Q.The product obtained when silica reacts with hydrogen fluoride is -
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What Makes a Halogen a Halogen?
Imagine you are a fluorine atom. You have seven electrons in your outermost shell. You are one electron short of a full, stable octet — that perfect, noble-gas configuration that every atom craves. That single missing electron makes you intensely hungry. You will grab an electron from almost anything that moves.
That hunger is the entire story of the halogens.
The Intuition: The One-Electron Gap
All elements in Group 17 — fluorine, chlorine, bromine, iodine, and astatine — share the same electronic signature: their outermost shell has seven electrons. The nearest noble gas has eight. So each halogen is exactly one electron short of stability.
This is not a small difference. It is the most powerful chemical drive in the periodic table. It means:
- Halogens are the most reactive non-metals in their respective periods.
- They exist naturally as diatomic molecules (F2, Cl2, Br2, I2) because two halogen atoms can share one electron each, giving both a pseudo-octet.
- When they react with metals, they gain one electron to become halide ions (F−, Cl−, Br−, I−), achieving the noble-gas configuration. The resulting compounds are called salts — sodium chloride, potassium iodide, calcium fluoride.
The name "halogen" comes from Greek: hals (salt) + gen (producer). Halogens literally produce salts when they react with metals.
The Precise Statement
Group 17 elements (fluorine, chlorine, bromine, iodine, astatine) are highly reactive non-metals with the general electronic configuration ns2np5 in their valence shell. Their characteristic oxidation state is −1, achieved by gaining one electron to form a halide ion. In their elemental form, they exist as diatomic molecules (X2). Their reactivity decreases down the group: fluorine is the most reactive, iodine the least.
The Trend Down the Group
| Property | Fluorine | Chlorine | Bromine | Iodine |
|---|---|---|---|---|
| Physical state at room temp | Pale yellow gas | Greenish-yellow gas | Reddish-brown liquid | Violet-black solid |
| Bond dissociation energy (kJ mol−1) | 158.8 | 242.6 | 192.8 | 151.1 |
| Electron gain enthalpy (kJ mol−1) | −333 | −349 | −325 | −295 |
| Electronegativity (Pauling) | 4.0 | 3.2 | 3.0 | 2.7 |
Fluorine is an exception to the trend in bond dissociation energy. Its F–F bond is unexpectedly weak because of the small size of fluorine atoms — the lone pairs on each atom repel each other strongly, making the bond easier to break. This is why fluorine is so explosively reactive.
Why −1 and Not +1 or +7?
You might ask: if halogens have seven valence electrons, could they not lose seven electrons and show a +7 oxidation state? In principle, yes — chlorine, bromine, and iodine do show positive oxidation states (+1, +3, +5, +7) when bonded to more electronegative elements like oxygen. But the characteristic oxidation state, the one that defines their salt-forming behaviour, is −1.
The reason is simple: gaining one electron is energetically far cheaper than losing seven. The energy required to remove seven electrons is enormous; the energy released when one electron is added is substantial. So whenever a halogen meets a metal, the metal loses electrons and the halogen gains one. That is the fundamental exchange. …
Silica is attacked by hydrogen fluoride; with excess HF the product is stable hexafluorosilicic acid: SiO₂ + 6HF → H₂SiF₆ + 2H₂O (SiF₄ forms first, then adds 2HF). …
SiO₂ + 6HF → H₂SiF₆ + 2H₂O.
HF is the one acid that attacks glass/silica. It first gives silicon tetrafluoride, SiO₂ + 4HF → SiF₄ + 2H₂O, and with excess HF this further reacts, SiF₄ + 2HF …
- CBSE 2022Set HE2181 markQ.Fill in the blank: The formula of Fluorspar is ______.
›Reveal solutionSolution
Fluorspar is calcium fluoride, CaF2, the chief natural ore/source of fluorine.
Fluorspar (also called fluorite) is a naturally occurring mineral with the formula CaF2. It is industrially important as the principal source of fluorine: treating fluorspar with concentrated sulphuric acid releases hydrogen fluoride gas,
CaF2 + H2SO4 -> CaSO4 + 2HF …
- CBSE 2020Set 56/1/11 markMCQQ.Assertion (A) : F – F bond in F2 molecule is weak. Reason (R) : F atom is small in size. (A) Both Assertion (A) and Reason (R) are correct statements, and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are correct statements, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is correct, but Reason (R) is incorrect statement. (D) Assertion (A) is incorrect, but Reason (R) is correct statement.
›Reveal solutionSolution
The F–F bond in F2 is weak mainly due to lone pair–lone pair repulsion between the small fluorine atoms, not simply because the atom is small. Both statements are correct, but the reason does not correctly explain the assertion.
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Understanding the Assertion (A): The F–F bond in F2 is indeed weak. Its bond dissociation energy is only about 159 kJ/mol, which is much lower than the Cl–Cl bond (243 kJ/mol) or the Br–Br bond (193 kJ/mol). This is a well-known anomaly in the halogen family.
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Understanding the Reason (R): Fluorine is the smallest halogen atom. Its atomic radius is about 71 pm, compared to chlorine (99 pm), bromine (114 pm), and iodine (133 pm). So the statement "F atom is small in size" is factually correct.
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Why the bond is weak — the real explanation: The weakness of the F–F bond arises from lone pair–lone pair repulsion. Each fluorine atom has three lone pairs of electrons. Because the atoms are so small, these lone pairs are forced very close together when the bond forms. The resulting electrostatic repulsion between the non-bonding electron clouds partially cancels the bonding attraction, making the bond weaker than expected.
Watch outA common mistake is to think that a smaller atom always forms a stronger bond. In fact, bond strength depends on a balance of factors: orbital overlap (which improves with smaller size) and electron–electron repulsion (which worsens with smaller size). For fluorine, repulsion wins. …
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- CBSE 2020Set HE8221 markQ.Write True or False: One halogen atom combined with another halogen and form Inter-halogen compound.
›Reveal solutionSolution
Interhalogen compounds (e.g. ClF, BrF₃, IF₅, ICl) form when two different halogens combine, the larger/heavier halogen usually being the central atom.
The statement is True.
When two different halogen atoms combine (rather than a halogen combining with a non-halogen element), the product is called an interhalogen compound. General formulas are XX′, XX′3, XX′5, XX′7, where X is the larger, less electronegative halogen (central atom) and X′ is the smaller, more electronegative halogen (mostly …
- CBSE 2019Set HE1 markQ.Match the following. Column A term: bleaching powder. Column B options to match from:(a) buna-rubber(b) thermosetting plastic(c) Isoprene(d) CaOCl2(e) lyophilic colloid(f) NaCl
›Reveal solutionSolution
Bleaching powder is calcium oxychloride, CaOCl₂, made by passing chlorine gas over slaked lime, Ca(OH)₂.
Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O
It is used as a bleaching agent (for cotton, linen, wood pulp), a disinfectant/oxidising agent, and for water treatment, because it slowly liberates chlorine/hypochlorite, which is the active bleaching/oxidising species.
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- CBSE 2018Set ANNUAL1 markMCQQ.The product obtained when silica reacts with hydrogen fluoride is -(a) SiF4(b) H2SiF6(c) H2SiF4(d) H2SiF3
›Reveal solutionSolution
SiO₂ + 6HF → H₂SiF₆ + 2H₂O.
HF is the one acid that attacks glass/silica. It first gives silicon tetrafluoride, SiO₂ + 4HF → SiF₄ + 2H₂O, and with excess HF this further reacts, SiF₄ + 2HF …
- CBSE 2016Set ANNUAL1 markQ.Identify X from the following reactions: Br2+X−oxdnBr−+X2 and X2 cannot oxidise other halide ions but it can oxidise KClO3 as X2+2KClO3Δlittle conc. HNO32KXO3+Cl2
›Reveal solutionSolution
X is iodine, identified from its relative oxidising power among the halogens.
Oxidising power of halogens follows F2>Cl2>Br2>I2. Since Br2 oxidises X− to X2, X2 must be a weaker oxidising agent than Br2, so X = Iodine:
Br2+2I−→2Br−+I2 …
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