p-Block Trends

8 MCQs9-step worked example
Source: NCERT p-Block ElementsPYQ coverage: NEET 2021, 2022, 2023, 2024Official key: NTA-verifiedLast updated: 24 Sep 2026

p-Block Trends, explained for NEET

The trap that costs the most marks here is treating molar mass as the sole driver of boiling point. In Groups 15, 16 and 17, the first-row hydride breaks the trend: H₂O, NH₃ and HF all sit at the top of their group's boiling-point order, not the bottom, because extensive hydrogen bonding must be overcome before the molecule can leave the liquid. Only after H₂O is removed does the remaining series behave monotonically: H₂S < H₂Se < H₂Te. Any option placing H₂O lowest is wrong on sight.

A second reversal runs the other way. Noble gases are chemically inert — very high ionisation enthalpy, large positive electron-gain enthalpy — yet their melting and boiling points are very low, because the only forces between atoms are weak London dispersion forces. NCERT Class 12 Chemistry (pre-2023 edition), Chapter 7 sets this out on page 200. Chemical inertness and thermal stability are independent properties; a statement claiming noble gases have high m.p./b.p. is the canonical false one.

Across period 3, ionisation enthalpy rises overall but reverses at Mg/Al: IE₁(Mg) > IE₁(Al), because the 3p electron removed from Al is higher in energy and shielded by the 3s pair. The period-3 order is Na < Al < Mg < Si < P > S < Cl < Ar.

Down a group, oxidation-state exceptions sit at the heaviest congener, not the lightest. Po, not O or Te, is the Group 16 element that does not show −2 — the inert-pair effect is a 6p phenomenon (Tl, Pb, Bi, Po). NCERT Class 11 Chemistry (pre-2023 edition), Chapter 11 covers the Group 13 inert-pair case on page 319.

Hydride classification follows group number directly: Group 13 electron-deficient (B₂H₆), Group 14 electron-precise (GeH₄), Groups 15–17 electron-rich (HF). MgH₂ is the ionic one — HF's high bond polarity is not ionicity.

Watch-out: in Assertion–Reason items, verify the Reason's chemistry on its own before judging whether it explains the Assertion.

Can you answer these p-Block Trends MCQs?

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

Which of the following correctly describes the melting and boiling points of the noble gases?

Show answer and why every option is right or wrong

Answer: A. A is correct. Noble gas atoms are monatomic with closed shells, so the only intermolecular attraction available is weak London dispersion — giving very low m.p. and b.p., as stated in NCERT Class 12 Chemistry (pre-2023 edition), Chapter 7, page 200.

Why B is wrong: B is wrong because chemical inertness describes resistance to reaction, not resistance to melting; the two properties are independent (trap: inertness read as thermal stability).

Why C is wrong: C is wrong because high ionisation enthalpy explains why noble gases do not react, not how strongly their atoms attract one another in the liquid or solid state (trap: inertness read as thermal stability).

Why D is wrong: D is wrong because noble gases do not form covalent dimers in the condensed phase; the condensed phase is held together by dispersion forces alone.

MCQ 2Direct ApplicationPractice

Arrange the Group 16 hydrides H₂O, H₂S, H₂Se and H₂Te in order of increasing boiling point.

Show answer and why every option is right or wrong

Answer: B. B is correct. H₂S, H₂Se and H₂Te follow the molar-mass trend, but H₂O's extensive hydrogen bonding lifts it above all three, making it the highest-boiling member of the group.

Why A is wrong: A is wrong because it places H₂O lowest on molar mass alone, ignoring hydrogen bonding (trap: molar-mass trend applied to the first-row hydride).

Why C is wrong: C is wrong because it inverts the molar-mass ordering for the three heavier hydrides; H₂Te boils higher than H₂S, not lower.

Why D is wrong: D is wrong because it slots H₂O between H₂S and H₂Se, treating hydrogen bonding as a small correction rather than the dominant term (trap: molar-mass trend applied to the first-row hydride).

MCQ 3Direct ApplicationPractice

Which Group 16 element does not exhibit the −2 oxidation state?

Show answer and why every option is right or wrong

Answer: D. D is correct. Po is the heaviest, most metallic Group 16 congener; the inert-pair tendency at 6p removes the −2 state that O, S, Se and Te all show.

Why A is wrong: A is wrong because oxygen shows −2 as its standard state; oxygen's anomalies (small size, no d-orbitals, restricted maximum covalence) are unrelated to this trend (trap: exception located at the lightest congener).

Why B is wrong: B is wrong because Se is a 4p element that readily forms selenides in the −2 state (trap: inert-pair reasoning applied mid-group).

Why C is wrong: C is wrong because Te is a 5p element and still forms tellurides in the −2 state; the exception sits one place lower (trap: second-heaviest congener chosen).

MCQ 4Direct ApplicationPractice

In the series MgH₂, B₂H₆, GeH₄ and HF, which compound is electron-deficient?

Show answer and why every option is right or wrong

Answer: C. C is correct. Boron contributes three valence electrons, leaving B₂H₆ short of enough electrons for conventional two-centre bonds; it uses three-centre two-electron bridge bonds instead.

Why A is wrong: A is wrong because MgH₂ is an ionic hydride — Mg transfers electrons to hydrogen, giving Mg²⁺ and H⁻ in a lattice (trap: ionic hydride misassigned).

Why B is wrong: B is wrong because germanium has four valence electrons and forms exactly four two-centre two-electron bonds, making GeH₄ electron-precise; its metalloid character does not make it deficient (trap: Group 13 and Group 14 hydrides confused).

Why D is wrong: D is wrong because HF carries three lone pairs on fluorine, making it electron-rich; its high bond polarity is not electron deficiency (trap: bond polarity read as ionic or deficient character).

MCQ 5Concept TrapPractice

Which statement about the first ionisation enthalpies of Na, Mg, Al and Si is correct?

Show answer and why every option is right or wrong

Answer: D. D is correct. The 3p orbital of Al lies higher in energy and is shielded by the filled 3s pair, so IE₁(Al) falls below IE₁(Mg) and breaks the otherwise rising period-3 trend.

Why A is wrong: A is wrong because it applies the blanket 'IE rises across a period' rule without the Mg/Al reversal (trap: period-3 IE anomaly ignored).

Why B is wrong: B is wrong because Na, not Al, has the largest radius and lowest IE₁ of the four; Al sits between Na and Mg.

Why C is wrong: C is wrong because atomic size decreases, not increases, across a period, so the whole ordering is inverted.

MCQ 6Concept TrapPractice

According to NCERT, ICl is more reactive than I₂. Which reason does NCERT give?

Show answer and why every option is right or wrong

Answer: B. B is correct. NCERT (Class 12 Chemistry, Chapter 7, interhalogen compounds) states that interhalogens are more reactive than halogens (except fluorine) because the X–X′ bond is weaker than the X–X bond, and gives ICl and I₂ as its example; a weaker bond breaks more easily. (Gas-phase bond-enthalpy tables list I–Cl, about 211 kJ/mol, above I–I, about 151 kJ/mol; the polarity of I–Cl, which lets it break heterolytically, is the fuller reason. In NEET, answer with NCERT's statement.)

Why A is wrong: A is wrong because it inverts the relationship: a stronger bond is harder to break and lowers reactivity (trap: bond strength and reactivity direction inverted).

Why C is wrong: C is wrong because ICl (about 162 g/mol) has a lower molar mass than I₂ (about 254 g/mol), and collision energetics are not what distinguishes their reactivity here.

Why D is wrong: D is wrong because ICl is polar, not non-polar — that polarity is part of why ICl reacts so readily.

MCQ 7CalculationPractice

Which statement about oxygen-family diatomic species is incorrect?

Show answer and why every option is right or wrong

Answer: C. C is the incorrect statement. The bond order 2.5 is right, but O₂⁺ has 15 electrons — removing one π*2p electron from O₂ leaves one unpaired electron, so O₂⁺ is paramagnetic, not diamagnetic.

Why A is wrong: A is wrong as an answer because the statement is factually correct: O₂ has 16 electrons with two unpaired electrons in the degenerate π*2p orbitals, giving bond order 2 and paramagnetism.

Why B is wrong: B is wrong as an answer because the statement is factually correct: O₂⁻ has 17 electrons, three in π*2p, leaving one unpaired — bond order 1.5 and paramagnetic.

Why D is wrong: D is wrong as an answer because the statement is factually correct: O₂²⁻ has 18 electrons with π*2p completely filled, so no unpaired electrons remain — bond order 1 and diamagnetic.

MCQ 8CalculationPractice

For C₂, how many electrons occupy the π2p molecular orbitals, and what is the molecule's magnetic character?

Show answer and why every option is right or wrong

Answer: A. A is correct. For second-period diatomics up to N₂, π2p lies below σ2p, so C₂'s four valence 2p electrons fill the two degenerate π2p orbitals completely — all paired, hence diamagnetic.

Why B is wrong: B is wrong because it also assumes σ2p fills first; with two π electrons in degenerate orbitals Hund's rule would in any case leave them unpaired, so the magnetic claim is inconsistent too (trap: O₂ energy ordering applied to C₂).

Why C is wrong: C is wrong because it uses the O₂-type ordering with σ2p below π2p, placing only two electrons in π2p (trap: O₂ energy ordering applied to C₂).

Why D is wrong: D is wrong because four electrons in two degenerate π2p orbitals pair up completely (two per orbital), leaving no unpaired electrons (trap: bond order checked without an independent magnetism check).

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How do you solve a p-Block Trends question? A worked example

  1. 1

    Given

    The species O₂, O₂⁺, O₂⁻ and O₂²⁻. Oxygen's atomic number is 8 (exact, a counting integer).

  2. 2

    Required

    Bond order and magnetic character for each species, and identification of which species is diamagnetic.

  3. 3

    Concept

    Molecular orbital theory. Electrons fill MOs in order of increasing energy; bond order is set by the imbalance between bonding and antibonding occupancy, and magnetic character by whether any electrons remain unpaired. These are two separate readings of the same configuration and must be checked independently.

  4. 4

    Formula

    Bond order = (number of bonding electrons − number of antibonding electrons) / 2. A species is paramagnetic if one or more electrons are unpaired, diamagnetic if all are paired.

  5. 5

    Substitution

    For O₂ (16 electrons, σ2p below π2p since Z = 8): bonding = 10, antibonding = 6. For O₂⁺, remove one π*2p electron: bonding = 10, antibonding = 5. For O₂⁻, add one: bonding = 10, antibonding = 7. For O₂²⁻, add two: bonding = 10, antibonding = 8.

  6. 6

    Calculation

    • O₂: (10 − 6)/2 = 2. π*2p holds 2 electrons in two degenerate orbitals → 2 unpaired → paramagnetic.• O₂⁺: (10 − 5)/2 = 2.5. π*2p holds 1 electron → 1 unpaired → paramagnetic.• O₂⁻: (10 − 7)/2 = 1.5. π*2p holds 3 electrons → 1 unpaired → paramagnetic.• O₂²⁻: (10 − 8)/2 = 1. π*2p holds 4 electrons → 0 unpaired → diamagnetic.
    All electron counts here are exact counting integers, and the divisor 2 in the bond-order expression is exact; none of them contribute to significant-figure limits, so the bond orders 2.5 and 1.5 are exact half-integers rather than two-significant-figure measurements.

  7. 7

    Final answer

    Bond orders O₂⁺ (2.5) > O₂ (2) > O₂⁻ (1.5) > O₂²⁻ (1). Only O₂²⁻ is diamagnetic; the other three are paramagnetic.

  8. 8

    Common trap

    Carrying O₂'s two-unpaired-electron result across to O₂⁺. The cation has one fewer electron, so it retains one unpaired electron — still paramagnetic. Any statement calling O₂⁺ diamagnetic is false. The related error is confirming the familiar bond-order value in an option and accepting the whole statement without re-deriving the magnetism from the configuration.

  9. 9

    Similar NEET-style question

    Among N₂, N₂⁺, O₂ and O₂²⁻, which species has the highest bond order, and is it paramagnetic or diamagnetic? (Note the σ/π ordering switches between the nitrogen and oxygen species.)

What to remember before solving p-Block Trends questions

In the NEET syllabus; removed from current NCERT.

11.1.6 Chemical Properties Oxidation state and trends in chemical reactivity Due to small size of boron, the sum of its first three ionization enthalpies is very high. This prevents it to form +3 ions and forces it to form only covalent compounds. But as we move from B to Al, the sum of the first three ionisation enthalpies of Al considerably decreases, and is therefore able to form Al3+ ions. In fact, aluminium is a highly electropositive metal. However, down the group, due to poor shielding effect of intervening d and f orbitals, the increased effective nuclear charge holds ns electrons tightly (responsible for inert pair effect) and thereby, restricting their participation in bonding. As a result of this, only p-orbital electron may be involved in bonding. In fact in Ga, In and Tl, both +1 and +3 oxidation states are observed. The relative stability of +1 oxidation state progressively increases for heavier elements: Al<Ga<In<Tl. In thallium +1 oxidation state is predominant whereas the +3 oxidation state is highly oxidising in character. The compounds in +1 oxidation state, as expected from energy considerations, are more ionic than those in +3 oxidation state. In trivalent state, the number of electrons around the central atom in a molecule of the compounds of these elements (e.g., boron in BF3) will be only six. Such electron deficient molecules have tendency to accept a pair of electrons to achieve stable electronic configuration and thus, behave as Lewis acids. The tendency to behave as Lewis acid decreases with the increase in the size down the group. BCl3 easily accepts a lone pair of electrons from ammonia to form BCl3⋅NH3.

-- NCERT Class 11 Chemistry (pre-2023 edition), Chapter 11, p. 318

In the NEET syllabus; removed from current NCERT.

In tetravalent state the number of electrons around the central atom in a molecule (e.g., carbon in CCl4) is eight. Being electron precise molecules, they are normally not expected to act as electron acceptor or electron donor species. Although carbon cannot exceed its covalence more than 4, other elements of the group can do so. It is because of the presence of d orbital in them. Due to this, their halides undergo hydrolysis and have tendency to form complexes by accepting electron pairs from donor species. For example, the species like, SiF6 2–, [GeCl6]2–, [Sn(OH)6]2– exist where the hybridisation of the central atom is sp3d2.

-- NCERT Class 11 Chemistry (pre-2023 edition), Chapter 11, p. 324

In the NEET syllabus; removed from current NCERT.

11.7 ALLOTROPES OF CARBON Carbon exhibits many allotropic forms; both crystalline as well as amorphous. Diamond and graphite are two well-known crystalline forms of carbon. In 1985, third form of carbon known as fullerenes was discovered by H.W.Kroto, E.Smalley and R.F.Curl. For this discovery they were awarded the Nobel Prize in 1996. 11.7.1 Diamond It has a crystalline lattice. In diamond each carbon atom undergoes sp3 hybridisation and linked to four other carbon atoms by using hybridised orbitals in tetrahedral fashion. The C–C bond length is 154 pm. The structure extends in space and produces a rigid three-dimensional network of carbon atoms. In this structure (Fig. 11.3) directional covalent bonds are present throughout the lattice. It is very difficult to break extended covalent bonding and, therefore, diamond is a hardest substance on the earth. It is used as an abrasive for sharpening hard tools, in making dyes and in the manufacture of tungsten filaments for electric light bulbs. 11.7.2 Graphite Graphite has layered structure (Fig.11.4). Layers are held by van der Waals forces and distance between two layers is 340 pm. Each layer is composed of planar hexagonal rings of carbon atoms. C—C bond length within the layer is 141.5 pm. Each carbon atom in hexagonal ring undergoes sp2 hybridisation and makes three sigma bonds with three neighbouring carbon atoms. Fourth electron forms a π bond. The electrons are delocalised over the whole sheet. Electrons are mobile and, therefore, graphite conducts electricity along the sheet. Graphite cleaves easily between the layers and, therefore, it is very soft and slippery. For this reason graphite is used as a dry lubricant in machines running at high temperature, where oil cannot be used as a lubricant. 11.7.3 Fullerenes Fullerenes are made by the heating of graphite in an electric arc in the presence of inert gases such as helium or argon. The sooty material formed by condensation of vapourised Cn small molecules consists of mainly C60 with smaller quantity of C70 and traces of fullerenes consisting of even number of carbon atoms up to 350 or above. Fullerenes are the only pure form of carbon because they have smooth structure without having ‘dangling’ bonds. Fullerenes are cage like molecules. C60 molecule has a shape like soccer ball and called Buckminsterfullerene (Fig. 11.5). It contains twenty six- membered rings and twelve five-membered rings. A six membered ring is fused with six or five membered rings but a five membered ring can only fuse with six membered rings. All the carbon atoms are equal and they undergo sp2 hybridisation. Each carbon atom forms three sigma bonds with other three carbon atoms. The remaining electron at each carbon is delocalised in molecular orbitals, which in turn give aromatic character to molecule. This ball shaped molecule has 60 vertices and each one is occupied by one carbon atom and it also contains both single and double bonds with C–C distances of 143.5 pm and 138.3 pm respectively. Spherical fullerenes are also called bucky balls in short.

-- NCERT Class 11 Chemistry (pre-2023 edition), Chapter 11, p. 325

In the NEET syllabus; removed from current NCERT.

Properties Ammonia is a colourless gas with a pungent odour. Its freezing and boiling points are 198.4 and 239.7 K respectively. In the solid and liquid states, it is associated through hydrogen bonds as in the case of water and that accounts for its higher melting and boiling points than expected on the basis of its molecular mass. The ammonia molecule is trigonal pyramidal with the nitrogen atom at the apex. It has three bond pairs and one lone pair of electrons as shown in the structure. Ammonia gas is highly soluble in water. Its aqueous solution is weakly basic due to the formation of OH– ions. NH3(g) + H2O(l) ⇌ NH4+ (aq) + OH– (aq) … On a large scale, ammonia is manufactured by Haber’s process. N2(g) + 3H2(g) ⇌ 2NH3(g); ∆fH0 = – 46.1 kJ mol–1 In accordance with Le Chatelier’s principle, high pressure would favour the formation of ammonia. The optimum conditions for the production of ammonia are a pressure of 200 × 10^5 Pa (about 200 atm), a temperature of ~ 700 K and the use of a catalyst such as iron oxide with small amounts of K2O and Al2O3 to increase the rate of attainment of equilibrium. The flow chart for the production of ammonia is shown in Fig. 7.1. Earlier, iron was used as a catalyst with molybdenum as a promoter. … The presence of a lone pair of electrons on the nitrogen atom of the ammonia molecule makes it a Lewis base. It donates the electron pair and forms linkage with metal ions and the formation of such complex compounds finds applications in detection of metal ions such as Cu2+, Ag+: Cu2+ (aq) + 4 NH3(aq) ⇌ [Cu(NH3)4]2+(aq)

-- NCERT Class 12 Chemistry (pre-2023 edition), Chapter 7, p. 176

In the NEET syllabus; removed from current NCERT.

(i) Reactivity towards hydrogen: They all react with hydrogen to give hydrogen halides but affinity for hydrogen decreases from fluorine to iodine. Hydrogen halides dissolve in water to form hydrohalic acids. Some of the properties of hydrogen halides are given in Table 7.9. The acidic strength of these acids varies in the order: HF < HCl < HBr < HI. The stability of these halides decreases down the group due to decrease in bond (H–X) dissociation enthalpy in the order: H–F > H–Cl > H–Br > H–I. Table 7.9: Properties of Hydrogen Halides (HF, HCl, HBr, HI) — Melting point/K: 190, 159, 185, 222; Boiling point/K: 293, 189, 206, 238; Bond length (H – X)/pm: 91.7, 127.4, 141.4, 160.9; ΔdissH/kJ mol–1: 574, 432, 363, 295; pKa: 3.2, –7.0, –9.5, –10.0.

-- NCERT Class 12 Chemistry (pre-2023 edition), Chapter 7, p. 201

In the NEET syllabus; removed from current NCERT.

Ionisation enthalpy decreases down the group. It is due to increase in size. However, the elements of this group have lower ionisation enthalpy values compared to those of Group15 in the corresponding periods. This is due to the fact that Group 15 elements have extra stable half-filled p orbitals electronic configurations. Because of the compact nature of oxygen atom, it has less negative electron gain enthalpy than sulphur. However, from sulphur onwards the value again becomes less negative upto polonium. Next to fluorine, oxygen has the highest electronegativity value amongst the elements. Within the group, electronegativity decreases with an increase in atomic number. This implies that the metallic character increases from oxygen to polonium. Oxidation states and trends in chemical reactivity The elements of Group 16 exhibit a number of oxidation states (Table 7.6). The stability of -2 oxidation state decreases down the group. Polonium hardly shows –2 oxidation state. Since electronegativity of oxygen is very high, it shows only negative oxidation state as –2 except in the case of OF2 where its oxidation state is + 2. Other elements of the group exhibit + 2, + 4, + 6 oxidation states but + 4 and + 6 are more common. Anomalous behaviour of oxygen The anomalous behaviour of oxygen, like other members of p-block present in second period is due to its small size and high electronegativity. One typical example of effects of small size and high electronegativity is the presence of strong hydrogen bonding in H2O which is not found in H2S.

-- NCERT Class 12 Chemistry (pre-2023 edition), Chapter 7, p. 187

In the NEET syllabus; removed from current NCERT.

All noble gases have general electronic configuration ns2np6 except helium which has 1s2 (Table 7.12). Many of the properties of noble gases including their inactive nature are ascribed to their closed shell structures. Due to stable electronic configuration these gases exhibit very high ionisation enthalpy. However, it decreases down the group with increase in atomic size. Atomic radii increase down the group with increase in atomic number. Since noble gases have stable electronic configurations, they have no tendency to accept the electron and therefore, have large positive values of electron gain enthalpy. Physical Properties All the noble gases are monoatomic. They are colourless, odourless and tasteless. They are sparingly soluble in water. They have very low melting and boiling points because the only type of interatomic interaction in these elements is weak dispersion forces. Helium has the lowest boiling point (4.2 K) of any known substance. It has an unusual property of diffusing through most commonly used laboratory materials such as rubber, glass or plastics. Chemical Properties In general, noble gases are least reactive. Their inertness to chemical reactivity is attributed to the following reasons: (i) The noble gases except helium (1s2) have completely filled ns2np6 electronic configuration in their valence shell. (ii) They have high ionisation enthalpy and more positive electron gain enthalpy. … After this discovery, a number of xenon compounds mainly with most electronegative elements like fluorine and oxygen, have been synthesised. … The structures of the three xenon fluorides can be deduced from VSEPR and these are shown in Fig. 7.9. XeF2 and XeF 4 have linear and square planar structures respectively. XeF6 has seven electron pairs (6 bonding pairs and one lone pair) and would, thus, have a distorted octahedral structure as found experimentally in the gas phase. Fig. 7.9 The structures of (a) XeF2 (b) XeF4 (c) XeF6 (d) XeOF4 and (e) XeO3: (a) Linear (b) Square planar (c) Distorted octahedral (d) Square pyramidal (e) Pyramidal

-- NCERT Class 12 Chemistry (pre-2023 edition), Chapter 7, p. 209

In the NEET syllabus; removed from current NCERT.

The absence of d orbitals in oxygen limits its covalency to four and in practice, rarely exceeds two. On the other hand, in case of other elements of the group, the valence shells can be expanded and covalence exceeds four. (i) Reactivity with hydrogen: All the elements of Group 16 form hydrides of the type H2E (E = O, S, Se, Te, Po). Some properties of hydrides are given in Table 7.7. Their acidic character increases from H2O to H2Te. The increase in acidic character can be explained in terms of decrease in bond enthalpy for the dissociation of H–E bond down the group. Owing to the decrease in enthalpy for the dissociation of H–E bond down the group, the thermal stability of hydrides also decreases from H2O to H2Po. All the hydrides except water possess reducing property and this character increases from H2S to H2Te. Table 7.7: Properties of Hydrides of Group 16 Elements (H2O, H2S, H2Se, H2Te) — m.p/K: 273, 188, 208, 222; b.p/K: 373, 213, 232, 269; H–E distance/pm: 96, 134, 146, 169; HEH angle (°): 104, 92, 91, 90; ΔfH/kJ mol–1: –286, –20, 73, 100; ΔdissH (H–E)/kJ mol–1: 463, 347, 276, 238; Dissociation constant (aqueous solution, 298 K): 1.8×10–16, 1.3×10–7, 1.3×10–4, 2.3×10–3.

-- NCERT Class 12 Chemistry (pre-2023 edition), Chapter 7, p. 188

In the NEET syllabus; removed from current NCERT.

7.22 Interhalogen Compounds When two different halogens react with each other, interhalogen compounds are formed. They can be assigned general compositions as XX′ , XX3′, XX5′ and XX7′ where X is halogen of larger size and X′ of smaller size and X is more electropositive than X′. As the ratio between radii of X and X′ increases, the number of atoms per molecule also increases. Thus, iodine (VII) fluoride should have maximum number of atoms as the ratio of radii between I and F should be maximum. That is why its formula is IF7 (having maximum number of atoms). Preparation The interhalogen compounds can be prepared by the direct combination or by the action of halogen on lower interhalogen compounds. The product formed depends upon some specific conditions, For example, Cl2 + F2 (equal volume) →(437 K) 2ClF; Cl2 + 3F2 (excess) →(573 K) 2ClF3; I2 + Cl2 (equimolar) → 2ICl; I2 + 3Cl2 (excess) → 2ICl3; Br2 + 3F2 (diluted with water) → 2BrF3; Br2 + 5F2 (excess) → 2BrF5 Properties Some properties of interhalogen compounds are given in Table 7.11. Table 7.11: Some Properties of Interhalogen Compounds (Type; Formula; Physical state and colour; Structure): XX′ — ClF colourless gas; BrF pale brown gas; IF detected spectroscopically (very unstable); BrCl gas; ICl ruby red solid (α-form), brown red solid (β-form); IBr black solid. XX′3 — ClF3 colourless gas, Bent T-shaped; BrF3 yellow green liquid, Bent T-shaped; IF3 yellow powder, Bent T-shaped (?); ICl3 orange solid, Bent T-shaped (?) (dimerises as Cl–bridged dimer (I2Cl6)). XX′5 — IF5 colourless gas but solid below 77 K, Square pyramidal; BrF5 colourless liquid, Square pyramidal; ClF5 colourless liquid, Square pyramidal. XX′7 — IF7 colourless gas, Pentagonal bipyramidal. These are all covalent molecules and are diamagnetic in nature. They are volatile solids or liquids at 298 K except ClF which is a gas. Their physical properties are intermediate between those of constituent halogens except that their m.p. and b.p. are a little higher than expected. Their chemical reactions can be compared with the individual halogens. In general, interhalogen compounds are more reactive than halogens (except fluorine). This is because X–X′ bond in interhalogens is weaker than X–X bond in halogens except F–F bond. All these undergo hydrolysis giving halide ion derived from the smaller halogen and a hypohalite ( when XX′), halite ( when XX′3), halate (when XX′5) and perhalate (when XX′7) anion derived from the larger halogen. XX′ + H2O → HX′ + HOX Their molecular structures are very interesting which can be explained on the basis of VSEPR theory (Example 7.19). The XX3 compounds have the bent ‘T’ shape, XX5 compounds square pyramidal and IF7 has pentagonal bipyramidal structures (Table 7.11). Uses: These compounds can be used as non aqueous solvents. Interhalogen compounds are very useful fluorinating agents. ClF3 and BrF3 are used for the production of UF6 in the enrichment of 235U. U(s) + 3ClF3(l) → UF6(g) + 3ClF(g)

-- NCERT Class 12 Chemistry (pre-2023 edition), Chapter 7, p. 206

Where do students lose marks on p-Block Trends?

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Category: Inorganic Exception

Students assume boiling point increases monotonically with molar mass across a group's hydrides. The first-row hydrides H2O, NH3 and HF boil far higher than that trend predicts because of strong intermolecular H-bonding. H2O is the highest-boiling Group 16 hydride (NCERT Table 7.7: 373 K) and HF the highest-boiling hydrogen halide (Table 7.9: 293 K). NH3 is NOT the highest in Group 15: it boils above PH3 and AsH3 but below SbH3 and BiH3 (Table 7.2: 238.5, 185.5, 210.6, 254.6, 290 K). Any statement that places H2O or HF at the low end of their series is incorrect.

When it triggers

Question presents a boiling-point order (or a statement about order) for a set of Group 15, 16, or 17 hydrides and includes H2O, NH3, or HF. The trap fires when an option treats the first-row hydride as the lowest-boiling member.

How to avoid

Identify whether H2O, NH3 or HF is in the series. For H2O and HF, put the first-row hydride at the TOP of the boiling-point ranking; the heavier hydrides then follow molar mass (H2S < H2Se < H2Te; HCl < HBr < HI). For Group 15, NH3 only beats PH3 and AsH3: the order is PH3 < AsH3 < NH3 < SbH3 < BiH3.

Category: Similar Terms

Noble gases are chemically inert (very high ionization enthalpy, negligible tendency to react), but this chemical stability does not carry over into thermal stability. Because the only intermolecular forces present are weak London dispersion forces, noble gases have very LOW melting and boiling points. Students who associate 'inert' or 'stable' with high thermal resistance will select a statement claiming high m.p./b.p. as correct, when in fact it is the canonical incorrect statement about noble gases.

When it triggers

Question asks about physical properties (melting point, boiling point) of noble gases, or asks to identify an incorrect statement about them. The distractor option asserts that noble gases have high melting or boiling points.

How to avoid

Separate chemical reactivity from physical state properties. Noble gases: chemically unreactive (true), large positive electron-gain enthalpy (true), sparingly soluble in water (true), weak dispersion forces (true), very LOW m.p. and b.p. (true). High m.p./b.p. is the false claim.

Category: Inorganic Exception

Across period 3, first ionization enthalpy generally increases from Na to Ar, but there is a reversal at Mg/Al: IE1(Mg) > IE1(Al). Removing a 3p electron (Al) requires less energy than removing a 3s electron (Mg) because 3p orbitals are higher in energy and experience more shielding from the inner 3s electrons. Students who apply the blanket 'IE increases uniformly across a period' rule expect Al's IE to exceed Mg's, leading them to accept an incorrect statement or order.

When it triggers

Question presents the first ionization enthalpy order for Na, Mg, Al, Si (or a subset). Any option giving a strictly monotonic order Na < Mg < Al < Si is wrong; the correct order has Mg > Al.

How to avoid

Memorise the two period-3 IE1 anomalies: (1) Mg > Al, because a 3p electron is more shielded and easier to remove than a 3s electron (NCERT Class XI Unit 3 explains the same s-versus-p effect for Be > B); (2) P > S, because of the extra-stable half-filled 3p3 configuration (Class XII Unit 7 page 18). Correct period-3 order: Na < Al < Mg < Si < S < P < Cl < Ar.

Category: Negative Marking

NEET Assertion-Reason questions require evaluating A and R as independent statements and then determining whether R actually explains A. Students who read a plausible-sounding R alongside a correct A default to option 2 (both correct, R explains A) without separately verifying that R's specific chemical claim is factually accurate. When R is false or irrelevant, selecting option 2 instead of option 3 or 4 costs 1 negative mark.

When it triggers

Any Assertion-Reason format question in chemistry where R invokes a chemical mechanism or property that sounds consistent with A. The trap fires when students accept R without checking whether its chemistry is correct.

How to avoid

Evaluate A and R as completely separate statements before considering their relationship. Protocol: (1) Is A true on its own? (2) Is R's specific chemical claim factually correct on its own? (3) If both true, does R mechanistically explain A? Skipping step 2 is the trap.

Category: Similar Terms

A weaker bond is easier to break (lower bond dissociation energy = lower activation barrier), so a compound with a weaker bond is MORE reactive under comparable conditions, not less. Students sometimes invert this, concluding that a compound with a weaker bond is less reactive because they associate lower energy content with lower driving force. The correct relationship is: bond strength down = reactivity up.

When it triggers

Question states or asks about the relative reactivity of two compounds where the reason involves comparing bond strengths (e.g. interhalogen vs parent dihalogen, or a bond in acid vs its conjugate). Distractor options invert the bond strength to reactivity mapping.

How to avoid

Bond strength up means harder to break, so reactivity down. Bond strength down means easier to break, so reactivity up. Apply to interhalogens: I-Cl is weaker than I-I, therefore ICl is more reactive than I2 (the R correctly explains the A).

Category: Inorganic Exception

In p-block groups, the element that deviates from the group standard oxidation state is almost always the HEAVIEST congener (due to metallic character, relativistic effects, and inert-pair tendency at high Z). Students make two opposite errors: (1) they pick the LIGHTEST member (e.g. O in Group 16) as the anomaly because oxygen is already known to have many unique properties; (2) they pick the second-heaviest (e.g. Te instead of Po) because the bottom congener is less familiar from standard syllabus coverage.

When it triggers

Question asks which element in a p-block group does NOT exhibit an oxidation state that all other members readily show. The group extends to a 5th- or 6th-period element (Po, At, Tl, Pb, Bi). Distractors offer the lightest or second-heaviest congener.

How to avoid

The exception is the HEAVIEST (most metallic) congener. Learn by group: Group 16 exception = Po; Group 15 exception = Bi (reduced -3 tendency); Group 17 exception = At. The lightest members (O, N, F) have different anomalies unrelated to this oxidation-state trend.

Category: Inorganic Exception

The inert pair effect is NCERT's name for the growing stability, down a group, of the oxidation state two units below the group oxidation state (Class XI Unit 11 page 1): Tl +1 over +3 (group 13), Pb +2 over +4 (group 14), Bi +3 over +5 (group 15). Students misapply it to negative oxidation states, e.g. explaining why polonium hardly shows -2. That is a different trend: NCERT states only that the stability of the -2 state decreases down Group 16 (Class XII Unit 7 page 18), alongside rising metallic character.

When it triggers

Question asks which element in a p-block group does not show an oxidation state (especially -2 in Group 16), or asks for the reason, and an option or explanation invokes the inert pair effect for a negative oxidation state or for a mid-group element.

How to avoid

Use the inert pair effect only for positive states two below the group state in the heavier members (Tl+, Sn2+/Pb2+, Bi3+). For the loss of the -2 state in Group 16, the answer is the bottom member, Po, and the reason is the decreasing stability of -2 down the group (increasing metallic character), not the inert pair.

Category: Similar Terms

B2H6 (Group 13) is electron-deficient: it 'has too few electrons for writing its conventional Lewis structure', and its two bridge B-H-B bonds are 3-centre-2-electron bonds. GeH4 (Group 14) is electron-precise: group 14 compounds 'have the required number of electrons to write their conventional Lewis structures'. The two hydrides sit side by side in a match-the-column list and look alike, so students carry the electron-deficient label from B2H6 over to GeH4.

When it triggers

Classification or match-the-column question presents a Group 13 hydride (B2H6, AlH3) and a Group 14 hydride (CH4, GeH4) alongside the labels electron-deficient and electron-precise. The trap fires when the electron-deficient label is placed on the Group 14 hydride.

How to avoid

The group number tells the category directly (NCERT Hydrogen unit 9.5.2): all group 13 elements form electron-deficient compounds (Lewis acids); all group 14 elements form electron-precise compounds, tetrahedral (e.g. CH4, GeH4); groups 15-17 form electron-rich hydrides with lone pairs (NH3 1, H2O 2, HF 3), which act as Lewis bases. Use the group number, not the element's metallic character.

Category: Similar Terms

Both graphite and fullerene are non-diamond carbon allotropes with sp2-hybridised carbon and delocalised electrons, so students who remember only 'non-diamond carbon allotrope' as the distinguishing property may swap their structural labels. Graphite has a 2D layered hexagonal structure; layers slide over each other making it a dry lubricant. Fullerene (C60) is a closed cage molecule (soccer-ball shaped); its defining NEET descriptor is cage-like structure, not lubricant. Assigning cage-structure to graphite, or lubricant to fullerene, is wrong.

When it triggers

Match-the-column or property-identification question lists graphite and fullerene alongside descriptors including cage-like molecule, dry lubricant, sp3 hybridised, or used as reducing agent.

How to avoid

Four-word anchors per allotrope: Diamond = sp3, hard; Graphite = sp2, layers, lubricant; Fullerene = sp2, cage (C60); Coke = amorphous, reducing agent. If asked about cage structure, the answer is always fullerene. If asked about lubricant, the answer is always graphite.

Category: Similar Terms

MO questions often bundle two distinct properties (bond order AND magnetic character) within a single statement about the same species. Students who have memorised the bond-order series for oxygen species (O2+: 2.5, O2: 2, O2-: 1.5, O22-: 1) confirm the bond-order part and then accept the entire multi-claim statement as correct, missing a separate magnetic-character error in the same statement. Bond order and magnetic character are derived from the same electron configuration but require independent checks: bond order from the difference of bonding vs antibonding occupancies; magnetic character from counting any residual unpaired electrons.

When it triggers

An incorrect-statement MO question includes an option that pairs a correct bond-order value with a wrong magnetic-character claim (or vice versa). Students who verify only the familiar bond-order fact accept the full option.

How to avoid

For every MO option: (1) write out the full electron configuration from scratch; (2) compute bond order = (bonding - antibonding)/2; (3) count unpaired electrons for magnetism. Treat bond order and magnetic character as two separate sub-checks on the same configuration, not as a single fact.

Category: Similar Terms

O2 having two unpaired electrons (paramagnetic) is a landmark MO result taught explicitly in NEET preparation. When O2+ appears in a question, students recall the O2 lesson and unconsciously carry over its configuration, missing that O2+ has one fewer electron. O2+ (15 electrons) has 1 unpaired electron in pi*2p, making it paramagnetic but with a different unpaired-electron count than O2. The canonical NEET trap is a statement claiming O2+ is diamagnetic, which students fail to challenge because they confuse it with O2.

When it triggers

Statement or incorrect-statement question about the magnetic character of oxygen species (O2, O2+, O2-, O22-). The option claiming O2+ is diamagnetic is the wrong one.

How to avoid

Do NOT reuse O2's configuration for O2+. Re-derive from scratch: O2 has 16 electrons (2 unpaired in pi*2p). O2+ has 15 electrons: remove 1 from pi*2p, leaving 1 unpaired. Still paramagnetic, NOT diamagnetic. O22- has 18 electrons: pi*2p fully filled, 0 unpaired, diamagnetic.

Category: Inorganic Exception

For diatomics from B2 to N2 (low-Z second-period elements), the pi2p MOs lie LOWER in energy than the sigma2p MO. This is the inversion relative to the O2-type diagram where sigma2p is below pi2p. Students who learn only the O2-type ordering misplace the sigma2p electrons in C2, arriving at wrong electron occupancy for pi2p and a wrong conclusion about C2's magnetic character or number of pi bonds. C2 has 4 electrons in the two degenerate pi2p MOs (all paired, diamagnetic) precisely because pi2p is filled before sigma2p.

When it triggers

Question about C2, B2, or N2 molecular properties (magnetic character, bond order, number of pi bonds, number of electrons in pi orbitals). Any derivation that uses sigma2p < pi2p ordering for these molecules gives a wrong result.

How to avoid

Memorise the crossover: for Z <= 7 (up to N2), energy order is: sigma1s < sigma*1s < sigma2s < sigma*2s < pi2p (x,y degenerate) < sigma2p < pi*2p < sigma*2p. The order flips at O2 (Z = 8): sigma2p moves below pi2p. When a question involves C2, always use the pi-first ordering.

p-Block Trends questions from past NEET papers

8 questions from NEET 2021, 2022, 2023, 2024. Answers verified against NTA official keys.

NEET 2024Revised key

Given below are two statements: Statement I: The boiling point of hydrides of Group 16 elements follow the order H O > H Te > H Se > H S. 2 2 2 2 Statement II: On the basis of molecular mass, H O is expected to have lower boiling point than the other 2 members of the group but due to the presence of extensive H-bonding in H O, it has higher boiling point. 2 In the light of the above statements, choose the correct answer from the options given below:

1Both Statement I and Statement II are true
2Both Statement I and Statement II are false
3Statement I is true but Statement II is false
4Statement I is false but Statement II is true
NTA Answer: Option 1(revised_final)
NEET 2022

Given below are two statements Statement I The boiling points of the following hydrides of group 16 elements increases in the order – H2O < H2S < H2Se < H2Te Statement II The boiling points of these hydrides increase with increase in molar mass. In the light of the above statements, choose the most appropriate answer from the options given below :

1Statement I is incorrect but Statement II is correct
2Both Statement I and Statement II are correct
3Both Statement I and Statement II are incorrect
4Statement I is correct but Statement II is incorrect
NTA Answer: Option 3(final)
NEET 2022

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): ICI is more reactive than I . 2 Reason (R): I-CI bond is weaker than I-I bond. In the light of the above statements, choose the most appropriate answer from the options given below:

1(A) is not correct but (R) is correct
2Both (A) and (R) are correct and (R) is the correct explanation of (A).
3Both (A) and (R) are correct but (R) is not the correct explanation of (A).
4(A) is correct but (R) is not correct
NTA Answer: Option 2(final)
NEET 2021

Statement I : Acid strength increases in the order given as HF << HCl << HBr << HI. Statement II : As the size of the elements F, Cl, Br, I increases down the group, the bond strength of HF, HCl, HBr and HI decreases and so the acid strength increases. In the light of the above statements, choose the correct answer from the options given below.

1Statement I is incorrect but Statement II is true
2Both statement I and Statement II are true
3Both Statement I and Statement II are false
4Statement I : correct but statement II is false
NTA Answer: Option 2(final)

All 18 past-paper questions from p-Block Elements →

Sources

Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.

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