Group 16 Elements General Introduction – Electronic configuration and general trends in physical and: NEET MCQs with Answers

12 free questions on Group 16 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.

Practise all 149 Group 16 Elements General Introduction – Electronic configuration and general trends in physical and questions free Take a free timed test

Question 1

Previous-year question
Which of the following oxoacids of sulphur has an –O–O– linkage? [NEET 2020]
  1. H₂SO₄, sulphuric acid
  2. H₂S₂O₈, peroxodisulphuric acid
  3. H₂S₂O₇, pyrosulphuric acid
  4. H₂SO₃, sulphurous acid
Show answer
Answer: (B) H₂S₂O₈, peroxodisulphuric acid
Explanation: Only peroxodisulphuric acid contains the peroxide bridge between the two sulphur-containing halves.

Question 2

Previous-year question
Which is the correct thermal stability order for H₂E (E = O, S, Se, Te and Po)? [NEET 2019]
  1. H₂S < H₂O < H₂Se < H₂Te < H₂Po
  2. H₂O < H₂S < H₂Se < H₂Te < H₂Po
  3. H₂Po < H₂Te < H₂Se < H₂S < H₂O
  4. H₂Se < H₂Te < H₂Po < H₂O < H₂S
Show answer
Answer: (C) H₂Po < H₂Te < H₂Se < H₂S < H₂O
Explanation: Bond strength and hence thermal stability decrease down the group, so H₂O is the most stable and H₂Po the least.

Question 3

Previous-year question
In which pair of ions do both the species contain an S–S bond? [NEET 2017]
  1. S₄O₆²⁻, S₂O₃²⁻
  2. S₂O₇²⁻, S₂O₈²⁻
  3. S₄O₆²⁻, S₂O₇²⁻
  4. S₂O₇²⁻, S₂O₃²⁻
Show answer
Answer: (A) S₄O₆²⁻, S₂O₃²⁻
Explanation: Tetrathionate has a chain of four sulphur atoms and thiosulphate has a central S bonded to a terminal S; the pyro and peroxodisulphate ions have S–O–S and O–O links instead.

Question 4

Previous-year question
Which of the statements given below is incorrect? [AIPMT 2015]
  1. Cl₂O₇ is an anhydride of perchloric acid
  2. O₃ molecule is bent
  3. ONF is isoelectronic with O₂N⁻
  4. OF₂ is an oxide of fluorine
Show answer
Answer: (D) OF₂ is an oxide of fluorine
Explanation: Since fluorine is more electronegative than oxygen, OF₂ is a fluoride of oxygen, not an oxide of fluorine.

Question 5

Previous-year question
The acidity of diprotic acids in aqueous solutions increases in the order [AIPMT 2014]
  1. H₂S < H₂Se < H₂Te
  2. H₂Se < H₂S < H₂Te
  3. H₂Te < H₂S < H₂Se
  4. H₂Se < H₂Te < H₂S
Show answer
Answer: (A) H₂S < H₂Se < H₂Te
Explanation: Acid strength increases as the H–E bond becomes longer and weaker down the group.

Question 6

Previous-year question
Name the gas that can readily decolourise acidified KMnO₄ solution: [2017]
  1. SO₂
  2. NO₂
  3. P₂O₅
  4. CO₂
Show answer
Answer: (A) SO₂
Explanation: SO₂ is a reducing agent and readily decolourises acidified KMnO₄ (Mn⁷⁺ → Mn²⁺).

Question 7

Previous-year question
Roasting of sulphides gives gas X (colourless, choking smell of burnt sulphur, causes acid rain, acidic aqueous solution acting as reducing agent, and its acid has never been isolated). X is: [NEET-UG, 2013]
  1. SO₃
  2. H₂S
  3. SO₂
  4. CO₂
Show answer
Answer: (C) SO₂
Explanation: Roasting metal sulphides releases SO₂ — a colourless, choking, reducing gas whose acid (H₂SO₃) exists only in solution and has never been isolated.
Doing well? Create a free account to take timed tests on this topic and see your weak areas.

Question 8

Statement I: The melting point of the group 16 elements rises from oxygen to tellurium. Statement II: Polonium has a higher melting point than tellurium.
  1. Both Statement I and Statement II are false
  2. Statement I is true but Statement II is false
  3. Both Statement I and Statement II are true
  4. Statement I is false but Statement II is true
Show answer
Answer: (B) Statement I is true but Statement II is false
Explanation: The melting and boiling point order in group 16 is O Po. The values rise as far as tellurium because of increasing size and intermolecular forces. Polonium breaks the trend and lies below tellurium, so the rise does not continue to the last member of the group.

Question 9

Which of the following statements about the halides of group 16 elements is incorrect?
  1. The stability of the halides increases in the order fluoride < chloride < bromide < iodide
  2. Among the hexahalides, only the hexafluorides are stable
  3. The well-known monohalides, such as S₂Cl₂, exist as dimers
  4. The tetrafluorides have sp³d hybridisation and a see-saw shape
Show answer
Answer: (A) The stability of the halides increases in the order fluoride < chloride < bromide < iodide
Explanation: The stability of these halides falls in the order F⁻ > Cl⁻ > Br⁻ > I⁻, which is the reverse of the order stated. This is why hexahalides are limited to the hexafluorides. The remaining statements are correct: monohalides occur as dimers such as S₂Cl₂, and tetrafluorides are sp³d see-saw molecules.

Question 10

The physical states of SF₄, SeF₄ and TeF₄ at room temperature are, respectively,
  1. liquid, liquid, solid
  2. gas, gas, liquid
  3. gas, liquid, solid
  4. solid, liquid, gas
Show answer
Answer: (C) gas, liquid, solid
Explanation: SF₄ is a gas, SeF₄ a liquid and TeF₄ a solid. All three have sp³d hybridisation and a see-saw shape, so shape does not explain the change. Intermolecular attraction grows as the central atom and the molecule become larger, so the fluoride passes from gas to liquid to solid down the group.

Question 11

Which option correctly gives the change in reducing character from SO₂ to TeO₂, along with its cause?
  1. It decreases, because the +4 state becomes more stable relative to +6 down the group owing to the inert pair effect
  2. It increases, because the +6 state becomes more stable down the group
  3. It does not change, because the element is in the +4 state in all these dioxides
  4. It decreases, because the +6 state becomes more stable relative to +4 down the group
Show answer
Answer: (A) It decreases, because the +4 state becomes more stable relative to +6 down the group owing to the inert pair effect
Explanation: A dioxide acts as a reducing agent when the element passes from +4 to +6. Down group 16, the stability of +6 falls and that of +4 rises because of the inert pair effect. So reducing character decreases from SO₂ to TeO₂. SO₂ is a reducing agent, whereas TeO₂ acts as an oxidising agent.

Question 12

Zinc gives hydrogen with dilute sulphuric acid but sulphur dioxide with hot concentrated sulphuric acid. Copper gives no gas with dilute acid but gives sulphur dioxide with the hot concentrated acid. What is the correct inference?
  1. H⁺ is the oxidising species in both acids, and copper reacts only because the concentrated acid has more H⁺
  2. Sulphate ion is the oxidising species in the dilute acid, whereas H⁺ is the oxidising species in the hot concentrated acid
  3. Sulphur(VI) is the oxidising species in both acids, and hydrogen forms from zinc by decomposition of sulphur dioxide
  4. H⁺ is the oxidising species in the dilute acid, whereas sulphur(VI) is the oxidising species in the hot concentrated acid
Show answer
Answer: (D) H⁺ is the oxidising species in the dilute acid, whereas sulphur(VI) is the oxidising species in the hot concentrated acid
Explanation: In dilute acid, H⁺ is reduced to H₂, and only metals below hydrogen in reduction potential, such as zinc, can do this. Copper has a positive E° and cannot reduce H⁺. Hot concentrated acid is a strong oxidising agent in which sulphur is reduced from +6 to +4, as in Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O.

Want more Group 16 Elements General Introduction – Electronic configuration and general trends in physical and questions?

E-GYAN NIDHI has 149 questions on this topic and 63,000+ in total, all free: timed CBT tests, NEET marking and instant analysis.

Register free Free Sunday Mock Test

More topics in P- BLOCK ELEMENTS

↑