(a) Account for the following: (i) Tendency to show -3 oxidation state decreases from N to Bi in group 15. (ii

Question

Q. (a) Account for the following:

(i) Tendency to show

-3

oxidation state decreases from N to Bi in group 15.

(ii) Acidic character increases from

H_2O

H_2Te

(iii)

F_2

is more reactive than

ClF_3

, whereas

ClF_3

is more reactive than

Cl_2

(b) Draw the structure of

(i)

XeF_2

(ii)

H_4P_2O_7

(OR)

(a) Give one example to show the anomalous reaction of fluorine.

(b) What is the structural difference between white phosphorus and red phosphorus?

XeF_6

reacts with NaF?

(d) Why is

H_2S

a better reducing agent than

H_2O

(e) Arrange the following acids in the increasing order of their acidic character: HF, HCl, HBr and HI

CBSECBSE Class XII Board 2019Subjective· 5mImportance★★★★★

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Part (a)Concept understanding — Group 15 Nitrogen Family

The Nitrogen Family – From Air to Metal

Think about the air you breathe. Nearly 78% of it is nitrogen gas — colourless, odourless, and so unreactive that it just sits there. Now imagine a grey, shiny solid that glows in the dark. That's phosphorus. And then there's arsenic, once used as a poison; antimony, used in fireproofing; and bismuth, a heavy metal with a rainbow oxide layer.

These five elements — nitrogen, phosphorus, arsenic, antimony, bismuth — make up Group 15 of the periodic table. They start as a gas, become a non-metal, then a metalloid, and finally a metal. That's the single most important pattern: metallic character increases as you go down the group.

The Common Thread: Five Valence Electrons

Every element in Group 15 has the electron configuration $ns^2 np^3$ in its outermost shell. Five valence electrons means they need three more to complete an octet — hence the common oxidation state of −3 (as in $\ce{NH3}$ , ammonia, or $\ce{PH3}$ , phosphine).

But because the $ns^2$ pair can also be lost or shared, you get +3 (e.g., $\ce{NCl3}$ , $\ce{PCl3}$ ) and +5 (e.g., $\ce{HNO3}$ , $\ce{PCl5}$ ). The +5 state becomes less stable as you go down — bismuth hardly ever shows +5 because its 6s electrons are too tightly held (the inert pair effect).

\text{General outer configuration: } ns^2 np^3

The Downward Trend: From Gas to Metal

Property	N	P	As	Sb	Bi
Nature	Non-metal	Non-metal	Metalloid	Metalloid	Metal
State at room temp	Gas	Solid (white/red)	Solid	Solid	Solid
Electronegativity	3.0	2.1	2.0	1.9	1.8
Ionisation enthalpy	High	↓	↓	↓	Lowest

Why does metallic character increase? As you go down, atomic size increases. The valence electrons are farther from the nucleus and more shielded. They become easier to lose (lower ionisation enthalpy) and harder to gain. So the elements start behaving less like electron-grabbers (non-metals) and more like electron-donors (metals). …

Part (b)Concept understanding — Group 17 Halogens

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 ( $\text{F}_2$ , $\text{Cl}_2$ , $\text{Br}_2$ , $\text{I}_2$ ) 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 ( $\text{F}^-$ , $\text{Cl}^-$ , $\text{Br}^-$ , $\text{I}^-$ ), achieving the noble-gas configuration. The resulting compounds are called salts — sodium chloride, potassium iodide, calcium fluoride.

Important

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 $ns^2 np^5$ 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 ( $\text{X}_2$ ). 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 ( $\text{kJ mol}^{-1}$ )	158.8	242.6	192.8	151.1
Electron gain enthalpy ( $\text{kJ mol}^{-1}$ )	−333	−349	−325	−295
Electronegativity (Pauling)	4.0	3.2	3.0	2.7

Note

Fluorine is an exception to the trend in bond dissociation energy. Its $\text{F–F}$ 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. …

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Figure — Lewis/structural electron-dot formula of pyrophosphoric acid H4P2O7. TWO phosphorus atoms each at th

Part (a): (i) inert-pair effect + falling electronegativity make −3 rarer down group 15; (ii) weaker H–E bonds and more stable conjugate bases raise acidity $H_2O\to H_2Te$ ; (iii) reactivity $F_2>ClF_3>Cl_2$ from weak F–F bond and polar labile Cl–F bonds; (b) XeF₂ linear, H₄P₂O₇ two bridged PO₄ tetrahedra. Part (b): F₂ oxidises water in the dark; white P = P₄, red P = polymeric; $XeF_6+NaF\to Na[XeF_7]$ ; H₂S is a better reductant than H₂O; acidity $HF<HCl<HBr<HI$ .

Part (a)

(i) Tendency to show −3 decreases from N to Bi. The −3 state needs the atom to gain three electrons. Down group 15 atomic size rises and electronegativity falls sharply (N 3.0 → Bi 2.0), so the pull on extra electrons weakens. The inert-pair effect (the $ns^2$ pair is reluctant to engage in bonding for heavy elements) makes As, Sb, Bi favour +3/+5 instead; Bi shows essentially no −3 chemistry.

(ii) Acidic character increases $\ce{H2O -> H2Te}$ . Acidity depends on how easily the H–E bond ionises. Bond strengths fall H–O (464) > H–S (347) > H–Se (276) > H–Te (238 kJ mol⁻¹) because orbital overlap worsens with the larger central atom. Weaker bonds release $\ce{H+}$ more readily, and the bigger conjugate base ( $\ce{Te^2-} > \ce{O^2-}$ ) disperses charge better. So $\ce{H2O}$ is neutral while acidity rises to $\ce{H2Te}$ .

(iii) $\ce{F2 > ClF3 > Cl2}$ . $\ce{F2}$ has the weakest halogen bond (lone-pair repulsion in the tiny molecule) and highest electronegativity → extremely reactive. The interhalogen $\ce{ClF3}$ has polar, weak, labile Cl–F bonds and a T-shape that readily delivers fluorine, so it exceeds $\ce{Cl2}$ , whose strong, non-polar Cl–Cl bond makes it least reactive of the three.

(b) Structures.

(i) $\ce{XeF2}$ — 5 electron pairs on Xe (2 bonds + 3 lone pairs) in a trigonal-bipyramidal set; the 3 lone pairs take equatorial sites, giving a linear molecule, F–Xe–F = 180°. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 13 on this concept.

CBSE 2022Set HE2181 mark
$Q. Fill in the blank: The formula of Fluorspar is ______.$
›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 …

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CBSE 2022Set ANNUAL1 mark
$PCl_5$
›Reveal solutionSolution

Forming a pentahalide requires expanding the central atom's octet using d-orbitals; phosphorus (n=3) has vacant 3d orbitals available for this, while nitrogen (n=2) has no d-orbitals at all in its valence shell, so it is structurally incapable of forming $NCl_5$ .

Octet expansion and $d$ -orbital availability

To form $PCl_5$ , the central phosphorus atom must bond to 5 chlorine atoms, requiring 5 hybrid orbitals ( $sp^3d$ , trigonal bipyramidal geometry) — i.e. an expanded octet of 10 electrons around P. This is possible because phosphorus is in period 3 and its valence shell ( $n=3$ ) includes empty $3d$ orbitals that are energetically accessible enough to participate in hybridisation, accommodating the extra electron pairs.

Why nitrogen cannot do the same

…

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CBSE 2020Set 56/1/11 markMCQ
$F_2$
›Reveal solutionSolution
The F–F bond in $F_2$ 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.

Understanding the Assertion (A): The F–F bond in $F_2$ 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.

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.

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 out

A 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 mark
$Q. Write True or False: One halogen atom combined with another halogen and form Inter-halogen compound.$
›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 …

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CBSE 2019Set HE1 mark
$Q. Fill in the blank: White phosphorus possesses ______ composition in vapour state.$
›Reveal solutionSolution

White phosphorus is molecular; in the vapour state it exists as P₄ molecules, with the four phosphorus atoms arranged at the corners of a regular tetrahedron.

White (yellow) phosphorus is the most reactive allotrope of phosphorus and consists of discrete P₄ tetrahedral units, both in the solid and in the vapour state. Each P atom is bonded to three other P atoms by single P–P bonds, with a P–P–P bond angle of about 60° (strained), which is one rea …

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CBSE 2019Set HE1 mark
$Q. Match the following. Column A term: bleaching powder. Column B options to match from:$
(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 2019Set ANNUAL1 markMCQ
$Q. Second period of periodic table contains elements :$
(a) 2
(b) 8
(c) 18
(d) 32
›Reveal solutionSolution

Second period = 8 elements (Li → Ne).

The second period corresponds to filling the n = 2 shell (2s and 2p orbitals), which hold a maximum of 8 electrons. The elements are Li, Be, B, C, N, O, F, Ne — a total o …

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CBSE 2019Set ANNUAL1 markMCQ
$Q. Group I elements of Periodic Table are known as -$
(a) Alkali metals
(b) Alkaline earth metals
(c) Inert gases
(d) None
›Reveal solutionSolution

Group 1 elements are the alkali metals.

The elements of Group 1 (Li, Na, K, Rb, Cs, Fr) are highly reactive metals whose hydroxides are strong bases (alkalis) that dissolve readily in water — hence 'alkali metals'. Group 2 elements a …

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CBSE 2018Set ANNUAL1 markMCQ
$Q. The molecule which has zero dipole moment is -$
(a) NF3
(b) BF3
(c) ClO2
(d) CH2Cl2
›Reveal solutionSolution
Only the symmetric trigonal-planar BF₃ has μ = 0.

BF₃: sp², trigonal planar, F at 120°; the three equal B→F dipoles cancel → μ = 0.

NF₃: sp³ pyramidal with a lone pair → μ ≠ 0. …
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CBSE 2018Set ANNUAL1 markMCQ
$Q. The product obtained when silica reacts with hydrogen fluoride is -$
(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 …

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CBSE 2018Set ANNUAL1 markMCQ
$Q. Which of the following has the most stable +2 oxidation state?$
(a) Sn
(b) Ag
(c) Fe
(d) Pb
›Reveal solutionSolution

Inert-pair effect makes Pb²⁺ the most stable +2 state.

For heavy p-block elements the ns² pair becomes reluctant to bond (inert-pair effect), stabilising the lower (+2) oxidation state. In group 14 this is strongest for lead, so Pb²⁺ is very stable while Pb⁴⁺ is oxidising. …

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CBSE 2017Set ANNUAL1 mark
$Q. What is laughing gas?$
›Reveal solutionSolution

N2O is called laughing gas because inhaling it produces a mild euphoric/intoxicating sensation, and it is used medically as a weak anaesthetic (e.g., in dentistry).

Laughing gas is the common name for nitrous oxide, $\text{N}_2\text{O}$ , an oxide of nitrogen in which nitrogen is in the +1 oxidation state.

…

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Q.(a) Account for the following:

Part (a)

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