; thirteen cover interhalogen shapes, halogen oxoacids and oxides, and the xenon fluorides.

Concept 1 of 3: Halogen properties: bond enthalpy, electron gain enthalpy and hydrogen halides

Most halogen properties change smoothly down the group, but fluorine breaks two of them because it is so small. Its lone pairs crowd each other in F₂, so the F–F bond is weaker than Cl–Cl and even Br–Br. And an incoming electron is squeezed into fluorine's compact 2p shell, so chlorine, not fluorine, releases the most energy on gaining an electron. The hydrogen halides follow size, except that HF is lifted by hydrogen bonding: it boils highest, though HI still melts highest.

Definition

Bond enthalpy: $\mathrm{Cl_2 > Br_2 > F_2 > I_2}$ .
Electron gain enthalpy (most negative first): $\mathrm{Cl > F > Br > I}$ . Of covalent radius, ionic radius, ionisation enthalpy and electron gain enthalpy, only the last is irregular for F, Cl, Br, I.
Fluorine shows only −1, being the most electronegative element with no d orbitals.
HX boiling point: $\mathrm{HCl < HBr < HI < HF}$ . Melting point: $\mathrm{HCl < HBr < HF < HI}$ .
Covalent character of a metal halide rises with the metal's oxidation state: $\mathrm{SnCl_4 > SnCl_2}$ , $\mathrm{PbCl_4 > PbCl_2}$ , $\mathrm{UF_6 > UF_4}$ .
With oxygen: halogens form oxides, but most are unstable; they do not combine easily and directly with oxygen.

Halogen (hydride)	X–X bond enthalpy (kJ/mol)	Electron gain enthalpy (kJ/mol)	HX boiling point (K)	HX melting point (K)
F (HF)	158.8	$-333$	293	190 Weak F–F bond and a less negative electron gain enthalpy than Cl: both from fluorine's small size.
Cl (HCl)	242.6	$-349$	189	159
Br (HBr)	192.8	$-325$	206	185
I (HI)	151.1	$-296$	238	222

Chlorine leads in both bond enthalpy and electron gain enthalpy. HF boils highest; HI melts highest.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 2 · Q27Moderate

Example 1 · The p-Block Elements · Groups 17 and 18: Halogens and Noble Gases

Given below are two statements : Statement I : The increasing order of boiling point of hydrogen halides is

HCl < HBr < HI < HF

. Statement II : The increasing order of melting point of hydrogen halides is

HCl < HBr < HF < HI

. In the light of the above statements, choose the correct answer from the options given below :

F₂ does not have the highest bond enthalpy

Lone-pair repulsion between the two small fluorine atoms weakens the F–F bond. The order is

\mathrm{Cl_2 > Br_2 > F_2 > I_2}

, so chlorine is highest.

HF boils highest but does not melt highest

Hydrogen bonding lifts HF's boiling point above HI's. For melting points the larger dispersion forces in HI win, so HI melts highest:

\mathrm{HCl < HBr < HF < HI}

Chlorine, not fluorine, has the most negative electron gain enthalpy

The order of the magnitudes is

\mathrm{Cl > F > Br > I}

. A statement that it is F > Cl > Br > I is false.

Drill 8 more on halogen properties: bond enthalpy, electron gain enthalpy and hydrogen halides

Concept 2 of 3: Oxidising power and disproportionation of the halogens

A halogen's oxidising power is its hunger for an electron in water, measured by its reduction potential, and it falls from F₂ to I₂. So a halogen higher in the group pushes a lower one out of its salt: chlorine water releases bromine from bromide and iodine from iodide. Iodide sits at the bottom, so it is the best reducing agent: it alone reduces Cu²⁺ and Fe³⁺, and it alone is oxidised by air in acid. A species can disproportionate only if the halogen can go both up and down from where it is, so F₂ (which can only go down) and perhalates (which can only go down) never do.

Definition

Oxidising power: $\mathrm{F_2 > Cl_2 > Br_2 > I_2}$ ; $E^\circ$ = 2.87, 1.36, 1.09 and 0.54 V. This is the basis of the layer test.
Disproportionation: $\mathrm{Cl_2}$ , $\mathrm{Br_2}$ , $\mathrm{I_2}$ do; $\mathrm{F_2}$ does not. $\mathrm{ClO^-}$ , $\mathrm{ClO_2^-}$ , $\mathrm{ClO_3^-}$ can; $\mathrm{ClO_4^-}$ and $\mathrm{BrO_4^-}$ , at +7, cannot.
Chlorine with alkali: cold and dilute gives chloride and hypochlorite (1 : 1); hot and concentrated gives chloride and chlorate.
Iodide as reductant: $\mathrm{2Cu^{2+} + 4I^- \rightarrow Cu_2I_2 + I_2}$ ; $\mathrm{4I^- + 4H^+ + O_2 \rightarrow 2I_2 + 2H_2O}$ . $\mathrm{FeX_2}$ is known for all four halogens, $\mathrm{FeX_3}$ for F, Cl and Br only.
Iodine with concentrated nitric acid: $\mathrm{I_2 + 10HNO_3 \rightarrow 2HIO_3 + 10NO_2 + 4H_2O}$ .
Chloride test: $\mathrm{4NaCl + MnO_2 + 4H_2SO_4 \rightarrow MnCl_2 + 4NaHSO_4 + 2H_2O + Cl_2}$ , a greenish-yellow gas.
Silver halides: AgCl white, soluble in $\mathrm{NH_4OH}$ ; AgBr pale yellow, sparingly soluble; AgI yellow, insoluble.
Concentrated $\mathrm{H_2SO_4}$ gives coloured vapours with bromide ( $\mathrm{Br_2}$ ), iodide ( $\mathrm{I_2}$ ) and nitrate ( $\mathrm{NO_2}$ ), but only colourless HF with fluoride.

Chlorine with alkali

\mathrm{Cl_2 + 2OH^- \xrightarrow{\text{cold, dilute}} Cl^- + ClO^- + H_2O} \qquad \mathrm{3Cl_2 + 6OH^- \xrightarrow{\text{hot, conc.}} 5Cl^- + ClO_3^- + 3H_2O}

Worked example

0.30 mol of chlorine gas is passed into 1.0 L of cold 1.0 M NaOH. Find the concentrations of

\mathrm{Cl^-}

\mathrm{ClO^-}

and

\mathrm{OH^-}

after the reaction, taking the volume as constant.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 2 · Q33Moderate

Example 2 · The p-Block Elements · Groups 17 and 18: Halogens and Noble Gases

One mole of

Cl_{2}(\text{ }g)

was passed into 2 L of cold 2 M KOH solution. After the reaction, the concentrations of

Cl^{-},ClO^{-}

and

OH^{-}

are respectively (assume volume remains constant)

Cold dilute alkali gives hypochlorite, not chlorate

Chlorine with cold, dilute alkali gives

\mathrm{Cl^-}

and

\mathrm{ClO^-}

in a 1 : 1 ratio. Chlorate,

\mathrm{ClO_3^-}

, forms only with hot, concentrated alkali.

A +7 oxoanion cannot disproportionate

Disproportionation needs an intermediate state, so the halogen can be both oxidised and reduced. In

\mathrm{ClO_4^-}

\mathrm{BrO_4^-}

the halogen is already at +7, its highest state.

FeI₃ does not exist

Iron(III) is a strong enough oxidant to turn iodide into iodine, so it cannot sit beside three iodides.

\mathrm{FeI_2}

is known, but

\mathrm{FeX_3}

exists only for F, Cl and Br.

Drill 12 more on oxidising power and disproportionation of the halogens

Concept 3 of 3: Interhalogen shapes, halogen oxoacids and xenon fluorides

In an interhalogen XX′ₙ the central halogen X has seven valence electrons. It uses n of them to bond, so the other 7 − n form (7 − n)/2 lone pairs; n is always odd so that every electron pairs. Lone pairs plus bonds give the electron-pair geometry, and the lone pairs take the positions that leave the shape: 3 lone pairs make XX′ linear, 2 make XX′₃ T-shaped, 1 makes XX′₅ square pyramidal, and 0 makes IF₇ pentagonal bipyramidal. Xenon has eight valence electrons, so the same count gives (8 − n)/2 lone pairs for XeFₙ.

Definition

Shapes: XX′ linear; XX′₃ T-shaped ( $sp^3d$ ); XX′₅ square pyramidal ( $sp^3d^2$ : $\mathrm{ClF_5}$ , $\mathrm{BrF_5}$ , $\mathrm{IF_5}$ ); $\mathrm{IF_7}$ pentagonal bipyramidal ( $sp^3d^3$ ).
Bromine with excess fluorine gives $\mathrm{BrF_5}$ , an interhalogen with bromine at +5.
Halogen oxoacids: fluorine forms only $\mathrm{HOF}$ . Halic(V) acids $\mathrm{HXO_3}$ exist for Cl, Br and I. Cl=O bonds: $\mathrm{HClO_2}$ 1, $\mathrm{HClO_3}$ 2, $\mathrm{HClO_4}$ 3.
Halogen oxides: higher oxides are more stable than lower ones; stability I > Cl > Br. $\mathrm{O_2F_2}$ removes plutonium from spent fuel as $\mathrm{PuF_6}$ .
Noble gases are monatomic, held only by weak dispersion forces, so they have very LOW boiling points.
Xenon fluorides: $\mathrm{XeF_2}$ linear, $\mathrm{XeF_4}$ square planar, $\mathrm{XeF_6}$ distorted octahedral. $\mathrm{XeF_4 + SbF_5 \rightarrow [XeF_3]^+[SbF_6]^-}$ ; $\mathrm{6XeF_4 + 12H_2O \rightarrow 4Xe + 2XeO_3 + 24HF + 3O_2}$ .

Lone pairs on the central atom

\mathrm{XX'_n}:\ \text{lone pairs on X} = \dfrac{7-n}{2},\ n = 1, 3, 5, 7 \qquad \mathrm{XeF_n}:\ \text{lone pairs on Xe} = \dfrac{8-n}{2}

Worked example

Find the number of lone pairs on the central atom and the shape of

\mathrm{IF_3}

and of

\mathrm{BrCl}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 28 July 2022 · Q57Moderate

Example 3 · The p-Block Elements · Groups 17 and 18: Halogens and Noble Gases

The number of interhalogens from the following having square pyramidal structure is:

ClF_{3},IF_{7},BrF_{5},BrF_{3},I_{2}Cl_{6},IF_{5},ClF,ClF_{5}

XX′₅ is square pyramidal, not trigonal bipyramidal

Five bonds and one lone pair make six electron pairs. The pairs point to the corners of an octahedron, and with one corner held by the lone pair the atoms form a square pyramid.

An interhalogen is not a halate

\mathrm{BrF_5}

has bromine at +5, the same oxidation state as bromate,

\mathrm{BrO_3^-}

, but it is an interhalogen compound, not an oxoanion.

Noble gases have very low boiling points

Their atoms attract each other only by weak dispersion forces. That is why they liquefy only at very low temperatures, so any statement that they have high boiling points is false.

Drill 12 more on interhalogen shapes, halogen oxoacids and xenon fluorides

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Oxidising power and disproportionation of the halogens
Chlorine with alkali
$\mathrm{Cl_2 + 2OH^- \xrightarrow{\text{cold, dilute}} Cl^- + ClO^- + H_2O} \qquad \mathrm{3Cl_2 + 6OH^- \xrightarrow{\text{hot, conc.}} 5Cl^- + ClO_3^- + 3H_2O}$
Interhalogen shapes, halogen oxoacids and xenon fluorides
Lone pairs on the central atom
$\mathrm{XX'_n}:\ \text{lone pairs on X} = \dfrac{7-n}{2},\ n = 1, 3, 5, 7 \qquad \mathrm{XeF_n}:\ \text{lone pairs on Xe} = \dfrac{8-n}{2}$

Reference tables (1)

Halogen properties: bond enthalpy, electron gain enthalpy and hydrogen halides4 rows

Halogen (hydride)	X–X bond enthalpy (kJ/mol)	Electron gain enthalpy (kJ/mol)	HX boiling point (K)	HX melting point (K)
F (HF)	158.8	$-333$	293	190 Weak F–F bond and a less negative electron gain enthalpy than Cl: both from fluorine's small size.
Cl (HCl)	242.6	$-349$	189	159
Br (HBr)	192.8	$-325$	206	185
I (HI)	151.1	$-296$	238	222

Chlorine leads in both bond enthalpy and electron gain enthalpy. HF boils highest; HI melts highest.

Watch out for (9)

F₂ does not have the highest bond enthalpy
→ Halogen properties: bond enthalpy, electron gain enthalpy and hydrogen halides
HF boils highest but does not melt highest
→ Halogen properties: bond enthalpy, electron gain enthalpy and hydrogen halides
Chlorine, not fluorine, has the most negative electron gain enthalpy
→ Halogen properties: bond enthalpy, electron gain enthalpy and hydrogen halides
Cold dilute alkali gives hypochlorite, not chlorate
→ Oxidising power and disproportionation of the halogens
A +7 oxoanion cannot disproportionate
→ Oxidising power and disproportionation of the halogens
FeI₃ does not exist
→ Oxidising power and disproportionation of the halogens
XX′₅ is square pyramidal, not trigonal bipyramidal
→ Interhalogen shapes, halogen oxoacids and xenon fluorides
An interhalogen is not a halate
→ Interhalogen shapes, halogen oxoacids and xenon fluorides
Noble gases have very low boiling points
→ Interhalogen shapes, halogen oxoacids and xenon fluorides

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