p-Block Elements (Groups 13 to 18)

What to remember

• In the p-block the last electron enters a p orbital. The general outer configuration is ns²np¹ to ns²np⁶ (groups 13 to 18). Non-metallic character increases across the block and metallic character increases down each group.
• The inert pair effect makes the oxidation state that is two units below the group state more stable for the heavier elements: Tl⁺ over Tl³⁺, Pb²⁺ over Pb⁴⁺, Bi³⁺ over Bi⁵⁺.
• The first element of each group (B, C, N, O, F) is anomalous because of small size, high electronegativity, no d orbitals (maximum covalence 4) and the ability to form pπ–pπ multiple bonds.
• Hydrogen bonding makes NH₃, H₂O and HF boil at higher temperatures than the hydrides of the elements below them. Thermal stability of hydrides decreases down a group, while acidity and reducing power increase.

General introduction

The p-block has six groups. Each group begins with a second-period element, and the last electron enters a 2p, 3p, 4p, 5p or 6p orbital.

• The p-block contains non-metals (top right), metalloids (along the diagonal line) and metals (bottom left).
• Across a period the atomic radius decreases and ionisation enthalpy and electronegativity increase. Down a group the radius increases and ionisation enthalpy and electronegativity generally decrease.

Oxidation states and the inert pair effect

The valence electrons are ns² and np. The elements show a group oxidation state in which all valence electrons are used (+3, +4, +5, +6, +7 for groups 13 to 17) and a state two units lower in which the ns² pair stays unused (+1, +2, +3, +4, +5).

The inert pair effect is the reluctance of the ns² electrons to take part in bonding. It becomes strong for heavy elements because the d and f electrons shield the nucleus poorly, and so the ns electrons are held tightly.

Elements of group 15 to 17 also show negative states: −3 (N, P), −2 (O, S) and −1 (halogens). Oxygen shows −2 mostly, and fluorine shows only −1 because it is the most electronegative element.

Group 13: boron family

• Ga is smaller than Al because the 3d¹⁰ electrons of Ga shield the nucleus poorly. This also makes the ionisation enthalpy and electronegativity trends irregular.
• Boron is a non-metal (metalloid) and the others are metals.
• Oxidation states: +3 and +1. B is always +3. The +1 state becomes more stable down the group because of the inert pair effect, so Tl⁺ is more stable than Tl³⁺.
• Oxides: B₂O₃ is acidic, Al₂O₃ and Ga₂O₃ are amphoteric, and In₂O₃ and Tl₂O are basic.
• Halides: the trihalides (BX₃) are electron-deficient and act as Lewis acids because boron has an empty 2p orbital.
• Lewis acid strength of boron trihalides: BF₃ < BCl₃ < BBr₃ < BI₃. In BF₃ the filled 2p orbital of F overlaps well with the empty 2p orbital of B (pπ–pπ back bonding, because both are 2p). The overlap is less effective with the larger orbitals of Cl, Br and I, so the electron deficiency of boron is least reduced in BI₃.
• Anomaly of boron: small size, high ionisation enthalpy and electronegativity, and no d orbitals. Boron therefore forms covalent compounds, whereas aluminium forms Al³⁺ ions in solution. The maximum covalence of boron is 4 (as in BF₄⁻), while aluminium can reach 6 (as in AlF₆³⁻).
• Hydrides of group 13 are electron-deficient. The hydrides of group 14 are precise (exact number of electrons), and of groups 15 to 17 are electron-rich (they have lone pairs).

Group 14: carbon family

• C is a non-metal, Si and Ge are metalloids, and Sn and Pb are metals.
• Oxidation states: +4 and +2. Carbon shows mainly +4, but also −4 in carbides. Down the group the +2 state becomes more stable: Ge(II) < Sn(II) < Pb(II) in stability. Pb(II) is more stable than Pb(IV).
• Catenation is the self-linking of atoms of an element to form chains and rings. The order is C >> Si > Ge ≈ Sn > Pb. The bond enthalpy of C–C (348 kJ mol⁻¹) is higher than that of Si–Si (297 kJ mol⁻¹), Ge–Ge (260) or Sn–Sn (240).
• Allotropes of carbon:

• Oxides: CO is neutral. CO₂, SiO₂ and GeO₂ are acidic. SnO₂ and PbO₂ (and SnO, PbO) are amphoteric.
• Anomaly of carbon: small size, high electronegativity, and no d orbitals. Carbon shows maximum covalence of 4, whereas silicon forms SiF₆²⁻ using vacant 3d orbitals. Carbon forms stable pπ–pπ multiple bonds (C=C, C≡C, C=O). Silicon does not, which is why CO₂ is a gas and SiO₂ is a network solid.
• Halides: CCl₄ is not hydrolysed by water because carbon has no d orbitals to accept lone pairs of water. SiCl₄ is easily hydrolysed because Si has vacant 3d orbitals.
• Hydrides (MH₄): thermal stability falls down the group: CH₄ > SiH₄ > GeH₄ > SnH₄ > PbH₄. Boiling points rise down the group (CH₄ < SiH₄ < GeH₄ < SnH₄) because the molecules get larger. There are no hydrogen bonds between these molecules.

Group 15: nitrogen family

• N and P are non-metals, As and Sb are metalloids, and Bi is a metal.
• Oxidation states: −3, +3 and +5. The +3 state becomes more stable down the group (inert pair effect). Bi(V) is a strong oxidising agent. Nitrogen cannot form pentahalides (no d orbitals).
• Anomaly of nitrogen: small size, high electronegativity, high ionisation enthalpy and no d orbitals. Nitrogen is a diatomic gas because it forms a strong pπ–pπ triple bond (N≡N). The heavier members do not form stable pπ–pπ bonds and exist as P–P single-bonded P₄, As₄ units or solids. Nitrogen has a maximum covalence of 4 (as in NH₄⁺), whereas P can form PCl₅ and PF₆⁻.
• The N–N single bond is weak because of lone pair repulsion, but the P–P single bond is stronger. This is why nitrogen prefers N≡N and phosphorus prefers P₄.
• Allotropes of phosphorus:

• Hydrides (MH₃): NH₃, PH₃, AsH₃, SbH₃, BiH₃.

• Down the group: thermal stability falls (NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃), reducing character rises, basicity falls, and the M–H bond length increases.
• Boiling point anomaly: NH₃ is higher than PH₃ and AsH₃ because of hydrogen bonding, but lower than SbH₃ because SbH₃ is much larger. The order is PH₃ < AsH₃ < NH₃ < SbH₃.
• Oxides: acidic character falls and basic character rises down the group. N₂O₃ and N₂O₅ are acidic, P₄O₆ and P₄O₁₀ are acidic, As₂O₃ and Sb₂O₃ are amphoteric, and Bi₂O₃ is basic.

Group 16: oxygen family

• O and S are non-metals, Se and Te are metalloids, and Po is a metal (radioactive).
• Oxidation states: −2, +2, +4 and +6. Oxygen shows −2 almost always (−1 in peroxides). Because of its high electronegativity it shows positive states only with fluorine. S, Se and Te show +4 and +6. Down the group the +4 state becomes more stable than +6.
• Anomaly of oxygen: small size, high electronegativity, and no d orbitals. Oxygen is a diatomic gas, O=O, with a pπ–pπ double bond, while sulphur forms S₈ rings by single bonds, showing strong catenation. Oxygen cannot expand its octet, so it does not form OF₄ or OF₆, but sulphur forms SF₄ and SF₆.
• Allotropes of sulphur: rhombic sulphur (α-sulphur) is stable below 369 K. Monoclinic sulphur (β-sulphur) is stable above 369 K. Both contain puckered S₈ rings. At 369 K the two forms are in equilibrium (transition temperature). In the vapour state at high temperature, S₂ is paramagnetic like O₂.
• Hydrides (H₂M): H₂O, H₂S, H₂Se, H₂Te.

• Down the group: thermal stability decreases (H₂O > H₂S > H₂Se > H₂Te), acidic character increases (H₂O < H₂S < H₂Se < H₂Te), and reducing power increases. The acidity increases because the M–H bond gets weaker as the size of M grows.
• Boiling points: H₂S < H₂Se < H₂Te < H₂O. Water is abnormally high because of hydrogen bonding.
• Oxides (MO₂ and MO₃): the dioxides and trioxides of S, Se and Te are acidic, and the acidic character falls down the group. SO₂ is a reducing agent, whereas TeO₂ shows oxidising behaviour.

Group 17: halogens

• Colour and state: F₂ pale yellow gas, Cl₂ greenish yellow gas, Br₂ red-brown liquid, I₂ violet solid.
• ΔegH order: Cl > F > Br > I. Bond enthalpy order: Cl₂ > Br₂ > F₂ > I₂. F₂ has a low bond enthalpy because of strong lone pair repulsion in the small F–F molecule.
• Oxidising power: F₂ > Cl₂ > Br₂ > I₂. A halogen with a higher E° oxidises the halide ions of lower E°.
• Oxidation states: −1 for all. Chlorine, bromine and iodine also show +1, +3, +5 and +7 with more electronegative atoms (O, F). Fluorine shows only −1 since it is the most electronegative and has no d orbitals.
• Anomaly of fluorine: small size, highest electronegativity, low F–F bond enthalpy, no d orbitals and strong hydrogen bonding in HF. Fluorine shows only the −1 state and forms the strongest bonds with other elements, such as in HF and fluorides.
• Hydrogen halides (HX):

• Acid strength: HF < HCl < HBr < HI. Thermal stability: HF > HCl > HBr > HI. Reducing power: HF < HCl < HBr < HI. Boiling points: HCl < HBr < HI < HF (HF is hydrogen bonded).
• Oxoacids of chlorine: HOCl < HOClO < HOClO₂ < HOClO₃, so acid strength rises with oxidation state of the halogen: HClO < HClO₂ < HClO₃ < HClO₄. The conjugate anions become more stable because the negative charge spreads over more oxygen atoms. For hypohalous acids: HOCl > HOBr > HOI.
• Inter-halogen compounds contain two different halogens. The general formulae are XX′, XX′₃, XX′₅ and XX′₇, in which X is the larger atom. Examples: ClF, BrF₃, ClF₃, IF₅ and IF₇. Shapes: ClF₃ is T-shaped, IF₅ is square pyramidal and IF₇ is pentagonal bipyramidal. They are more reactive than the halogens (except F₂) because the X–X′ bond is weaker than the X–X bond (except F–F).

Group 18: noble gases

• They have completely filled valence shells (ns²np⁶; He 1s²). Their ionisation enthalpies are the highest in each period and fall down the group. Their electron gain enthalpies are large and positive.
• The atomic radius quoted for noble gases is the van der Waals radius, so it is larger than the covalent radii of the neighbouring halogens. Boiling points rise down the group because dispersion forces grow with size.
• Xenon, with the lowest ionisation enthalpy of the stable noble gases, forms compounds with F and O only. Examples with shapes from VSEPR theory:

• The oxidation states of Xe in XeF₂, XeF₄ and XeF₆ are +2, +4 and +6. Xenon forms compounds with fluorine and oxygen because they are the most electronegative elements.

Oxoacids and oxides across the block

• Oxides become more acidic across a period (Al₂O₃ amphoteric; SiO₂, P₄O₁₀, SO₃ and Cl₂O₇ acidic) and more basic down a group. Acidity of hydrides also increases across a period: CH₄ < NH₃ < H₂O < HF.
• Acidity of oxoacids increases across a period: H₄SiO₄ < H₃PO₄ < H₂SO₄ < HClO₄, as the central atom becomes more electronegative.

NEET traps

• 1. Boron trihalides: Lewis acidity increases from BF₃ to BI₃, which is the opposite of what the electronegativity order suggests.
• 2. Tl⁺ is more stable than Tl³⁺ and Pb²⁺ is more stable than Pb⁴⁺. Do not apply the group oxidation state as the stable one for heavy elements.
• 3. NH₃ has a higher boiling point than PH₃ and AsH₃, but not higher than SbH₃. Check every member before choosing the order.
• 4. H₂O is the least acidic and the most stable hydride of group 16. H₂Te is the most acidic and the strongest reducing agent.
• 5. HF is the weakest acid but has the highest boiling point among HX. HI is the strongest acid and the best reducing agent.
• 6. Cl has a more negative electron gain enthalpy than F, and the bond enthalpy of Cl₂ is higher than that of F₂, but F₂ is the strongest oxidising agent.
• 7. Carbon has the highest catenation tendency in group 14. Do not extend it to N–N or O–O single bonds, which are weak.
• 8. CCl₄ is not hydrolysed but SiCl₄ is. The reason is vacant d orbitals in silicon, not bond strength alone.
• 9. Fluorine shows only −1; chlorine, bromine and iodine show higher states.
• 10. Xenon forms compounds with F and O only. XeF₂ is linear (2 bond pairs, 3 lone pairs), not bent.

One-liners

• 1. The p-block runs from group 13 to group 18.
• 2. Boron is a non-metal; all other group 13 members are metals.
• 3. Ga has a smaller radius than expected because of poorly shielding 3d electrons.
• 4. The inert pair effect makes Tl⁺ more stable than Tl³⁺.
• 5. Catenation order in group 14 is C >> Si > Ge ≈ Sn > Pb.
• 6. Diamond is sp³ and graphite is sp²; fullerene C₆₀ has 20 hexagons and 12 pentagons.
• 7. White phosphorus is tetrahedral P₄ and glows in the dark.
• 8. Basic strength of group 15 hydrides falls from NH₃ to BiH₃.
• 9. Oxygen does not show positive oxidation states with elements other than fluorine.
• 10. Rhombic sulphur changes into monoclinic sulphur at 369 K.
• 11. H₂O has the highest boiling point among group 16 hydrides.
• 12. Fluorine shows only the −1 oxidation state.
• 13. Order of acid strength of HX: HF < HCl < HBr < HI.
• 14. Acid strength of chlorine oxoacids increases from HClO to HClO₄.
• 15. XeF₂ is linear, XeF₄ is square planar and XeF₆ is distorted octahedral.

Practice questions

1. Which p-block element shows +1 as its most stable oxidation state?

   1. Boron
   2. Aluminium
   3. Gallium
   4. Thallium
   Answer

   D. Thallium

   Because of the inert pair effect, Tl⁺ is more stable than Tl³⁺.

2. Which is the correct order of Lewis acid strength of boron trihalides?

   1. BCl₃ < BF₃ < BBr₃ < BI₃
   2. BI₃ < BBr₃ < BCl₃ < BF₃
   3. BF₃ < BCl₃ < BBr₃ < BI₃
   4. BF₃ < BBr₃ < BCl₃ < BI₃
   Answer

   C. BF₃ < BCl₃ < BBr₃ < BI₃

   pπ–pπ back bonding is strongest in BF₃ (2p–2p) and weakest in BI₃, so acidity increases from BF₃ to BI₃.

3. Which element of group 14 shows the greatest tendency for catenation?

   1. Silicon
   2. Carbon
   3. Tin
   4. Germanium
   Answer

   B. Carbon

   The C–C bond is the strongest homonuclear bond in the group (348 kJ mol⁻¹).

4. Which is the correct order of thermal stability of group 15 hydrides?

   1. NH₃ > PH₃ > AsH₃ > SbH₃
   2. PH₃ > NH₃ > AsH₃ > SbH₃
   3. NH₃ > AsH₃ > PH₃ > SbH₃
   4. SbH₃ > AsH₃ > PH₃ > NH₃
   Answer

   A. NH₃ > PH₃ > AsH₃ > SbH₃

   Stability falls as the M–H bond lengthens and weakens down the group.

5. Which is the correct order of boiling points of the hydrides of group 16?

   1. H₂Te < H₂Se < H₂S < H₂O
   2. H₂S < H₂Se < H₂Te < H₂O
   3. H₂S < H₂O < H₂Se < H₂Te
   4. H₂O < H₂S < H₂Se < H₂Te
   Answer

   B. H₂S < H₂Se < H₂Te < H₂O

   Boiling points rise from H₂S to H₂Te with size, but H₂O is highest due to hydrogen bonding.

6. Which allotrope of phosphorus glows in the dark and is kept under water?

   1. White phosphorus
   2. Red phosphorus
   3. Black phosphorus
   4. Violet phosphorus
   Answer

   A. White phosphorus

   White phosphorus (P₄) is very reactive and glows in the dark.

7. The shape of XeF₄ is

   1. see-saw
   2. tetrahedral
   3. square planar
   4. octahedral
   Answer

   C. square planar

   Xe has 4 bond pairs and 2 lone pairs (sp³d²), giving a square planar shape.

8. Which halogen shows only the −1 oxidation state in its compounds?

   1. Chlorine
   2. Bromine
   3. Iodine
   4. Fluorine
   Answer

   D. Fluorine

   F is the most electronegative element and has no d orbitals.

9. Which hydrogen halide is the strongest acid in water?

   1. HBr
   2. HF
   3. HI
   4. HCl
   Answer

   C. HI

   The H–I bond is the weakest, so HI ionises most easily.

10. Which hydride of group 16 is the strongest reducing agent?

    1. H₂Te
    2. H₂Se
    3. H₂O
    4. H₂S
    Answer

    A. H₂Te

    The H–Te bond is the weakest, and reducing power rises down the group.

11. Which halogen molecule has the highest bond dissociation enthalpy?

    1. Br₂
    2. F₂
    3. I₂
    4. Cl₂
    Answer

    D. Cl₂

    Cl–Cl is 242.6 kJ mol⁻¹; F₂ is lower because of lone pair repulsion.

12. In which allotrope of carbon is every carbon atom sp³ hybridised?

    1. Graphite
    2. Diamond
    3. Fullerene (C₆₀)
    4. Carbon dioxide
    Answer

    B. Diamond

    Each C in diamond is bonded to four other C atoms by sp³ bonds.

13. Which of the following compounds does NOT exist?

    1. SF₆
    2. NF₅
    3. AsF₅
    4. PF₅
    Answer

    B. NF₅

    Nitrogen has no d orbitals, so it cannot expand its octet to form a pentahalide.

14. Which of the following species is the strongest oxidising agent?

    1. Pb⁴⁺
    2. Ge⁴⁺
    3. Sn⁴⁺
    4. Si⁴⁺
    Answer

    A. Pb⁴⁺

    Because of the inert pair effect, the +2 state is most stable for Pb, so Pb(IV) readily gains electrons.

15. The shape of IF₇ is

    1. octahedral
    2. trigonal bipyramidal
    3. square pyramidal
    4. pentagonal bipyramidal
    Answer

    D. pentagonal bipyramidal

    IF₇ has 7 bond pairs and no lone pairs on I (sp³d³).

16. Given below are two statements: Statement I: Boron trihalides act as Lewis acids. Statement II: Boron has an empty 2p orbital in BX₃.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Both correct; the empty 2p orbital accepts an electron pair.

17. Given below are two statements: Statement I: Pb(IV) is more stable than Pb(II). Statement II: Sn(II) is a reducing agent.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    Pb(II) is more stable because of the inert pair effect, so I is incorrect. Sn²⁺ is easily oxidised to Sn⁴⁺, so II is correct.

18. Given below are two statements: Statement I: NH₃ has a higher boiling point than PH₃. Statement II: NH₃ has a higher boiling point than SbH₃.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    I is correct (hydrogen bonding). SbH₃ is much larger and boils at 254.6 K, above NH₃ (238.5 K), so II is incorrect.

19. Which of the following statements are correct? 1. H₂Te is the most acidic hydride of group 16. 2. Thermal stability of group 16 hydrides decreases down the group. 3. H₂O is a stronger reducing agent than H₂S.

    1. 1 and 3 only
    2. 1 and 2 only
    3. 2 and 3 only
    4. 1, 2 and 3
    Answer

    B. 1 and 2 only

    Reducing power increases down the group, so H₂S > H₂O; 3 is wrong.

20. Which of the following statements are correct? 1. HF has the highest boiling point among the hydrogen halides. 2. HI is the weakest reducing agent among the hydrogen halides. 3. HClO₄ is the strongest oxoacid of chlorine.

    1. 2 and 3 only
    2. 1, 2 and 3
    3. 1 and 2 only
    4. 1 and 3 only
    Answer

    D. 1 and 3 only

    HI is the strongest reducing agent, so 2 is wrong. 1 and 3 are correct.

21. Given below are two statements: Statement I: Diamond conducts electricity. Statement II: The carbon atoms in graphite are sp³ hybridised.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    Diamond is an insulator; graphite carbons are sp². Both are incorrect.

22. Given below are two statements: Statement I: XeF₂ is a linear molecule. Statement II: Xenon in XeF₂ has three lone pairs.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    2 bond pairs + 3 lone pairs (sp³d) give a linear shape. Both correct.

23. Given below are two statements: Statement I: CCl₄ is readily hydrolysed by water. Statement II: Carbon has vacant d orbitals.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    Carbon has no d orbitals, so CCl₄ is not hydrolysed. Both are incorrect.

24. Which of the following statements are correct? 1. Rhombic sulphur is stable below 369 K. 2. Monoclinic sulphur contains S₈ rings. 3. White phosphorus is more reactive than red phosphorus.

    1. 1, 2 and 3
    2. 1 and 2 only
    3. 2 and 3 only
    4. 1 and 3 only
    Answer

    A. 1, 2 and 3

    All three statements are correct.

25. Given below are two statements: Statement I: The basic strength of group 15 hydrides increases down the group. Statement II: The size of the central atom increases down the group.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    Basic strength falls from NH₃ to BiH₃, so I is incorrect. II is correct.

26. Assertion (A): SiCl₄ is hydrolysed by water but CCl₄ is not. Reason (R): Silicon has vacant 3d orbitals that can accept lone pairs from water.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    Carbon has no d orbitals. The d orbitals in Si allow attack by water.

27. Assertion (A): Fluorine has a less negative electron gain enthalpy than chlorine. Reason (R): Fluorine has the highest electronegativity.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    Both true. The low ΔegH is due to repulsion in the small 2p subshell, not to high electronegativity.

28. Assertion (A): Graphite is soft and conducts electricity. Reason (R): The carbon atoms in graphite are sp³ hybridised.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    C. A is true but R is false

    A is true; R is false, since graphite carbons are sp² and have a free electron per atom.

29. Assertion (A): BF₃ is the strongest Lewis acid among the boron trihalides. Reason (R): The pπ–pπ back bonding is most effective in BF₃.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    D. A is false but R is true

    A is false (BI₃ is strongest). R is true, and it is the reason BF₃ is the weakest.

30. Assertion (A): Tl(III) compounds are strong oxidising agents. Reason (R): Tl⁺ is more stable than Tl³⁺ because of the inert pair effect.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    Tl³⁺ readily gains electrons to reach the more stable Tl⁺.

31. Assertion (A): Rhombic sulphur is stable below 369 K. Reason (R): Monoclinic sulphur contains S₈ rings.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    Both true. The ring structure of monoclinic sulphur does not explain the stability range of rhombic sulphur.

32. Match List I with List II. List I: a. Fullerene b. White phosphorus c. Diamond d. Graphite. List II: i. sp³ three-dimensional network ii. Layered sp² structure that conducts electricity iii. 20 six-membered and 12 five-membered rings iv. Discrete tetrahedral P₄ molecules.

    1. a-iv, b-iii, c-i, d-ii
    2. a-iii, b-iv, c-ii, d-i
    3. a-iii, b-iv, c-i, d-ii
    4. a-i, b-iv, c-iii, d-ii
    Answer

    C. a-iii, b-iv, c-i, d-ii

    Fullerene has 20 hexagons and 12 pentagons; white P is P₄; diamond is sp³; graphite is layered sp².

33. Match List I with List II. List I: a. XeF₂ b. XeF₄ c. XeF₆ d. IF₅. List II: i. Distorted octahedral ii. Linear iii. Square pyramidal iv. Square planar.

    1. a-ii, b-iv, c-iii, d-i
    2. a-ii, b-i, c-iv, d-iii
    3. a-ii, b-iv, c-i, d-iii
    4. a-iv, b-ii, c-i, d-iii
    Answer

    C. a-ii, b-iv, c-i, d-iii

    XeF₂ is linear, XeF₄ square planar, XeF₆ distorted octahedral, IF₅ square pyramidal.

34. Match List I with List II. List I: a. Group 13 b. Group 15 c. Group 16 d. Group 17. List II: i. ns²np³ ii. ns²np¹ iii. ns²np⁵ iv. ns²np⁴.

    1. a-i, b-ii, c-iv, d-iii
    2. a-ii, b-i, c-iii, d-iv
    3. a-ii, b-i, c-iv, d-iii
    4. a-iv, b-i, c-ii, d-iii
    Answer

    C. a-ii, b-i, c-iv, d-iii

    Group 13 = ns²np¹, group 15 = ns²np³, group 16 = ns²np⁴, group 17 = ns²np⁵.

35. Match List I with List II. List I: a. CO b. Tl₂O c. B₂O₃ d. Al₂O₃. List II: i. Acidic ii. Amphoteric iii. Basic iv. Neutral.

    1. a-iv, b-iii, c-ii, d-i
    2. a-i, b-iii, c-iv, d-ii
    3. a-iv, b-iii, c-i, d-ii
    4. a-iv, b-ii, c-i, d-iii
    Answer

    C. a-iv, b-iii, c-i, d-ii

    CO is neutral, Tl₂O basic, B₂O₃ acidic, Al₂O₃ amphoteric.

36. Match List I with List II. List I: a. NH₃ b. PH₃ c. SbH₃ d. AsH₃. List II (boiling point, K): i. 185.5 ii. 254.6 iii. 238.5 iv. 210.6.

    1. a-iii, b-iv, c-ii, d-i
    2. a-iii, b-i, c-iv, d-ii
    3. a-ii, b-i, c-iii, d-iv
    4. a-iii, b-i, c-ii, d-iv
    Answer

    D. a-iii, b-i, c-ii, d-iv

    NH₃ 238.5, PH₃ 185.5, SbH₃ 254.6, AsH₃ 210.6.

37. Match List I with List II. List I: a. F₂ b. Cl₂ c. Br₂ d. I₂. List II: i. Violet solid ii. Highest bond enthalpy iii. Red-brown liquid iv. Strongest oxidising agent.

    1. a-iv, b-ii, c-i, d-iii
    2. a-ii, b-iv, c-iii, d-i
    3. a-iii, b-ii, c-iv, d-i
    4. a-iv, b-ii, c-iii, d-i
    Answer

    D. a-iv, b-ii, c-iii, d-i

    F₂ is the strongest oxidising agent, Cl₂ has the highest bond enthalpy, Br₂ is a liquid, I₂ is a violet solid.

38. How many lone pairs are present on the central atom in XeF₄?

    1. 1
    2. 2
    3. 3
    4. 0
    Answer

    B. 2

    Xe has 8 valence electrons; 4 are used in bonds, leaving 4 electrons = 2 lone pairs.

39. The sum of the numbers of lone pairs on the xenon atom in XeF₂ and XeF₄ is

    1. 5
    2. 4
    3. 6
    4. 7
    Answer

    A. 5

    XeF₂: 3 lone pairs; XeF₄: 2 lone pairs; total 5.

40. The sum of the oxidation states of iodine in IF₅ and IF₇ is

    1. 10
    2. 14
    3. 12
    4. 2
    Answer

    C. 12

    Fluorine is −1: IF₅ gives +5 and IF₇ gives +7; 5 + 7 = 12.

41. A group 15 element forms an oxide that is basic in nature. The element is

    1. P
    2. Bi
    3. N
    4. As
    Answer

    B. Bi

    Basic character of oxides increases down the group; Bi₂O₃ is basic.

42. On heating rhombic sulphur above 369 K it changes slowly into

    1. S₆ rings
    2. S₂ molecules
    3. monoclinic sulphur
    4. white phosphorus
    Answer

    C. monoclinic sulphur

    369 K is the transition temperature between rhombic and monoclinic sulphur.

43. Which of the following is the strongest acid?

    1. HOCl
    2. HOI
    3. HOBr
    4. HOClO (HClO₂)
    Answer

    D. HOClO (HClO₂)

    Acid strength rises with oxidation state of the halogen (HClO₂ > HOCl), and for hypohalous acids falls from Cl to I.

44. The Cl–Cl bond enthalpy is 242.6 kJ mol⁻¹. The energy needed to dissociate 7.1 g of Cl₂ (molar mass 71 g mol⁻¹) into atoms is

    1. 24.26 kJ
    2. 2.426 kJ
    3. 242.6 kJ
    4. 12.13 kJ
    Answer

    A. 24.26 kJ

    7.1 g = 0.1 mol; energy = 0.1 × 242.6 = 24.26 kJ.

45. A molecule has a central atom with 3 bond pairs and 2 lone pairs, as in ClF₃. Its shape is

    1. trigonal pyramidal
    2. see-saw
    3. T-shaped
    4. bent
    Answer

    C. T-shaped

    Five electron pairs give a trigonal bipyramidal arrangement. Lone pairs occupy equatorial positions, leaving a T shape.