P Block Elements MCQ | Class 12 | Chemistry | Chapter-7

P Block Elements MCQ Chapter 7

The elements that can be found in the periodic table from group 13 to group 18 are called the P Block elements.

Below are some of the very important NCERT P Block Elements MCQ Class 12 Chemistry Chapter 7 with Answers. These P Block Elements MCQs have been prepared by expert teachers and subject experts based on the latest syllabus and pattern of CBSE Term 1 examination.

We have given these P Block Elements Class 12 Chemistry MCQs Questions with Answers to help students understand the concept.

MCQ Questions for Class 12 Chemistry are very important for the latest CBSE Term 1 and Term 2 pattern. These MCQs are very important for students who want to score high in CBSE Board, NEET and JEE exam. 

We have put together these NCERT MCQ Questions of P Block Elements for Class 12 Chemistry Chapter 7 with Answers for the practice on a regular basis to score high in exams. Refer to these MCQs Questions with Answers here along with a detailed explanation.

Note : – Answers after every 10 questions

p block elements mcq

P Block Elements MCQ (1-10)

  1. Which one of the following oxides of nitrogen is a solid?
    1. NO2
    2. N2O
    3. N2O3
    4. N2O5
  2. Which one of the following is stable?
    1. NCl3
    2. NBr3
    3. NI3
    4. NF3
  3. The BCl3 is a planar molecule, whereas NCl3 is pyramidal because
    1. N-Cl bond is more covalent than B-Cl bond
    2. B-Cl bond is more polar than N-Cl bond 
    3. BCl3 has no lone pair but NCl3 is a lone pair of electrons
    4. Nitrogen atom is smaller than boron
  4. Which of the following is tetrabasic acid?
    1. Metaphosphoric acid
    2. Orthophosphoric acid
    3. Hypophosphoric acid
    4. Hypophosphorous acid
  5. The reagent used to distinguish between hydrogen peroxide and ozone is
    1. PbS
    2. Starch and Iodine
    3. KMnO4
    4. Bleaching powder
  6. Ozone can be detected by using
    1. mercury
    2. silver
    3. sodium
    4. none of these 
  7. The correct sequence of decrease in the bond angle of following hydride is 
    1. NH3 > PH3 > AsH3 > SbH3
    2. NH3 > AsH3 > PH3 > SbH3
    3. SbH3 > AsH3 > PH3 > NH3
    4. PH3 > NH3 > AsH3 > SbH3
  8. Among the following oxides, the least acidic is
    1. P4O6
    2. P4O10
    3. As4O6
    4. As4O10
  9. Which of the following is paramagnetic?
    1. N2O3
    2. NO2
    3. N2O5
    4. All of these
  10. Concentrated H2SO4 is not used to prepare HBr from KBr because it 
    1. oxidises HBr
    2. reduces HBr
    3. causes disproportionation of HBr
    4. reacts to slowly with KBr

Answers (1-10)

1 . (4)

N2O5 exists as a solid under the temperature of 273 K. After this temperature, it starts too decompose.

2. (4)

The unstable nature of NBr3, NCl3 and NI3 is due to low polarity of the N-X bond and large size difference between nitrogen and halogen atoms. The size of nitrogen and fluorine are comparable.

3. (3)

Presence of bond pairs and lone pairs affect the shape of the molecule. This is due to the fact that lone pairs and bond pairs repel each other and affect the shape. 

BCl3 has 3 bond pairs and 0 lone pair

NCl3 has 3 bond pairs and 1 lone pair

4. (3)

Tetrabasic acid means an acid having four hydrogens that can be replaced.

Hypophosphorous acid – H3PO2

Metaphosphoric acid – HPO3

Orthophosphoric acid – H3PO4

Hypophosphoric acid – H4P2O6

Hypophosphoric acid has 4 replaceable hydrogens.

5. (2)

Iodine is used in the quantitative estimation of ozone.

6. (1)

Ozone reacts with mercury to form mercurous oxide, which sticks to the walls of the container. Therefore, the presence of ozone can be detected by mercury.

7. (1)

As we move down the group, the radius of elements increases and electronegativity decreases so bond angle decreases.

8. (3)

Acidic character of oxides decreases with the decrease in the oxidation state and also decreases down the group.

9. (2)

NO2 is paramagnetic due to the presence of an unpaired electron in the nitrogen.

10. (1)

Conc. H2SO4 can not be used to prepare HBr from KBr because conc. H2SO4 oxidises HBr to bromine.

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P Block Elements MCQ (11-20)

  1. Which of the following is thermally most stable ?
    1. H2S
    2. H2O
    3. H2Se
    4. H2Te
  2. Which of the following oxides react with both HCl and NaOH?
    1. N2O5
    2. ZnO
    3. CaO
    4. CO2
  3. Which gas evolves when urea is treated with sodium hydroxide? 
    1. Nitrogen
    2. Laughing gas
    3. Ammonia
    4. NO
  4. Which of the following is an amphoteric oxide?
    1. Cr2O3
    2. Cl2O7
    3. SnO2
    4. None of these
  5. Which of the following allotropic forms of sulphur is thermodynamically most stable?
    1. β-monoclinic
    2. ℽ-monoclinic
    3. Plastic sulphur
    4. Orthorhombic
  6. Which of the following does not have a p-o-p bond?
    1. P4O6
    2. P4O10
    3. H4P2O6
    4. H4P2O7
  7. Which acid on heating produces phosphine?
    1. Phosphoric acid 
    2. Phosphorous acid 
    3. Peroxymonophosphoric acid
    4. Metaphosphoric acid 
  8. Xenon difluoride is
    1. linear
    2. angular
    3. trogonal
    4. pyramidal
  9. Which of the following is not a reducing agent 
    1. Sulphur Dioxide
    2. Hydrogen Peroxide 
    3. Carbon Dioxide
    4. Nitrogen Dioxide
  10. Which of the following molecules does not possess a permanent dipole moment 
    1. H2S
    2. SO2
    3. SO3
    4. CS2

Answers (11-20)

11. (2)

H2O is the most stable as there is a formation of hydrogen bonds. 

The order goes like : H2Te < H2Se < H2S < H2O

12. (2)

ZnO is an amphoteric oxide, therefore it reacts with both HCl and NaOH.

13. (3)

NH2CONH2 + 2NaOH → 2NH3 + NH2CO3

14. (1)

Cr2O3 is an amphoteric oxide.

15. (4)

Orthorhomic sulphur is the most stable form of sulphur having eight sulphur atoms arranged in an octahedron.

16. (3)

H4P2O6 (Hypophosphoric acid) doesn’t have any P-O-P bond.

17. (3)

H3PO5 (Peroxymonophosphoric acid) gives H3PO4 (orthophoric acid) and PH3 (phosphine).

18. (1)

XeF2 has a linear shape.

19. (2)

Carbon dioxide is an oxidising agent.

20. (4)

CS2 has a linear shape, hence doesn’t have a permanent dipole moment.

P Block Elements MCQ (21-30)

  1. Potassium chlorate on heating with concentrated sulphuric acid gives
    1. Chlorine dioxide 
    2. HClO4
    3. KHSO4 
    4. All of these
  2. The most abundant and common oxidation state of sulphur is 
    1. -2
    2. +4
    3. +2
    4. +6
  3. Ozone is 
    1. an unstable, dark blue, diamagnetic gas
    2. an unstable, dark blue, paramagnetic gas
    3. a stable, dark blue, paramagnetic gas
    4. found in the upper atmosphere where it absorbs UV radiation
  4. Structure of ClF3 is
    1. T-shape
    2. Octahedral 
    3. Tetrahedral
    4. None of these 
  5. In two forms NCl3 whereas P can form both PCl3 and PCl5. Why? 
    1. N atom has is larger than P in size
    2. P is more reactive towards Cl than N 
    3. P has d orbitals which can be used for bonding but N2 does not have d orbitals
    4. None of these 
  6. Which characteristic is not correct about sulphuric acid?
    1. Sulphonating agent
    2. Reducing agent
    3. Oxidizing agent
    4. Highly Vicious
  7. Which of the following is planar?
    1. XeF4
    2. XeO4
    3. XeO2F2
    4. XeOF4
  8. Which of the following halide ions is the most basic?
    1. Cl
    2. Br
    3. I
    4. F
  9. Which of the following gives nitrogen on heating?
    1. Ammonium nitride 
    2. Ammonium nitrate
    3. Ammonium nitrite
    4. Ammonium hydroxide
  10. One of the reasons why F-F bond is weak is that 
    1. The repulsion between the non bonding pairs of electrons of the two fluorine atoms is high
    2. The F-F Bond distance is small and hence intermolecular repulsion is small
    3. The ionization enthalpy of the fluorine atom is very high 
    4. The F-F Bond distances large

Click Below To Learn Chemistry Term-1 Syllabus Chapters MCQs

Answers (21-30)

21. (4)

3KClO3 + 3H2SO4 → 2KHSO4 + HClO4 + 2ClO2 + H2O

22. (2)

Sulphur’s most common oxidation state is +4.

23. (1)

Ozone is an unstable, dark blue, diamagnetic gas.

24. (1)

ClF3 is of T-shape

25. (3)

There is no vacant d-orbital in the outermost orbit of nitrogen. Thus nitrogen show valency only 3. There are vacant d-orbital is in the outermost orbit of phosphorus and hence it shows variable covalency 3 and 5 in ground state and excited state respectively. Hence, nitrogen forms only NCl3 but phosphorus forms PCl3 and PCl5 both.

26. (2)

H2SO4 is a strong oxidising agent and can oxidise both metals and non-metals. It’s not a reducing agent.

27. (1)

XeF4 has a square planar structure.

28. (4)

F ion is the most electronegative, most unstable and the most basic .

29. (3)

Ammonium nitrite on thermal decomposition gives ammonia.

30. (2)

The F-F Bond distance is small and hence intermolecular repulsion is small.

P Block Elements MCQ (31-40)

  1. The mixture of NaI and NaIO3 is treated with hot conc. H2SO4. The iodine-containing product formed is
    1. HIO3
    2. NaIO4
    3. I2
    4. HIO4
  2. When bleaching powder is treated with carbon dioxide
    1. Calcium chloride is formed
    2. No reaction occurs
    3. Chlorine is evolved
    4. It absorbs the gas
  3. PtF6 converts O2 to
    1. O2PtF6+
    2. O2+PtF6
    3. F2+PtO2
    4. F2O+Pt
  4. Which of the following has the greatest reducing power?
    1. HBr
    2. HI
    3. HCl
    4. HF
  5. XeF6 on complete hydrolysis gives
    1. XeO3
    2. Xe
    3. XeO2
    4. XeO4
  6. Which of the following statements is not correct?
    1. Krypton is obtained during radioactive disintegration
    2. Argon is used in electric bulbs 
    3. Half life of radon is only 3.8 days
    4. Helium is used in producing very low temperatures 
  7. Which of the following noble gas is not found in the atmosphere?
    1. Rn
    2. Kr
    3. Ne
    4. Ar
  8. What is the colour produced when ammonia is passed through a solution of copper sulphate?
    1. Red
    2. Orange
    3. Yellow
    4. Blue
  9. What is the correct order of electron affinities among halogens?
    1. F2 > Cl2 > Br2 > I2
    2. Cl2 > Br2 > F2 > I2
    3. F2 < Cl2 < Br2 < I2
    4. I2 < Br2 < F2 < Cl2
  10. What is the hybridisation of Xe in XeF2?
    1. sp3
    2. sp2
    3. sp2d
    4. sp3d

Answers (31-40)

31. (3)

5 NaI + NaIO3 + 3H2SO4 → 3Na2SO4 + 3I2 +3H2O

32. (3)

When bleaching powder is treated with carbon dioxide chlorine is evolved.

33. (2)


34. (2)

HI has the strongest reducing agent among halogen acids because of lowest bond dissociation enthalpy.

35. (1)

XeF2 + 3H2O → 6HF + XeO3

36. (1)

Krypton is not obtained during radioactive disintegration.

37. (1)

Radon and helium is not found in the atmosphere.

38. (4)

The complex formed is [Cu(NH3)4]2+ which is square planar and paramagnetic and has a deep blue colour.

39. (2)

The electron affinity of chlorine is maximum in the periodic table and so the bond dissociation enthalpy. Fluorine has lower bond dissociation enthalpy than Br2 and Cl2. Due to small size electronic repulsion is very high. I2 has lowest bond dissociation enthalpy due to its quite larger size it is easiest to break the bond.

40. (4)

Xe in XeF2 has sp3d hybridisation.

P Block Elements MCQ (41-50)

  1. Which one of the following undergoes thermal decomposition?
    1. XeF2
    2. XeF4
    3. ZeF6
    4. None of these
  2. Which of the following is the least oxidising in nature?
    1. HClO3
    2. HClO4
    3. HClO2
    4. HClO
  3. Which one of the following is the most stable thermally?
    1. HI
    2. HBr
    3. HCl
    4. HF
  4. The crystals of ferrous sulphate on heating give
    1. FeO + SO2 + H2O
    2. FeO + SO3 + H2SO4 + H2O
    3. Fe2O3 + SO2 + SO3 + H2O
    4. Fe2O3 + H2SO4 + H2O
  5. Pure chlorine is obtained 
    1. By heating MnO2 with HCl
    2. By heating bleaching powder with HCl
    3. By heating PtCl4
    4. By heating a mixture of NaCl and MnO2 with conc. H2SO4
  6. The element which forms oxides in all the oxidation states from +1 to +5 is
    1. N
    2. P
    3. As
    4. Sb
  7. Which one of the following acts as an antichlor?
    1. MnO2
    2. Na2S2O3
    3. K2Cr2O7
    4. Na2SO4
  8. N2O3 is
    1. An acidic oxide and the anhydride of HNO2
    2. An acidic oxide and the anhydride of H2N2O2
    3. A neutral oxide and the anhydride of HNO3
    4. A basic oxide and the anhydride of HNO2
  9. (NH4)2CrO7 on heating liberates a gas. The same gas will be obtained by
    1. Heating NH4NO3
    2. Treating Mg3N2 with H2O
    3. Heating NH4NO2
    4. Heating H2O2 on NaNO2
  10. According to Le Chatlier’s principle, low temperature is required for more yield of ammonia in Haber’s process. What is the temperature commercially kept for the process?
    1. 350-450℃
    2. 450-550℃
    3. 250-350℃
    4. 150-250℃

Answers (41-50)

41. (3)

XeF6 on thermal decomposition gives XeF2, XeF4 and F2

42. (2)

The oxidation state of chlorine in HClO4, HClO3, HClO2 and HClO is 7, 5, 3 and 1. The chlorine with the highest oxidation state has the lowest potential of getting oxidised.

43. (4)

The H-X bond strength decreases from HF to HI.

44. (3)

FeSO4 on heating gives Fe2O3 + SO2 + SO3 + H2O.

45. (3)

Pure chlorine is obtained on heating PtCl4.

46. (1)

Nitrogen has an oxidation state varying from +1 to +5.

47. (2)

An antichlor is a substance used to decompose residual HCl or chlorine after chlorine-based bleaching in order to prevent ongoing reaction. Na2S2O3 acts as an antichlor.

48. (1)

Nitrogen in N2O3 has an oxidation state of +3 and so does in HNO2. Hence, N2O3 is an anhydride of HNO2. N2O3 is an acidic oxide.

49. (3)

(NH4)2CrO7 on heating liberates nitrogen gas which can also be obtained from heating NH4NO2.

50. (2)

The temperature required for Haber’s process is 450-550℃.

Assertion and Reasoning MCQs


(a) Both A and R are true and R is the correct explanation of A

(b) Both A and R are true and but R is not a correct explanation of A

(c) A is true but R is false

(d) A is false, but R is true

1. Assertion (A) Valency of noble gas is 0.

Reason (R) Noble gases possess complete octet.

2. Assertion (A)  F2 has lower bond dissociation enthalpy then Cl2.

Reason (R) Fluorine is more electronegative than chlorine.

3. Assertion (A) Acidic character of group 16 hydrides increases from H2O to H2Te.

Reason (R) Thermal stability of hydrides decreases down the group.

4. Assertion (A)  Interhalogen compounds are more reactive than halogens (except chlorine).

Reason (R) They all undergo hydrolysis giving halide ion derived from the smaller halogen.

5. Assertion (A) Halogens are not found in free state in nature.

Reason (R) Halogens are highly reactive compound.

6. Assertion (A) F2 has low reactivity.

Reason (R) F-F bonds ΔbondH.

7. Assertion (A) Ozone layer in the upper region of atmosphere protect earth from UV radiation.

Reason (R) Ozone is a powerful oxidizing agent as compared to oxygen.

8. Assertion (A) S shows paramagnetic nature, when present in vapour state.

Reason (R) S exists as S2 in vapour state.

9. Assertion (A) F2 is a strong oxidizing agent.

Reason (R) Electron gain enthalpy of fluorine is less negative. 

10. Assertion (A) PbI4 is not a stable compound.

Reason  (R) Iodide stabilizes higher oxidizing state.

11. Assertion (A) Solubility of noble gases in water decreases with increasing size of the noble gas.

Reason (R) Solubility of noble gases in water is due to dipole-induced dipole interaction. 

Assertion and Reasoning MCQ Answers

1. (a)

Noble gases possess the electronic configuration ns1np6 and has 8 electrons in their outer shell, hence there valency is 0.

2. (b)

Flourine has low bond dissociation enthalpy due to small atomic size number of electrons create large repulsion in bonded electron.

3. (b)

The acidic character increases down the group and thermal stability of hydrides decreases down the group due to decrease in bond (H-E) dissociation enthalpy down the group.

4. (b)

Interhalogen compounds are more reactive than halogens because X-X’ bond in interhalogens in weaker than X-X bond in halogen (except F-F bond).

5. (a)

Halogens are highly reactive as they have seven electrons in their outermost orbit and they want to stabilize by acquiring an electron. therefore, they do not occur in free state.

6. (d)

F2 is more reactive that other halogens because its valence electrons are more closer to nucleus and its more electronegative so, bonded electrons repel each other causing low bond dissociation enthalpy.

7. (b)

Ozone layer filters the radiation coming from sun, hence serves as the protective layers.

8. (a)

In vapor state S exists as S2 and it behaves like O2. It has 2 unpaired electrons in antibonding pi orbitals. Presence of these unpaired electrons make S paramagnetic.

9. (b)

As F2 has low bond dissociation energy and high hydration energy than Cl2 but less electron gain enthalpy due to its smaller size. Due to these factors F2 wins in getting reduced fastly.

10. (c)

Small highly elctronegative atoms such as F can stabilise higher oxidation state.

11. (d)

PbI4 is not a stable compound because Pb shows (II) oxidation state more frequently than Pb (IV) due to inert pait effect. Iodide cannot stabilize higher oxidation states.

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Chemistry Term-1 Syllabus Chapters MCQs and Assertion/Reasoning

Final Words

From the above article, you have practiced P Block Elements MCQ of Class 12 Chemistry Chapter 7. We hope that the above mentioned latest MCQs for Term 1 of Chapter 7 P Block Elements will surely help you in your exam. If you have any doubts or queries regarding the P Block Elements Multiple Choice Questions with Answers of CBSE Class 12 Chemistry, feel free to reach us and we will get back to you as early as possible.

Frequently Asked Questions (FAQs)

Q1. Why P Block elements called the P Block elements?

They are called so because their valence electrons are in the p-orbital.

Q2. Why group 16 elements called chalcogens?

They are called so because most ores of copper (Greek Chalcos) are oxides or sulphides and also traces of selenium and tellurium.

Q3. Why is nitrogen so inert?

It is due to the presence of the very strong triple bond present between the molecular state of nitrogen N2.

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