p-Block Elements (Groups 15-18)

10 NEET-UG practice questions on p-Block Elements (Groups 15-18), part of Chemistry. The first 10 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

1. Question 1 · Medium

   Given below are two statements: Statement I: The two O–O bonds in ozone have different lengths. Statement II: Ozone is thermodynamically unstable with respect to oxygen because its decomposition is exothermic and increases the entropy. Choose the correct answer.

   1. ABoth Statement I and Statement II are correct
   2. BStatement I is correct but Statement II is incorrect
   3. CBoth Statement I and Statement II are incorrect
   4. DStatement I is incorrect but Statement II is correct

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   Answer: D · Statement I is incorrect but Statement II is correct

   Ozone is a resonance hybrid, so both O–O bonds have the same length of 128 pm, and Statement I is false. Its decomposition releases heat and raises the entropy, so ΔG is negative and Statement II is true.

   • A: Statement I is false; resonance makes the bonds equal.
   • B: The statements are reversed; I is false and II is true.
   • C: Statement II is true; decomposition of O₃ has a negative ΔG.

   Remember this

   Resonance makes the two O–O bonds equal, between a single and a double bond in length.

2. Question 2 · Hard

   Complete hydrolysis of XeF₆ gives a xenon compound. Which option correctly names that compound and its shape?

   1. AXeOF₄, square pyramidal
   2. BXeO₃, pyramidal
   3. CXeO₂F₂, see-saw
   4. DXeO₃, trigonal planar

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   Answer: B · XeO₃, pyramidal

   XeF₆ + 3H₂O → XeO₃ + 6HF. In XeO₃, xenon has three bond pairs and one lone pair on its sp³ arrangement, so the molecule is pyramidal.

   • A: XeOF₄ comes from partial hydrolysis.
   • C: XeO₂F₂ also comes from partial hydrolysis.
   • D: Right compound, but it ignores the lone pair on xenon.

   Remember this

   XeF₆ gives XeOF₄ then XeO₂F₂ on partial hydrolysis, and XeO₃ on complete hydrolysis.

3. Question 3 · Hard

   On heating, phosphorous acid disproportionates: H₃PO₃ → H₃PO₄ + PH₃ (unbalanced). In the balanced reaction, what is the ratio of phosphorus atoms oxidised to those reduced?

   1. A1 : 3
   2. B2 : 1
   3. C1 : 1
   4. D3 : 1

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   Answer: D · 3 : 1

   P goes from +3 to +5 in H₃PO₄ (loses 2 e⁻) and from +3 to −3 in PH₃ (gains 6 e⁻). Electron balance needs three oxidised per one reduced, so 4H₃PO₃ → 3H₃PO₄ + PH₃.

   • A: Inverts the ratio, mixing up oxidation and reduction.
   • B: Takes the change to PH₃ as 4 electrons rather than 6.
   • C: Assumes equal numbers without balancing electrons.

   Remember this

   In disproportionation, balance the electrons lost and gained; the oxidation-state changes set the ratio.

4. Question 4 · Medium

   Arrange the steps of the contact process in the correct order. A. Diluting oleum with water B. Catalytic oxidation of SO₂ over V₂O₅ C. Burning sulphur or a sulphide ore to give SO₂ D. Absorbing SO₃ in concentrated H₂SO₄

   1. AC, B, D, A
   2. BC, B, A, D
   3. CB, C, D, A
   4. DC, D, B, A

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   Answer: A · C, B, D, A

   SO₂ is produced first, then oxidised over V₂O₅ to SO₃. SO₃ is absorbed in sulphuric acid to give oleum, H₂S₂O₇, which is diluted to the required strength.

   • B: Dilutes oleum before it has been formed.
   • C: Oxidises SO₂ before it has been made.
   • D: Absorbs SO₃ before it has been made.

   Remember this

   SO₃ is not dissolved directly in water; it goes through oleum first.

5. Question 5 · Medium

   Match List-I with List-II. List-I (interhalogen) A. ClF B. ClF₃ C. IF₅ D. IF₇ List-II (shape) I. Pentagonal bipyramidal II. Bent T-shaped III. Linear IV. Square pyramidal

   1. AA-III, B-II, C-I, D-IV
   2. BA-III, B-II, C-IV, D-I
   3. CA-III, B-IV, C-II, D-I
   4. DA-II, B-III, C-IV, D-I

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   Answer: B · A-III, B-II, C-IV, D-I

   ClF is diatomic, so linear. ClF₃ has 3 bond pairs and 2 lone pairs (bent T), IF₅ has 5 and 1 (square pyramidal), and IF₇ has 7 and 0 (pentagonal bipyramidal).

   • A: Swaps the shapes of IF₅ and IF₇.
   • C: Swaps the shapes of ClF₃ and IF₅.
   • D: Swaps the shapes of ClF and ClF₃.

   Remember this

   Interhalogen shapes follow VSEPR from the bond and lone pairs on the central, larger halogen.

6. Question 6 · Easy

   Copper reacts with dilute nitric acid. Identify the nitrogen-containing gas evolved.

   1. ANH₃
   2. BNO₂
   3. CN₂O
   4. DNO

   Answer and explanation in LearnFree account

7. Question 7 · Easy

   What is the shape of the XeF₄ molecule?

   1. ASquare planar
   2. BTetrahedral
   3. CSee-saw
   4. DSquare pyramidal

   Answer and explanation in LearnFree account

8. Question 8 · Medium

   Copper(II) sulphate solution is treated with excess aqueous ammonia and turns deep blue. Which species is responsible for this colour?

   1. ACu(OH)₂
   2. BCuO
   3. C[Cu(NH₃)₄]²⁺
   4. D[Cu(NH₃)₂]⁺

   Answer and explanation in LearnFree account

9. Question 9 · Medium

   Chlorine gas is passed into hot, concentrated NaOH solution. Which chlorine-containing salts are formed?

   1. ANaCl and NaClO₃
   2. BNaClO₃ and NaClO₄
   3. CNaCl and NaOCl
   4. DNaOCl and NaClO₄

   Answer and explanation in LearnFree account

10. Question 10 · Easy

    Hypophosphorous acid, H₃PO₂, has a basicity of

    1. A2
    2. B3
    3. C0
    4. D1

    Answer and explanation in LearnFree account