Chapter Exam

Unit 10: p-Block Elements — Online MCQ Test

Q: Which of the following oxoacid of sulphur has — O — O — linkage?

Answer: B. H2S2O8, peroxodisulphuric acid — Peroxodisulphuric acid (H2S2O8) contains a peroxide linkage (—O—O—).

Q: When copper is heated with conc. HNO3 it produces:

Answer: D. Cu(NO3)2 and NO2 — Copper reacts with concentrated nitric acid to produce copper(II) nitrate, water, and reddish-brown nitrogen dioxide gas: Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2.

Q: Match the compounds given in column I with the hybridisation and shape given in column II and mark the correct option. (a) XeF6 (b) XeO3 (c) XeOF4 (d) XeF4

Answer: D. (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii) — XeF6: distorted octahedral (i); XeO3: pyramidal (iii); XeOF4: square pyramidal (iv); XeF4: square planar (ii).

Q: The product obtained as a result of a reaction of nitrogen with CaC2 is

Answer: A. CaCN — The reaction of nitrogen with calcium carbide (CaC2) yields calcium cyanamide (CaCN2) and carbon: CaC2 + N2 -> CaCN2 + C. As the option CaCN2 is missing from the list, it is marked as a bonus.

Q: Given below are two statements: Statement I: The boiling point of hydrides of Group 16 elements follow the order H2O > H2Te > H2Se > H2S. Statement II: On the basis of molecular mass, H2O is expected to have lower boiling point than the other members of the group but due to the presence of extensive H-bonding in H2O, it has higher boiling point. In the light of the above statements, choose the correct answer from the options given below:

Answer: C. Both Statement I and Statement II are true. — The boiling point order is H2O > H2Te > H2Se > H2S, and H2O is anomalously high due to strong intermolecular hydrogen bonding.

Q: The correct order of N-compounds in its decreasing order of oxidation states is:

Answer: B. HNO3, NO, N2, NH4Cl — Oxidation states are: HNO3 (+5), NO (+2), N2 (0), NH4Cl (-3). Decreasing order: +5 > +2 > 0 > -3.

Q: The correct order of N-compounds in its decreasing order of oxidation states is

Answer: C. HNO₃, NO, N₂, NH₄Cl — Oxidation states of N: HNO₃ → +5; NO → +2; N₂ → 0; NH₄Cl → –3. Decreasing order: +5 > +2 > 0 > –3, i.e., HNO₃ > NO > N₂ > NH₄Cl.

CHEMISTRY · NEET (All) · NEET

Practice Unit 10: p-Block Elements with a free chapter-wise online MCQ test for NEET NEET (All) CHEMISTRY. This chapter covers: Group -13 to Group 18 Elements General Introduction: Electronic configuration and general trends in physical and chemical properties of elements across the periods and down the gro.... AI-generated questions from basic to board-exam level, with instant results and explanations.

Questions

20m

Time Limit

Attempts Left

10 random questions from this chapter (mixed difficulty)
Questions you've seen before won't repeat until the pool resets
You have 20 minutes — exam auto-submits when time is up
Maximum 3 attempts per chapter
Results and explanations shown immediately after submission

New here? Register free — includes 3 free chapter exams.

Unit 10: p-Block Elements — Important Questions & Answers (FAQ)

Frequently asked questions from NEET NEET (All) CHEMISTRY — Unit 10: p-Block Elements, with answers and explanations. These are sample questions; the exam has its own separate question set.

Which of the following oxoacid of sulphur has — O — O — linkage?

A. H2SO4, sulphuric acid
B. H2S2O8, peroxodisulphuric acid ✓
C. H2S2O7, pyrosulphuric acid
D. H2SO3, sulphurous acid

Answer: B. H2S2O8, peroxodisulphuric acid
Peroxodisulphuric acid (H2S2O8) contains a peroxide linkage (—O—O—).

Which of the following is not correct about carbon monoxide?

A. It reduces oxygen carrying ability of blood.
B. The carboxyhaemoglobin (haemoglobin bound to CO) is less stable than oxyhaemoglobin. ✓
C. It is produced due to incomplete combustion.
D. It forms carboxyhaemoglobin.

Answer: B. The carboxyhaemoglobin (haemoglobin bound to CO) is less stable than oxyhaemoglobin.
Carboxyhaemoglobin is about 300 times more stable than oxyhaemoglobin, which is why CO poisoning is dangerous.

Given below are two statements: Statement I: Like nitrogen that can form ammonia, arsenic can form arsine. Statement II: Antimony cannot form antimony pentoxide.

A. Both Statement I and Statement II are correct.
B. Both Statement I and Statement II are incorrect.
C. Statement I is correct but Statement II is incorrect. ✓
D. Statement I is incorrect but Statement II is correct.

Answer: C. Statement I is correct but Statement II is incorrect.
Statement I is correct (AsH₃ is arsine). Statement II is incorrect because Antimony forms Sb₂O₅ (antimony pentoxide).

When copper is heated with conc. HNO3 it produces:

A. Cu(NO3)2 and NO
B. Cu(NO3)2, NO and NO2
C. Cu(NO3)2 and N2O
D. Cu(NO3)2 and NO2 ✓

Answer: D. Cu(NO3)2 and NO2
Copper reacts with concentrated nitric acid to produce copper(II) nitrate, water, and reddish-brown nitrogen dioxide gas: Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2.

Match the compounds given in column I with the hybridisation and shape given in column II and mark the correct option. (a) XeF6 (b) XeO3 (c) XeOF4 (d) XeF4

A. (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)
B. (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
C. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
D. (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii) ✓

Answer: D. (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)
XeF6: distorted octahedral (i); XeO3: pyramidal (iii); XeOF4: square pyramidal (iv); XeF4: square planar (ii).

Which is the correct statement for the given acids?

A. Phosphinic acid is a monoprotic acid while phosphonic acid is a diprotic acid ✓
B. Phosphinic acid is a diprotic acid while phosphonic acid is a monoprotic acid
C. Both are triprotic acids
D. Both are diprotic acids

Answer: A. Phosphinic acid is a monoprotic acid while phosphonic acid is a diprotic acid
Phosphinic acid (H3PO2) has only one P-OH group (monoprotic). Phosphonic acid (H3PO3) has two P-OH groups (diprotic).

The product obtained as a result of a reaction of nitrogen with CaC2 is

A. CaCN ✓
B. CaCN3
C. Ca2CN
D. Ca(CN)2

Answer: A. CaCN
The reaction of nitrogen with calcium carbide (CaC2) yields calcium cyanamide (CaCN2) and carbon: CaC2 + N2 -> CaCN2 + C. As the option CaCN2 is missing from the list, it is marked as a bonus.

Given below are two statements: Statement I: The boiling point of hydrides of Group 16 elements follow the order H2O > H2Te > H2Se > H2S. Statement II: On the basis of molecular mass, H2O is expected to have lower boiling point than the other members of the group but due to the presence of extensive H-bonding in H2O, it has higher boiling point. In the light of the above statements, choose the correct answer from the options given below:

A. Statement I is true but Statement II is false.
B. Statement I is false but Statement II is true.
C. Both Statement I and Statement II are true. ✓
D. Both Statement I and Statement II are false.

Answer: C. Both Statement I and Statement II are true.
The boiling point order is H2O > H2Te > H2Se > H2S, and H2O is anomalously high due to strong intermolecular hydrogen bonding.

The correct order of N-compounds in its decreasing order of oxidation states is:

A. HNO3, NH4Cl, NO, N2
B. HNO3, NO, N2, NH4Cl ✓
C. NH4Cl, N2, NO, HNO3
D. HNO3, NO, NH4Cl, N2

Answer: B. HNO3, NO, N2, NH4Cl
Oxidation states are: HNO3 (+5), NO (+2), N2 (0), NH4Cl (-3). Decreasing order: +5 > +2 > 0 > -3.

The correct order of N-compounds in its decreasing order of oxidation states is

A. HNO₃, NO, NH₄Cl, N₂
B. HNO₃, NH₄Cl, NO, N₂
C. HNO₃, NO, N₂, NH₄Cl ✓
D. NH₄Cl, N₂, NO, HNO₃

Answer: C. HNO₃, NO, N₂, NH₄Cl
Oxidation states of N: HNO₃ → +5; NO → +2; N₂ → 0; NH₄Cl → –3. Decreasing order: +5 > +2 > 0 > –3, i.e., HNO₃ > NO > N₂ > NH₄Cl.

← Previous Chapter

Unit 9: Classification of Elements and Periodicity in Properties

Next Chapter →

Unit 11: d- and f-Block Elements