Group 15 Elements General Introduction – Electronic configuration and general trends in physical and: NEET MCQs with Answers

15 free questions on Group 15 Elements General Introduction – Electronic configuration and general trends in physical and for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Electronic configuration and general trends in physical and chemical properties of elements across the periods and down the groups; unique behaviour of the first element in each group.

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Question 1

Previous-year question
Identify the incorrect statement. [Re-NEET 2024]
  1. PEt₃ and AsPh₃ as ligands can form dπ–dπ bonds with transition metals.
  2. The N–N single bond is as strong as the P–P single bond.
  3. Nitrogen has a unique ability to form pπ–pπ multiple bonds with nitrogen, carbon and oxygen.
  4. Nitrogen cannot form dπ–pπ bonds like the heavier elements of its group.
Show answer
Answer: (B) The N–N single bond is as strong as the P–P single bond.
Explanation: The N–N single bond is weaker than the P–P single bond because the short N–N bond causes strong repulsion between the non-bonding electron pairs on the two nitrogen atoms, so the claim that they are equally strong is incorrect. P and As have vacant d orbitals and can form dπ–dπ bonds with transition metals, nitrogen forms pπ–pπ multiple bonds, and it cannot form dπ–pπ bonds since it lacks d orbitals.

Question 2

Previous-year question
Urea reacts with water to form A, which will decompose to form B. When B is passed through Cu²⁺(aq), a deep blue colour solution C is formed. What is the formula of C? [NEET 2020]
  1. [Cu(NH₃)₄]²⁺
  2. Cu(OH)₂
  3. CuCO₃·Cu(OH)₂
  4. CuSO₄
Show answer
Answer: (A) [Cu(NH₃)₄]²⁺
Explanation: Urea hydrolyses to ammonium carbamate/carbonate (A), which decomposes to ammonia (B); ammonia with Cu²⁺ gives the deep blue tetraamminecopper(II) ion.

Question 3

Previous-year question
Match the following: (a) Pure nitrogen – (b) Haber process – (c) Contact process – (d) Deacon's process with (i) chlorine, (ii) sulphuric acid, (iii) ammonia, (iv) sodium azide or barium azide. The correct option is [NEET 2019]
  1. a–(iv), b–(iii), c–(ii), d–(i)
  2. a–(ii), b–(iv), c–(i), d–(iii)
  3. a–(iii), b–(iv), c–(ii), d–(i)
  4. a–(iv), b–(iii), c–(i), d–(ii)
Show answer
Answer: (A) a–(iv), b–(iii), c–(ii), d–(i)
Explanation: Pure nitrogen comes from azides, the Haber process makes ammonia, the contact process makes sulphuric acid and Deacon's process makes chlorine.

Question 4

Previous-year question
The correct order of N-compounds in decreasing order of oxidation states is [NEET 2018]
  1. HNO₃, NO, N₂, NH₄Cl
  2. HNO₃, NO, NH₄Cl, N₂
  3. NH₄Cl, N₂, NO, HNO₃
  4. HNO₃, NH₄Cl, NO, N₂
Show answer
Answer: (A) HNO₃, NO, N₂, NH₄Cl
Explanation: The oxidation states are +5 in HNO₃, +2 in NO, 0 in N₂ and –3 in NH₄Cl.

Question 5

Previous-year question
Which is the correct statement for the given acids? [NEET 2016]
  1. Phosphinic acid is a diprotic acid while phosphonic acid is a monoprotic acid
  2. Phosphinic acid is a monoprotic acid while phosphonic acid is a diprotic acid
  3. Both are triprotic acids
  4. Both are diprotic acids
Show answer
Answer: (B) Phosphinic acid is a monoprotic acid while phosphonic acid is a diprotic acid
Explanation: Phosphinic (hypophosphorous) acid H₃PO₂ has one –OH and is monoprotic; phosphonic (phosphorous) acid H₃PO₃ has two –OH groups and is diprotic.

Question 6

Previous-year question
Nitrogen dioxide and sulphur dioxide have some properties in common. Which property is shown by one of these compounds, but not by the other? [AIPMT 2015]
  1. is a reducing agent
  2. is soluble in water
  3. is used as a food-preservative
  4. forms 'acid-rain'
Show answer
Answer: (C) is used as a food-preservative
Explanation: Only sulphur dioxide is used as a food preservative; both are reducing agents, soluble in water and contribute to acid rain.

Question 7

Previous-year question
Strong reducing behaviour of H₃PO₂ is due to [AIPMT 2015]
  1. presence of one –OH group and two P–H bonds
  2. high electron gain enthalpy of phosphorus
  3. high oxidation state of phosphorus
  4. presence of two –OH groups and one P–H bond
Show answer
Answer: (A) presence of one –OH group and two P–H bonds
Explanation: The two directly bonded hydrogen atoms are easily released, so H₃PO₂ is a strong reducing agent.
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Question 8

Previous-year question
Strong reducing behaviour of H₃PO₂ is due to: [Re-AIPMT, 2015]
  1. High oxidation state of phosphorus
  2. Presence of two –OH groups and one P–H bond
  3. Presence of one –OH group and two P–H bonds
  4. High electron gain enthalpy of phosphorus
Show answer
Answer: (C) Presence of one –OH group and two P–H bonds
Explanation: Hypophosphorous acid H₃PO₂ has one –OH group and two P–H bonds; the P–H bonds make it a strong reducing agent.

Question 9

Previous-year question
Which statement is not valid for oxy-acids of phosphorus? [AIPMT Pre, 2012]
  1. All oxyacids contain tetrahedral four-coordinated phosphorus
  2. All oxyacids contain at least one P=O unit and one P–OH group
  3. Orthophosphoric acid is used in the manufacture of triple superphosphate
  4. Hypophosphorous acid is a diprotic acid
Show answer
Answer: (D) Hypophosphorous acid is a diprotic acid
Explanation: H₃PO₂ has only one ionisable P–OH hydrogen; its other two hydrogens are bonded directly to phosphorus. It is therefore monoprotic, so calling it diprotic is the invalid statement.

Question 10

Previous-year question
How many bridging oxygen atoms are present in P₄O₁₀? [AIPMT Pre, 2010]
  1. 6
  2. 2
  3. 5
  4. 4
Show answer
Answer: (A) 6
Explanation: P₄O₁₀ has six bridging (P–O–P) oxygen atoms and four terminal (P=O) oxygen atoms.

Question 11

Previous-year question
Oxidation states of P in H₄P₂O₅, H₄P₂O₆, H₄P₂O₇ are respectively: [AIPMT Pre, 2010]
  1. +3, +4, +5
  2. +3, +5, +4
  3. +5, +3, +4
  4. +5, +4, +3
Show answer
Answer: (A) +3, +4, +5
Explanation: Calculating from H, O totals: P is +3 in H₄P₂O₅, +4 in H₄P₂O₆ and +5 in H₄P₂O₇.

Question 12

The melting points of the group 15 elements do not rise steadily down the group. Which of the following orders is correct?
  1. N < P < As < Sb < Bi
  2. N < P < Bi < Sb < As
  3. P < N < As < Bi < Sb
  4. Bi < Sb < As < P < N
Show answer
Answer: (B) N < P < Bi < Sb < As
Explanation: Melting point increases from nitrogen, a diatomic gas, up to arsenic, which has the highest value in the group. It then decreases through antimony to bismuth. The correct sequence is therefore N < P < Bi < Sb < As, with Bi falling below Sb and As.

Question 13

Down group 15, the covalent radius increases considerably from nitrogen to phosphorus, but only slightly from arsenic to bismuth. What accounts for the small increase among the heavier members?
  1. Heavier members lose their outer s electrons and so have fewer valence electrons
  2. The nuclear charge decreases from arsenic to bismuth
  3. The number of electron shells stops increasing after arsenic
  4. Completely filled d and/or f orbitals in the heavier members shield the nuclear charge poorly
Show answer
Answer: (D) Completely filled d and/or f orbitals in the heavier members shield the nuclear charge poorly
Explanation: Each step down the group adds a shell, which tends to increase the radius. In As, Sb and Bi, however, filled d and/or f orbitals lie inside the valence shell and screen the rising nuclear charge poorly. The outer electrons feel a stronger pull, so the radius grows only a little. Nitrogen to phosphorus has no such filled inner d subshell, so the increase there is large.

Question 14

Nitrogen forms a much stronger multiple bond than phosphorus, yet it shows a weaker tendency than phosphorus to form chains of like atoms. What is the best explanation?
  1. Vacant d orbitals on nitrogen withdraw electron density from the N–N bond
  2. The N–N single bond is weaker than the P–P single bond, because non-bonding electrons on the closely placed N atoms repel strongly
  3. The nitrogen atom is larger than the phosphorus atom, so N–N overlap is poor
  4. Nitrogen is less electronegative than phosphorus, so the N–N bond is less polar
Show answer
Answer: (B) The N–N single bond is weaker than the P–P single bond, because non-bonding electrons on the closely placed N atoms repel strongly
Explanation: Catenation depends on the strength of the single bond between like atoms. The N–N bond is very short, so the lone pairs on the two nitrogen atoms repel each other strongly and weaken it. The P–P single bond is longer, suffers less of this repulsion and is stronger, so phosphorus catenates more readily. Nitrogen's strong triple bond comes from pπ–pπ overlap and does not help chain formation.

Question 15

The H–N–H angle in NH₃ is about 107°, whereas the H–P–H angle in PH₃ is only about 93.6°. Which is the best explanation of this difference?
  1. In PH₃ the bond pairs lie closer to phosphorus than they do to nitrogen in NH₃, which increases bond pair repulsion
  2. In PH₃ the P–H bonds are stronger than the N–H bonds of NH₃, which pulls the hydrogen atoms together
  3. In PH₃ the P–H bonds use almost pure p orbitals of phosphorus, with the lone pair in an s-rich orbital
  4. In PH₃ the phosphorus atom is sp³d hybridised, which narrows the angle
Show answer
Answer: (C) In PH₃ the P–H bonds use almost pure p orbitals of phosphorus, with the lone pair in an s-rich orbital
Explanation: Pure p orbitals are mutually at 90°, so an angle close to 90° shows that the central atom uses almost unhybridised p orbitals for bonding. Phosphorus is larger and less electronegative than nitrogen, so its bond pairs sit farther from the centre and repel each other less. Hybridisation is therefore not favoured, and the bond angle falls towards 90° down the group.

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