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NCERT Exemplar Class 12 Chemistry Solutions Chapter 7 p block elements

NCERT Exemplar Class 12 Chemistry Solutions Chapter 7 p block elements

#CBSE Class 12th
Shivani PooniaUpdated on 08 Sep 2025, 10:33 PM IST

Do you know which element glows in the dark and plays a vital role in life and light? The answer to this is Phosphorus, which is an example of a p-block element. The elements present in the p-block have their last electron filled in the p-orbital. The p-block contains metals, non-metals as well as metalloids that have a wide range of real-life applications.

This Story also Contains

  1. NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: MCQ (Type 1)
  2. NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: MCQ (Type 2)
  3. NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: Short Answer Type
  4. NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: Matching Type
  5. NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: Assertion and Reason Type
  6. NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: Long Answer Type
  7. NCERT Exemplar Class 12 Chemistry Chapter 7: Higher Order Thinking Skills (HOTS) Questions
  8. Approach to Solve Questions of Class 12 Chemistry Chapter 7
  9. Topics and Subtopics Covered in the NCERT Exemplar Class 12 Chemistry Chapter 7
  10. NCERT Exemplar Class 12 Chemistry Chapter 7: Important Trends
  11. NCERT Exemplar Solutions Class 12 Chemistry Chapter-Wise
  12. NCERT Solutions for Class 12 Chemistry Chapter-Wise
  13. NCERT Solutions subject-wise
  14. NCERT Exemplar Class 12 Solutions subject-wise
  15. NCERT Class 12 subject-wise notes
  16. NCERT Books and NCERT Syllabus
NCERT Exemplar Class 12 Chemistry Solutions Chapter 7 p block elements
p-block elements

The important topics like the periodic trends, chemical and physical properties of the p-block elements, electronic configuration, and the chemistry of the important p- block compounds are all discussed in this chapter. The NCERT Exemplar Class 12 Chemistry Solutions are designed to offer a systematic approach to understand these topics through a series of solved questions. These NCERT exemplar solutions will improve your problem-solving skills and will prepare you for competitive exams like NEET, JEE Mains, etc. In NCERT Exemplar Class 12 Chemistry Solutions Chapter 7 p block elements, the higher-order thinking skill (HOTS) questions are included to test your critical thinking. We have also added some points that will help students build a good approach to solve the questions. Students can also refer to NCERT Solutions for more practice and conceptual clarity.

NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: MCQ (Type 1)

The MCQ questions are covered in the NCERT Exemplar Class 12 Chemistry Solutions Chapter 7 to enhance your knowledge. These questions help students test their conceptual understanding and prepare effectively for exams.

Question 1. On addition of conc. $\text {H}_{2}\text {SO}_{4}$ to a chloride salt, colourless fumes are evolved but in case of an iodide salt, violet fumes come out. This is because
(i) $\text {H}_{2}\text {SO}_{4}$ reduces $\text {HI to I}_{2}$
(ii) $\text {HI}$ is of violet colour
(iii) $\text {HI}$ gets oxidised to $\text {I}_{2}$
(iv) $\text {HI}$ changes to $\text {HIO}_{3}$
Answer:

The answer is option (iii).

When concentrated $\text {H}_{2}\text {SO}_{4}$ is added to iodine salt, HI, which is a strong reducing agent, is formed. During the reaction, it is oxidized to $\text {I}_{2}$ , which is violet in colour.
$\text {2NaCl + H}_{2}\text {SO}_{4}\rightarrow \text {Na}_{2}\text {SO}_{4}+\text {2HCl}$
In case of iodine the halogen acid $\text {HI}$ obtained reduces $\text {H}_{2}\text {SO}_{4}$ to $\text {SO}_{2}$ and itself is oxidised to free iodine.
$\text {2NaI + H}_{2}\text {SO}_{4}\rightarrow \text {Na}_{2}\text {SO}_{4}+ \text {2HI}\overset{\text {H}_{2}\text {SO}_{4}}{\longrightarrow} \text {2H}_{2}\text {O + SO}_{2}+\text {I}_{2}$

Question 2. In qualitative analysis when $\text {H}_{2}\text {S}$ is passed through an aqueous solution of salt acidified with dil.$\text {HCl}$, a black precipitate is obtained. On boiling the precipitate with dil. $\text {HNO}_{3}$, it forms a solution of blue colour. Addition of excess of aqueous solution of ammonia to this solution gives _________.
(i) a deep blue precipitate of $\text {Cu(OH)}_{2}$
(ii) a deep blue solution of $[Cu(NH_{3})_{4}]^{2+}$
(iii) a deep blue solution of $Cu(NO_{3})_{2}$
(iv) a deep blue solution of $Cu(OH)_{2}.Cu(NO_{3})_{2}$
Answer:

The answer is option (ii).

Black precipitate of copper sulphide is formed on passing $H_{2}S$ through acidified solution of salt. On boiling with dilute $HNO_{3}$, the precipitate gives blue colour of copper nitrate. On addition of excess aqueous solution of ammonia, deep blue coloured compound is found. The reactions involved are:

$Cu^{2+}+H_{2}S\rightarrow CuS (black)+2H^{+}$


$CuS+dil.HNO_{3}\rightarrow Cu\left ( NO_{3} \right )_{2}$


$Cu\left ( NO_{3} \right )_{2}+4NH_{3}\rightarrow \left [ Cu\left ( NH_{3} \right )_{4} \right ]^{2+}(Deep\; blue)$

Question 3. In a cyclotrimetaphosphoric acid molecule, how many single and double bonds are present?
(i) 3 double bonds; 9 single bonds
(ii) 6 double bonds; 6 single bonds
(iii) 3 double bonds; 12 single bonds
(iv) Zero double bonds; 12 single bonds

Answer:

The answer is option (i) 3 double bonds; 9 single bonds
In Cyclotrimetaphosphoric acid, there are 3 double bonds and 9 single bonds as present

Question 4. Which of the following elements can be involved in $p\pi -d\pi$ bonding?
(i) Carbon
(ii) Nitrogen
(iii) Phosphorus
(iv) Boron
Answer:

The answer is option (iii) Phosphorus
The involvement of phosphorous in pπ-dπ bonding is due to the presence of vacant d orbitals. The elements C, N, and B do not have vacant d orbitals.

Question 5. Which of the following pairs of ions are isoelectronic and isostructural?
(i) $CO_{3}^{2-}, NO^{-}_{3}$
(ii) $ClO_{3}^{-} , CO_{3}^{2-}$
(iii) $SO_{3}^{2-},NO^{-}_{3}$
(iv) $ClO^{-}_{3} ,SO_{3}^{2-}$
Answer:

The answer is the option (i).

Isoelectronic means that two ions or molecules have the same number of electrons. Isostructural means they have the same shape or geometry. The number of electrons in both the isoelectronic species is 32 and they exhibit $sp^{2}$ hybridisation and thus trigonal planar structure.

Question 6. Affinity for hydrogen decreases in the group from fluorine to iodine. Which of the halogen acids should have the highest bond dissociation enthalpy?
$(i) HF$
$(ii) HCl$
$(iii) HBr$
$(iv) HI$
Answer:

The answer is the option (i).
On moving down the group, there is an increase in the atomic radii and a decrease in bond dissociation enthalpy. That is why; $HF$ should have the highest bond dissociation enthalpy.

Question 7. Bond dissociation enthalpy of $E-H \left ( E = element \right )$ bonds are given below. Which of the compounds will act as the strongest reducing agent?

Compound
$\Delta _{diss} \left ( E-H \right )/kJ\; mol^{-1}$

$NH_{3}$
389

$PH_{3}$
322

$AsH_{3}$
297

$SbH_{3}$

255

(i) $NH_{3}$
(ii) $PH_{3}$
(iii) $AsH_{3}$
(iv) $SbH_{3}$

Answer:

The answer is the option (iv). The bond dissociation enthalpy decreases down the group. $SbH_{3}$ has the least bond enthalpy and thus acts as the strongest reducing agent.

Question 8. On heating with concentrated $\text {NaOH}$ solution in an inert atmosphere of $\text {CO}_{2}$, white phosphorus gives a gas. Which of the following statement is incorrect about the gas?
(i) It is highly poisonous and has smelled like rotten fish.
(ii) It’s a solution in water decomposes in the presence of light.
(iii) It is more basic than $\text {NH}_{3}$
(iv) It is less basic than $\text {NH}_{3}$
Answer:

The answer is option (c)

$P_{4}+3NaOH+3H_{2}O\rightarrow PH_{3} \left ( Phosphine \right )+3NaH_{2}PO_{2}$
$PH_{3}$ is less basic than $NH_{3}$

Question 9. Which of the following acids forms three series of salts?
(i) $H_{3}PO_{2}$
(ii) $H_{3}BO_{3}$
(iii) $H_{3}PO_{4}$
(iv) $H_{3}PO_{3}$
Answer:

The answer is the option (iii) $H_{3}PO_{4}$
There are 3 ionisable H-atoms in $H_{3}PO_{4}$ as it has 3- $OH$ groups.

Thus, it forms three series of salts are as follows: .
$NaH_{2}PO_{4}, NaHPO_{4}\; and \; Na_{3}PO_{4}$.

Question 10. Strong reducing behaviour of $H_{3}PO_{2}$ is due to
(i) The low oxidation state of phosphorus
(ii) Presence of two $-OH$ groups and one $P-H$ bond
(iii) Presence of one $-OH$ group and two $P-H$ bonds
(iv) High electron gain enthalpy of phosphorus

Answer:

The answer is option (iii). In the monobasic acid $H_{3}PO_{2}$, there are two $P-H$ bonds that impart reducing character to the acid.

Question 11. On heating lead, nitrate forms oxides of nitrogen and lead. The oxides formed are ______.
(i) $N_{2}O,PbO$
(ii) $NO_{2},PbO$
(iii) $NO,PbO$
(iv) $NO,PbO_{2}$
Answer:

The answer is the option (ii) $2Pb \left ( NO_{3} \right )_{2}\rightarrow 2PbO+4NO_{2}+O_{2}$

Question 12. Which of the following elements does not show allotropy?
(i) Nitrogen
(ii) Bismuth
(iii) Antimony
(iv) Arsenic
Answer:

The answer is option (i). Nitrogen has weak N – N single bond because of high inter-electronic repulsion of non-bonding electrons. This is why, nitrogen does not exhibit allotropy.

Question 13. Maximum covalency of nitrogen is ______________.
(i) 3
(ii) 5
(iii) 4
(iv) 6

Answer:

The answer is the option (iii). Maximum covalency of nitrogen is four as it cannot extend its valency beyond four $\left [ NH_{4},R_{4}N \right ]$ due to absence of vacant d orbitals.

Question 14. Which of the following statements is wrong?
(i) Single $N-N$ bond is stronger than the single $P-P$ bond.
(ii) $PH_{3}$ can act as a ligand in the formation of a coordination compound with transition elements.
(iii) $NO_{2}$ is paramagnetic.
(iv) Covalency of nitrogen in $N_{2}O_{5}$ is four
Answer:

The answer is option (i).Single $P-P$ bond is stronger than single $N-N$ bond, however, naturally occurring nitrogen forms $p(pi)-p(pi)$ multiple bonds and becomes much more stable than phosphorous.

Question 15. A brown ring is formed in the ring test for $NO_{3}^{-}$ ion. It is due to the formation of
(i) $\left [ Fe\left ( H_{2}O \right )_{5} \left ( NO \right ) \right ]^{2+}$
(ii) $FeSO_{4}.NO_{2}$
(iii) $\left [Fe\left ( H_{2}O \right )_{4} \left ( NO \right )_{2} \right]^{2+}$
(iv) $FeSO_{4}.HNO_{3}$

Answer:

The answer is the option (i). On addition of freshly prepared solution of $FeSO_{4}$ in a $NO_{3}^{-}$ ion containing solution, a brown coloured complex is formed. This is also known as brown ring test of nitrate.


$NO_{3}^{-}+3Fe^{2+}+4H^{+}\rightarrow NO+3Fe^{3+}+2H_{2}O$


$\left [ Fe\left ( H_{2}O \right )_{6} \right ]^{2+}+NO\rightarrow \left [ Fe\left ( H_{2}O \right )_{5}\left ( NO \right ) \right ]^{2+}\left ( Brown\; ring \right )+H_{2}O$

Hence, 2 moles of $NO$ are produced from 2 moles of ammonia.

Question 16. Elements of group-15 form compounds in +5 oxidation state. However, bismuth forms only one well characterised compound in +5 oxidation state. The compound is
(i) $Bi_{2}O_{5}$
(ii) $BiF_{5}$
(iii) $BiCl_{5}$
(iv) $Bi_{2}S_{5}$
Answer:

The answer is the option (ii). Due to inert pair effect, bismuth generally exhibits +3 oxidation state. However, due to the small size and high electronegativity of the fluoride ion it forms the compound $BiF_{3}$.

Question 17. On heating ammonium dichromate and barium azide separately we get
(i) $N_{2}$ in both cases
(ii) $N_{2}$ with ammonium dichromate and $NO$ with barium azide
(iii) $N_{2}O$ with ammonium dichromate and $N_{2}$ with barium azide
(iv) $N_{2}O$ with ammonium dichromate and $NO_{2}$ with barium azide
Answer:

The answer is option (i)
$\left ( NH_{4} \right )_{2}Cr_{2}O_{7}\rightarrow N_{2}+Cr_{2}O_{3}+4H_{2}O$
$Ba \left ( N_{3} \right )_{2}\rightarrow 3N_{2}+Ba$

Question 18. In the preparation of $HNO_{3}$, we get $NO$ gas by catalytic oxidation of ammonia. The moles of $NO$ produced by the oxidation of two moles of $NH_{3}$ will be ______.
(i) 2
(ii) 3
(iii) 4
(iv) 6
Answer:

The answer is the option (i). In the process of preparation of nitric acid, catalytic oxidation of 2 moles of ammonia produces 2 moles of $NO.$
$4NH_{3}+5O_{2}\overset{\Delta }{\rightarrow}4NO \left ( g \right )+6H_{2}O \left ( l \right )$
Hence, 2 moles of ammonia will produce 2 moles of $NO$.

Question 19. The oxidation state of a central atom in the anion of compound $NaH_{2}PO_{2}$ will be ______.
(i) +3
(ii) +5
(iii) +1
(iv) –3
Answer:

The answer is the option (c)

$NaH_{2}PO_{2} :$

$\left ( +1 \right )+\left ( +1 \times 2 \right )+x+\left ( -2 \times 2 \right )=0$
$x=+1$

Question 20. Which of the following is not tetrahedral in shape ?
(i) $NH^{+}_{4}$
(ii) $SiCl_{4}$
(iii) $SF_{4}$
(iv) $SO_{4}^{2-}$
Answer:

The answer is the option (iii). $SF_{4}$ is not tetrahedral in shape. It is a see saw structure.

Question 21. Which of the following are peroxoacids of sulphur?
(i) $H_{2}SO_{5}\; and \; H_{2}S_{2}O_{8}$
(ii) $H_{2}SO_{5}\; and \; H_{2}S_{2}O_{7}$
(iii) $H_{2}S_{2}O_{7} \; and \; H_{2}S_{2}O_{8}$
(iv) $H_{2}S_{2}O_{6} \; and \; H_{2}S_{2}O_{7}$
Answer:

The answer is the option (i).Peroxoacids of sulphur should contain one $-O-O-$ bond as shown below

Question 22. Hot conc. $H_{2}SO_{4}$ acts as a moderately strong oxidising agent. It oxidises both metals and nonmetals. Which of the following element is oxidised by conc. $H_{2}SO_{4}$ into two gaseous products?
(i) $Cu$
(ii) $S$
(iii) $C$
(iv) $Zn$
Answer:

The answer is the option (c)

$\left [ H_{2}SO_{4}\rightarrow H_{2}O+SO_{2}+O \right ]\times 2$
$C+2O\rightarrow CO_{2}$
$C+2H_{2}SO_{4}\rightarrow CO_{2}+2SO_{2}+2H_{2}O$
$CO_{2}$ and $SO_{2}$ are two gaseous products formed by oxidation of $C$ by $H_{2}SO_{4}$

Question 23. A black compound of manganese reacts with a halogen acid to give greenish yellow gas. When an excess of this gas reacts with $NH_{3}$ an unstable trihalide is formed. In this process the oxidation state of nitrogen changes from _________.
(i) – 3 to +3
(ii) – 3 to 0
(iii) – 3 to +5
(iv) 0 to – 3
Answer:

The answer is the option (i).

$MnO_{2}+4HCl\rightarrow MnCl_{2}+2H_{2}O+Cl$ (greenish yellow gas)

on reaction of excess of $Cl_{2}$ with $NH_{3}$ we obtained $NCl_{3}$ and $HCl.$


$NH_{3}+3Cl_{2}\rightarrow NCl_{3}+3HCl$
O.S. (-3) O.S.(+3)

Question 24. In the preparation of compounds of $Xe$, Bartlett had taken $O_{2}^+Pt F^{-}_{6}$ as a base compound. This is because
(i) both $O_{2}$ and $Xe$ has the same size.
(ii) both $O_{2}$ and $Xe$ has the same electron gain enthalpy.
(iii) both $O_{2}$ and $Xe$ has almost same ionisation enthalpy.
(iv) both $Xe$ and $O_{2}$ are gases
Answer:

The answer is the option (iii). The ionization enthalpy of $O_{2}$ is nearly equal to that of $Xe$, because of which Barlett chose $O_{2}$ compounds as base.

Question 25. In solid state $PCl_{5}$ is a _________.
(i) covalent solid
(ii) octahedral structure
(iii) ionic solid with $\left [ PCl_{6} \right ]^{+}$ octahedral and $\left [ PCl_{4} \right ]^{-}$ tetrahedra
(iv) ionic solid with $\left [ PCl_{4} \right ]^{+}$tetrahedral and $\left [ PCl_{6} \right ]^{-}$ octahedral
Answer:

The answer is the option (iv). Structure of $PCl_{5}$ in solid state

Question 26.

Reduction potentials of some ions are given below. Arrange them in decreasing order of oxidising power.

Ion
Reduction potential $E^{\ominus }/V$

$ClO_{4}^{-}$
$E^{\ominus }=1.19V$

$IO_{4}^{-}$
$E^{\ominus }=1.65V$

$BrO_{4}^{-}$
$E^{\ominus }=1.74V$

(i) $ClO_{4}^{-} > IO_{4}^{-} > BrO_{4}^{-}$
(ii) $IO_{4}^{-} > BrO_{4}^{-}>ClO_{4}^{-}$
(iii) $BrO_{4}^{-} > IO_{4}^{-} >ClO_{4}^{-}$
(iv) $BrO_{4}^{-} > ClO_{4}^{-} >IO_{4}^{-}$

Answer:

The answer is the option (iii).A higher reduction potential means that there is greater tendency to get reduced. Therefore, the order of oxidizing power is as follows
$BrO_{4}^{-} > IO_{4}^{-} >ClO_{4}^{-}$

Question 27. Which of the following is isoelectronic pair?
(i) $ICl_{2},ClO_{2}$
(ii) $BrO_{2}^{-}, BrF_{2}^{+}$
(iii) $ClO_{2}, BrF$
(iv) $CN^{-}, O_{3}$
Answer:

The answer is the option (ii). Isoelectronic pair have same number of electrons
$(i)ICl_{2}=53+2\times 17$

$=87$
$ClO_{2}=17+16$

$=33$
$(ii)BrO_{2}^{-}=35+2\times 8+1$

$=52$
$BrF_{2}^{+}=35+9\times2-1 $

$= 52$
$(iii)\; ClO_{2}=17+16$

$=33$
$BrF=35+9$

$=44$
$(iv)CN^{-}=6+7+1$

$=14$
$O_{3}=8\times 3$

$=24$
Hence, option b is correct.

NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: MCQ (Type 2)

The class 12 chemistry chapter 7 questions are provided here with simple explanations. Learn more through these advanced MCQs

Question 28. If chlorine gas is passed through a hot $NaOH$ solution, two changes are observed in the oxidation number of chlorine during the reaction. These are ________ and _________.
(i) 0 to +5
(ii) 0 to +3
(iii) 0 to –1
(iv) 0 to +1
Answer:

The answer is the option (i) and (iii). The oxidation state changes from 0 to +5 and 0 to -1 respectively.

Question 29. Which of the following options are not in accordance with the property mentioned against them?
(i) $F_{2}> Cl_{2}> Br_{2}> I_{2}$ Oxidising power.
(ii) $MI > MBr > MCl > MF$ Ionic character of metal halide.
(iii) $F_{2}> Cl_{2}> Br_{2}> I_{2}$ Bond dissociation enthalpy.
(iv) $HI < HBr < HCl < HF$ Hydrogen-halogen bond strength.
Answer:

The answer is option (ii, iii) The correct order of bond dissociation enthalpy is $Cl_{2}>Br_{2}>F_{2}>I_{2}$
There is an increase in the ionic character of metal halides with the electronegativity of halogen
$MI > MBr > MCI > MF$

Question 30. Which of the following is correct for P4 the molecule of white phosphorus?
(i) It has 6 lone pairs of electrons.
(ii) It has six P–P single bonds.
(iii) It has three P–P single bonds.
(iv) It has four lone pairs of electrons.
Answer:

The answer is the option (ii, iv) Structure of $P_{4}$ molecule can be represented as

There are 4 lone pairs of electrons over each P-atom. It has six P – P single bond.

Question 31. Which of the following statements are correct?
(i) Among halogens, radius ratio between iodine and fluorine is maximum.
(ii) Leaving $F-F$ bond, all halogens have weaker $X-X$ bond than $X-X'$ bond in interhalogens.
(iii) Among interhalogen compounds, the maximum number of atoms are present in iodine fluoride.
(iv) Interhalogen compounds are more reactive than halogen compounds.
Answer:

The answer is the option (i, iii, iv)
Iodine is the largest halogen, while fluorine is the smallest. $\frac{R_{Iodine}}{R_{fluorine}}$ is thus the maximum. A-B bond in interhalogen compound is weaker than A-A bond in halogen compounds; this also means that interhalogen compounds are more reactive.

Question 32. Which of the following statements are correct for $SO_{2}$ gas?
(i) It acts as a bleaching agent in moist conditions.
(ii) Its molecule has linear geometry.
(iii) Its dilute solution is used as a disinfectant.
(iv) It can be prepared by the reaction of dilute $H_{2}SO_{4}$ with metal sulphide.
Answer:

The answer is the option (i, iii). $SO_{2}$ is used as a bleaching agent in presence of moisture because of the reducing nature of $SO_{2}$
$SO_{2}+2H_{2}O\rightarrow H_{2}SO_{4}+2H$
It is also used as a preservative, disinfectant and anti-chlor.

Question 33. Which of the following statements are correct?
(i) All the three N—O bond lengths in $HNO_{3}$ are equal.
(ii) All P—Cl bond lengths in $PCl_{5}$ molecule in the gaseous state are equal.
(iii) $P_{4}$ molecule in white phosphorus have angular strain therefore white phosphorus is very reactive.
(iv) $PCl$ is ionic in the solid-state in which cation is tetrahedral and the anion is octahedral.
Answer:

The answer is the option (iii, iv).
(i) All the N – O bond in $HNO_{3}$ are not of the same bond length.
(ii) In gaseous state, all P – Cl bond lengths in $PCl_{5}$ molecule are not equal. Axial bond is longer than equatorial bond.
(iii) White phosphorous is more reactive than the other solid phases because of angular strain in the $P_{4}$ molecule under normal conditions.
(iv) In Solid state $PCl_{5}$ exists as ionic solid in which cation is tetrahedral and anion is octahedral.
$Cation-\left [ PCl_{4} \right ]^{+}\; \; \; \; \; \; Anion-\left [ PCl_{6} \right ]^{-}$

Question 34. Which of the following orders are correct as per the properties mentioned against each?
(i) $As_{2}O_{3}<SiO_{2}<P_{2}O_{3}<SO_{2}$ Acid strength.
(ii) $AsH_{3}<PH_{3}<NH_{3}$ Enthalpy of vaporization.
(iii) $S < O < Cl < F$ More negative electron gain enthalpy.
(iv) $H_{2}O >H_{2}S>H_{2}Se>H_{2}Te$ Thermal stability.

Answer:

The answer is the option (i, iv) There is an increase in the acidic strength of the group as we move up the group or left to right along a period. As we move down the group, thermal stability of hydrides decreases.

Question 35. Which of the following statements are correct?
(i) S–S bond is present in $H_{2}S_{2}O_{6}$
(ii) In peroxosulphuric acid $\left ( H_{2}SO_{5} \right )$ sulphur is in +6 oxidation state.
(iii) Iron powder along with $Al_{2}O_{3}$ and $K_{2}O$ is used as a catalyst in the preparation of $NH_{3}$ by Haber’s process.
(iv) Change in enthalpy is positive for the preparation of $SO_{3}$ by catalytic oxidation of $SO_{2}$
Answer:

The answer is the option (i, ii)
In $H_{2}SO_{5}$, there is a peroxo-linkage.

$O$ in peroxide linkage has oxidation state - 1

Question 36. In which of the following reactions conc. $H_{2}SO_{4}$ is used as an oxidising reagent?
(i) $CaF_{2}+ H_{2}SO_{4} \rightarrow CaSO_{4}+ 2HF$
(ii) $2HI + H_{2}SO_{4} \rightarrow I_{2} + SO_{2}+ 2H_{2}O$
(iii) $Cu + H_{2}SO_{4} \rightarrow CuSO_{4} + SO_{2}+ 2H_{2}O$
(iv) $NaCl + H_{2}SO_{4} \rightarrow NaHSO_{4} + HCl$
Answer:

The answer is option (ii, iii). Amongst the given options, (ii). and (iii). represent oxidizing behaviour of $H_{2}SO_{4}$. The oxidizing agent reduces itself in the process. In option b, it oxidises HI and gets reduced to $SO_{2}$ .Here,

Question 37. Which of the following statements are true?
(i) The only type of interactions between particles of noble gases is due to weak dispersion forces.
(ii) Ionisation enthalpy of molecular oxygen is very close to that of xenon.
(iii) Hydrolysis of XeF6 is a redox reaction.
(iv) Xenon fluorides are not reactive.
Answer:

The answer is the option (i, ii). Weak dispersion forces are responsible for attraction in noble gases. The ionisation enthalpy of molecular oxygen and xenon are very similar. Also, Xenon fluorides are reactive in nature.

NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: Short Answer Type

The short-answer type questions are also given here in the NCERT Exemplar Class 12 Chemistry Solutions Chapter 7 p block elements for practice. This section contains important questions that are asked in the exams. Practice short answer types from the questions below

Question 38. In the preparation of $H_{2}SO_{4}$ by Contact Process, why is $SO_{3}$ not absorbed directly in water to form $H_{2}SO_{4}$ ?
Answer:

Formation of acid fog takes place which is difficult to condense.

Question 39. Write a balanced chemical equation for the reaction showing catalytic oxidation of $NH_{3}$ by atmospheric oxygen.
Answer:

$4NH_{3}+5O_{2}\left ( from\; air \right )\overset{Pt\; gauge}{\rightarrow}4NO+6H_{2}O$

Question 40. Write the structure of pyrophosphoric acid.

Answer:

$H_{4}P_{2}O_{7}$ is pyrophosphoric acid. Its structure is

Question 41. $PH_{3}$ forms bubbles when passed slowly in water but $NH_{3}$ dissolves. Explain why?
Answer:

$NH_{3}$ dissolves in water as it forms hydrogen bonds with water but $PH_{3}$ cannot form hydrogen bonds with water so it escapes as gas.

Question 42. In $PCl_{5}$, phosphorus is in $sp^{3}d$ hybridised state, but all its five bonds are not equivalent. Justify your answer with reason.
Answer:

$PCl_{5}$ has a trigonal bipyramidal structure in liquid and gaseous phases.

The 3 equatorial P–Cl bonds are equivalent whereas 2 axial bonds are longer than equatorial bonds. This is because axial bond pairs suffer more repulsion compared to the equatorial bond pairs.

Question 43. Why is nitric oxide paramagnetic in gaseous state but the solid obtained on cooling it is diamagnetic ?
Answer:

In the gaseous state, nitric oxide $\left ( NO \right )$ exists as a monomer having one unpaired electron as it is paramagnetic in nature.

However, in solid state it dimerises to $N_{2}O_{2}$ which is diamagnetic in nature.

Question 44. Give reason to explain why $ClF_{3}$ exists but $FCl_{3}$ does not exist.
Answer:

Though both $Cl \; and \; F$ are halogens, $F$ has a configuration $2s^{2}2p^{5}$ and doesn’t have a d-orbital. Chlorine on the other hand can undergo $sp^{3}d$-hybridisation, where three orbitals will have unpaired electrons as shown below. Three fluorine atoms with a single unpaired electron each can pair up with the $sp^{3}d$-hybridised chlorine atom.

Question 45. Out of $H_{2}O$ and $H_{2}S$, which one has a higher bond angle and why?
Answer:

Due to higher electronegativity of Oxygen than Sulphur, O-H bond will have a higher electron density closer to the Oxygen atom. The two O-H bonds will repel each other more strongly than S-H bond leading to a larger bond angle in $H_{2}O$.

Question 46. $SF_{6}$ is known but $SCl_{6}$ is not. Why?
Answer:

There are two reasons as to why $SF_{6}$ exists in nature, but $SCl_{6}$ doesn’t:

  • Fluorine is smaller in size than Chlorine and is easily accommodated around Sulphur atom.

  • Due to higher electronegativity possessed by fluorine, it can unpair the Sulphur atom to the highest degree (where it attains the oxidation state of +6).

Question 47. On reaction with $Cl_{2}$, phosphorus forms two types of halides ‘A’ and ‘B’. Halide A is a yellowish-white powder, but halide ‘B’ is a colourless oily liquid. Identify A and B and write the formulas of their hydrolysis products.
Answer:

On passing dry chlorine over white phosphorus, first colourless oily phosphorus trichloride is formed. $PCl_{3}$ again reacts with $Cl_{2}$ and forms yellowish white powder like phosphorus pentachloride $PCl_{5}$

(i)

(ii)

$PCl_{5}$

$PCl_{3}$

Yellowish

Colorless Oily

White Powder

Liquid

$PCl_{3}$ reacts violently with water forming

phosphosphorous acid $\left ( H_{3}PO_{3} \right )$
$PCl_{3}+3H_{2}O\rightarrow H_{3}PO_{3}+3HCl$

$PCl_{5}$ undergoes a violent hydrolysis
$PCl_{5}+H_{2}O\rightarrow POCl_{3}+2HCl$
$PCl_{5}+4H_{2}O\rightarrow H_{3}PO_{4}+5HCl$

Question 48. In the ring test of $NO^{-}_{3}$ ion, $Fe^{2+}$ ion reduces nitrate ion to nitric oxide, which combines with $Fe^{2+}(aq)$ ion to form the brown complex. Write the reactions involved in the formation of a brown ring.
Answer:

$NO_{3}^{-}+3Fe^{2+}+4H^{+}\rightarrow NO+3Fe^{3+}+2H_{2}O$


$\left [ Fe\left ( H_{2}O \right )_{6} \right ]SO_{4}+NO\rightarrow \left [ Fe\left ( H_{2}O \right )_{5}NO \right ]SO_{4}\; \left ( Brown \; Complex \right )+H_{2}O$

Question 49. Explain why the stability of oxoacids of chlorine increases in the order given below:
$HClO< HClO_{2}< HClO_{3}< HClO_{4}$
Answer:

As number of oxygen atoms around chlorine increases, the electron density moves away from chlorine. It is so because, oxygen atoms are more electronegative than chlorine atoms.
The stability of ions will also increase with a higher number of Oxygen atoms:
$ClO^{-}<ClO_{2}^{-}<ClO_{3}^{-}<ClO_{4}^{-}$
Thus due to increase in stability of conjugate base, acidic strength of corresponding acid increases in the following order
$HClO< HClO_{2}< HClO_{3}< HClO_{4}$

Question 50. Explain why ozone is thermodynamically less stable than oxygen.
Answer:

Ozone is thermodynamically unstable with respect to oxygen because its decomposition results in liberation of heat $\left ( \Delta H=-ve \right )$ and increase in entropy $\left ( \Delta H=+ve \right )$
$2O_{3}\rightleftharpoons 3O_{2}$
These 2 factors reinforce each other resulting in large negative Gibbs free energy change $\left ( \Delta G=-ve \right )$ for its conversion into oxygen.
Therefore, the high concentration of ozone can result into dangerous explosion

Question 51. $P_{4}O_{6}$ reacts with water according to equation $P_{4}O_{6}+6H_{2}O\rightarrow 4H_{3}PO_{3}.$
Calculate the volume of 0.1 M $NaOH$ solution required to neutralise the acid formed by dissolving 1.1 g of $P_{4}O_{6}$ in $H_{2}O$.
Answer:

$P_{4}O_{6}+6H_{2}O\rightarrow 4H_{3}PO_{3}$


$4H_{3}PO_{3}+8NaOH\rightarrow 4Na_{2}HPO_{3}+8H_{2}O$

($H_{3}PO_{3}$ can be neutralised with $NaOH$ )


$P_{4}O_{6}(1 \; mol)+8NaOH (8\; mol)\rightarrow 4Na_{2}HPO_{3}+2H_{2}O$
Moles of $P_{4}O_{6} = \frac{1.1}{220}$

$=0.005\; mol$


Acid formed by 1 mol of $P_{4}O_{6}$ Require 8 mol $NaOH$


Acid formed by 0.005 mol of $P_{4}O_{6}$ required $=8\times 0.005$

$=0.04\; mol$


0.1 M $NaOH$ means 0.1 mol of $NaOH$ is present in 1000 mL solution


0.4 M of $NaOH$ is present in solution $=\frac{1000}{0.1}\times 0.04$

$=400mL$

Question 52. White phosphorus reacts with chlorine and the product hydrolyses in the presence of water. Calculate the mass of $HCl$ obtained by the hydrolysis of the product formed by the reaction of 62 g of white phosphorus with chlorine in the presence of water.
Answer:

Phosphorus reacts with $Cl_{2}$ as :

$P_{4}+6Cl_{2}\rightarrow 4PCl_{3}$


$PCl_{3}+3H_{2}O\rightarrow H_{3}PO_{3}+3HCl] \times 4 \left ( PCl_{3}\; is \; hydrolysed \right )$


$P_{4}+6Cl_{2}+12H_{2}O\rightarrow 4H_{3}PO_{3}+12HCl$


Moles of white $P=\frac{62}{124}$

$=0.5\; mol$


1 mol of white $P_{4}$ produce $HCl=12\; mol$


0.5 mol of white $P_{4}$ will produce $HCl=12\; \times 0.5$

$=6\; mol$


Mass of $HCl=6\times 36.5$

$=219.0\; g$

Question 53. Name three oxoacids of nitrogen. Write the disproportionation reaction of that oxoacid of nitrogen in which nitrogen is in +3 oxidation state.

Answer:

The 3 oxyacids of nitrogen are:

  1. Nitric acid $\left ( HNO_{3} \right )$

  2. Nitrous acid $\left ( HNO_{2} \right )$

  3. Hyponitrous acid $\left ( H_{2}N_{2}O_{2} \right )$

The disproportionation reaction wherein the nitrogen is in +3 oxidation state is:
$\\3HNO_{2}\rightarrow HNO_{3}+2NO+H_{2}O\\$

$+3\; \; \; \; \; \; \; \; \; \; \; \; +5\; \; \; \; \; \; \; \; \; +2$
Oxidation state changes from +3 to +5 and +2

Question 54. Nitric acid forms an oxide of nitrogen on reaction with $P_{4}O_{10}$. Write the reaction involved. Also, write the resonating structures of the oxide of nitrogen formed.

Answer:

$P_{4}O_{10}+4NHO_{3}\rightarrow 4HPO_{3}+2N_{2}O_{5}$

Reactivity: White phosphorus are more reactive compared to Red phosphorus as it undergoes more angular strain since the bonds are angled at 60°.

Question 55. Phosphorus has three allotropic forms — (i) white phosphorus (ii) red phosphorus and (iii) black phosphorus. Write the difference between white and red phosphorus on the basis of their structure and reactivity.
Answer:

White phosphorous exists as a single molecule where the phosphorous atoms are $sp^{3}$-hybridised and has a tetrahedral structure with 6 P-P bonds. It is highly reactive due to the strained bond angles (60o).

Red Phosphorous on the other hands also has a tetrahedral structure, but each tetrahedron is linked to two others as shown below. The bond angle is much more relaxed, leading to a lower reactivity than White phosphorus.

Question 56. Give an example to show the effect of concentration of nitric acid on the formation of oxidation product.

Answer:

We get different oxidation products on reaction of dilute and concentrated nitric acid with copper metals.

For example,

zinc with very dilute

$HNO_{3}$ ( 6%) forms $NH_{4}NO_{3},$

with dilute

$HNO_{3}$ ( 20%) forms $N_{2}O$

and with concentration.

$HNO_{3}$ ( 70%) forms $NO_{2}.$


(Very dilute) $HNO_{3}:4Zn+10HNO_{3}\rightarrow 4Zn(NO_{3})_{2}+NH_{4}NO_{3}+3H_{2}O$


(Dilute) $HNO_{3}:4Zn+10HNO_{3}\rightarrow 4Zn(NO_{3})_{2}+N_{2}O+5H_{2}O$


(Concentrated) $HNO_{3}:Zn+HNO_{3}\rightarrow Zn(NO_{3})_{2}+2NO_{2}+2H_{2}O$

Question 57. $PCl_{5}$ reacts with finely divided silver on heating and a white silver salt is obtained, which dissolves on adding excess aqueous $NH_{3}$ solution. Write the reactions involved to explain what happens.

Answer:

$PCl_{5}$ reacts with finely divided silver on heating to form white silver salt $(AgCl)$. This dissolves on addition of excess aqueous ammonia to form soluble complex.
$PCl_{5}+2Ag\rightarrow 2AgCl+PCl_{3}$
$AgCl+2NH _3(aq)\rightarrow [Ag(NH_3 )_ 2]+Cl ^{-}$

Question 58. Phosphorus forms a number of oxoacids. Out of these oxoacidsphosphinic acid has strong reducing property. Write its structure and also write a reaction showing its reducing behaviour.
Answer:

One of the oxoacids of phosphorous that has strong reducing property is phosphinic acid or hypophosphorus acid $\left ( H_{3}PO_{2} \right )$. This is due to possession of P – H bonds.

Its reducing properties are :


$4AgNO_{3}+H_{3}PO_{2}+2H_{2}O\rightarrow 4Ag+H_{3}PO_{4}+4HNO_{3}$


$2HgCl_{2}+H_{3}PO_{2}+2H_{2}O\rightarrow 2Hg+H_{3}PO_{4}+4HCl$


$H_{3}PO_{2}+2H_{2}O+2Cl_{2}\rightarrow H_{3}PO_{4}+4HCl$

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NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: Matching Type

The class 12 chemistry chapter 7 important questions are discussed below. These are generally asked in exams to test your knowledge. These solutions is quite helpful for competitive exams.

Question 59. Match the compounds given in Column I with the hybridization and shape given in Column II and mark the correct option.

Column I

Column II

(A)

$\text {XeF}_{6}$

(1)

$\text {sp}^{3}\text {d}^{2}$ - distorted octahedral

(B)

$\text {XeO}_{3}$

(2)

$\text {sp}^{3}\text {d}^{2}$-square planar

(C)

$\text {XeOF}_{4}$

(3)

$\text {sp}^{3}$ - pyramidal

(D)

$\text {XeF}_{4}$

(4)

$\text {sp}^{3}\text {d}^{2}$-square pyramidal

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(a) A (1) B(3) C(4) D(2)
(b) A (1) B(2) C(4) D(3)
(c) A (4) B(3) C(1) D(2)
(d) A (4) B(1) C(2) D(3)

Answer:

(a) $(A\rightarrow 1),(B\rightarrow 3),(C\rightarrow 4),(D\rightarrow 2)$

Compound

Hybridisation

(i).

(1)

$(\text {sp}^{3}\text {d}^{2})$- distorted octahedral

(ii).

(2)

$(\text {sp}^{3})$ - pyramidal

(iii).

(3)

$(\text {sp}^{3}\text {d}^{2})$- square pyramidal

(iv).

(4)

$(\text {sp}^{3}\text {d}^{2})$ - square planar

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Question 60. Match the formulas of oxides given in Column I with the type of oxide given in Column II and mark the correct option.

Column I

Column II

(A)

$Pb_{3}O_{4}$

(1)

Neutral Oxide

(B)

$N_{2}O$

(2)

Acidic Oxide

(C)

$Mn_{2}O_{7}$

(3)

Basic Oxide

(D)

$Bi_{2}O_{3}$

(4)

Mixed Oxide

(a) A (1) B(2) C(3) D(4)
(b) A (4) B(1) C(2) D(3)
(c) A (3) B(2) C(4) D(1)
(d) A (4) B(3) C(1) D(2)

Answer:

(b). (A $\rightarrow$ 4), (B $\rightarrow$ 1), (C $\rightarrow$ 2), (D $\rightarrow$ 3)

Formulas

Type of Oxide

(i)

$Pb_{3}O_{4}$ $(PbO.Pb_{2}O_{3})$

(4)

Mixed Oxide

(ii)

$N_{2}O$

(1)

Neutral Oxide

(iii)

$Mn_{2}O_{7}$

(2)

Acidic Oxide

(iv)

$Bi_{2}O_{3}$

(3)

Basic Oxide

Question 61. Match the items of Column I and Column II and mark the correct option.

Column I

Column II

(A)

$\text {H}_{2}\text {SO}_{4}$

(1)

Highest electron gain enthalpy

(B)

$\text {CCl}_{3}\text {NO}_{2}$

(2)

Chalcogen

(C)

$\text {Cl}_{2}$

(3)

Tear gas

(D)

$\text {Sulphur}$

(4)

storage batteries

(a) A (4) B(3) C(1) D(2)
(b) A (3) B(4) C(1) D(2)
(c) A (4) B(1) C(2) D(3)
(d) A (2) B(1) C(3) D(4)

Answer:

(a). (A $\rightarrow$ 4), (B $\rightarrow$ 3), (C $\rightarrow$ 1), (D $\rightarrow$ 2)

Question 62. Match the species given in Column I with the shape given in Column II and mark the correct option.

Column I

Column II

(A)

$\text {SF}_{4}$

(1)

Tetrahedral

(B)

$\text {BrF}_{3}$

(2)

Pyramidal

(C)

$\text {BrO}_{3}^{-}$

(3)

Sea-Saw shaped

(D)

$\text {NH}_{4}^{+}$

(4)

Bent-T-shaped

(a) A (3) B(2) C(1) D(4)
(b) A (3) B(4) C(2) D(1)
(c) A (1) B(2) C(3) D(4)
(d) A (1) B(4) C(3) D(2)

Answer:

(b) (A $\rightarrow$3), (B $\rightarrow$ 4), (C $\rightarrow$ 2), (D $\rightarrow$ 1)

Species

Shape

Structure

$\text {SF}_{4}$

Sea-Saw shaped

$\text {BrF}_{3}$

Bent-T-shaped

$\text {BrO}_{3}^{-}$

Pyramidal

$\text {NH}_{4}^{+}$

Tetrahedral

Question 63. Match the items of Column I and Column II and mark the correct option.

Column I

Column II

(A)

Its partial hydrolysis does not change oxidation state of central atom

(1)

$\text {He}$

(B)

It is used in modern diving apparatus

(2)

$\text {XeF}_{6}$

(C)

It is used to provide inert atmosphere for filling electrical bulbs

(3)

$\text {XeF}_{4}$

(D)

Its central atom is in $\text {sp}^{3}\text {d}^{2}$ hybridisation

(4)

$\text {Ar}$

(a) A (1) B(4) C(2) D(3)
(b) A (1) B(2) C(3) D(4)
(c) A (2) B(1) C(4) D(3)
(d) A (1) B(3) C(2) D(4)

Answer:

(c). (A $\rightarrow$ 2), (B $\rightarrow$ 1), (C $\rightarrow$ 4), (D $\rightarrow$ 3)

  1. Oxidation state of $\text {Xe}$ is unaffected by partial hydrolysis of $\text {XeF}_{6}$

  2. He is utilised in the modern driving apparatus.

  3. Air is used for filling electrical bulbs to provide an inert atmosphere.

  4. Xe has 4 bond pair and 2 lone pair and a hybridisation state of $\text {sp}^{3}\text {d}^{2}$ in $\text {XeF}_{4}$

NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: Assertion and Reason Type

This is one of the most important sections covered in the NCERT exemplar Class 12 chemistry solutions chapter 7. These questions will improve your critical thinking. The most typical and important section for exams

Question 64. In the following question, a statement of Assertion (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion (I): $N_{2}$ is less reactive than $P_{4}$
Reason (R): Nitrogen has more electron gain enthalpy than phosphorus.
(i) Both Assertion and Reason are correct statements and Reason is the correct explanation of Assertion.
(ii) Both Assertion and Reason are correct statements but Reason is not the correct explanation of Assertion.
(iii) Assertion is correct but the reason is wrong.
(iv) Assertion is wrong but the Reason is correct.
(v) Both Assertion and Reason are wrong.
Answer:

The answer is option (iii). Nitrogen has a higher electron gain enthalpy than Phosphorous. N-N atoms in Nitrogen are bonded by triple bond, whereas Phosphorous atoms are bonded by single P-P bond, due to which Nitrogen has a high bond dissociation energy and is less reactive.

Question 65. In the following question, a statement of Assertion (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion (I): $HNO_{3}$ makes iron passive.
Reason (R): $HNO_{3}$ forms a protective layer of ferric nitrate on the surface of iron.
(i) Both Assertion and Reason are correct statements and Reason is the correct explanation of Assertion.
(ii) Both Assertion and Reason are correct statements but Reason is not the correct explanation of Assertion.
(iii) Assertion is correct but Beason is wrong.
(iv) Assertion is wrong but Reason is correct.
(v) Both Assertion and Reason are wrong.
Answer:

The answer is option (iii). Iron attains passivity due to formation of a thin layer of oxide.

Question 66. In the following question, a statement of Assertion (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion (I): $HI$ cannot be prepared by the reaction of $KI$ with concentrated $H_{2}SO_{4}$.
Reason (R): $HI$ has lowest H – X bond strength among halogen acids.
(i) Both Assertion and Reason are correct statements and Reason is the correct explanation of Assertion.
(ii) Both Assertion and Reason are correct statements but Reason is not the correct explanation of Assertion.
(iii) Assertion is correct but Beason is wrong.
(iv) Assertion is wrong but Reason is correct.
(v) Both Assertion and Reason are wrong.
Answer:

The answer is option (ii). Though correct, both the statements are independent of each other.

Question 67. In the following question, a statement of Assertion (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion (i): Both rhombic and monoclinic sulphur exist as $S_{8}$ but oxygen exists as $O_{2}$.
Reason (R): Oxygen forms $p\pi -p\pi$ multiple bond due to small size and small bond length but $p\pi -p\pi$ bonding is not possible in sulphur.
(i) Both Assertion and Reason are correct statements and Reason is the correct explanation of Assertion.
(ii) Both Assertion and Reason are correct statements but Reason is not the correct explanation of Assertion.
(iii) Assertion is correct but Beason is wrong.
(iv) Assertion is wrong but Reason is correct.
(v) Both Assertion and Reason are wrong.
Answer:

The answer is option (i). Sulphur cannot form $p\pi -p\pi$ multiple bonds like oxygen, so it exists as $S_{8}$.

Question 68. In the following question, a statement of Assertion (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion (I): $NaCl$ reacts with concentrated $H_{2}SO_{4}$ to give colourless fumes with pungent smell. But on adding $MnO_{2}$ the fumes become greenish yellow.
Reason (R): $MnO_{2}$ oxidises $HCl$ to chlorine gas which is greenish yellow.
(i) Both Assertion and Reason are correct statements and Reason is the correct explanation of Assertion.
(ii) Both Assertion and Reason are correct statements but Reason is not the correct explanation of Assertion.
(iii) Assertion is correct but Beason is wrong.
(iv) Assertion is wrong but Reason is correct.
(v) Both Assertion and Reason are wrong.
Answer:

The answer is option (i). $MnO_{2}$ is a strong oxidising agent and it oxidises colourless fumes of $HCl$ to greenish yellow $Cl_{2}$

Question 69. In the following question, a statement of Assertion (i). followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion (i): $SF_{6}$ cannot be hydrolysed but $SF_{4}$ can be.
Reason (R): Six $F$ atoms in $SF_{6}$ prevent the attack of $H_{2}O$ on sulphur atom of $SF_{6}$ .
(i) Both Assertion and Reason are correct statements and Reason is the correct explanation of Assertion.
(ii) Both Assertion and Reason are correct statements but Reason is not the correct explanation of Assertion.
(iii) Assertion is correct but Beason is wrong.
(iv) Assertion is wrong but Reason is correct.
(v) Both Assertion and Reason are wrong.
Answer:

The answer is option (i). $H_{2}O$ can’t react with $SF_{6}$ because the fluorine atoms don’t allow $H_{2}O$ to attack $SF_{6}$, and thus, is sterically protected.

NCERT Exemplar Solutions Class 12 Chemistry Chapter 7: Long Answer Type

The following are the long-answer type questions that needs more practice and learning. These are the chemistry chapter 7 important questions that are asked in the exams.

Question 70. An amorphous solid “A” burns in air to form a gas “B” which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified aqueous $KMnO_{4}$ solution and reduces $Fe^{3+}$ to $Fe^{+2}$. Identify the solid “A” and the gas “B” and write the reactions involved.

Answer:

A is sulphur $(S_{8})$ while B is sulphur dioxide $(SO_{2})$.
$S_{8}+8O_{2}\rightarrow 8SO_{2}$


$2ZnS+3O_{2}\rightarrow 2ZnO+2SO_{2}$


$2KMnO_{4}+3H_{2}SO_{4}\rightarrow K_{2}SO_{4}+2MnSO_{4}+3H_{2}O+5\left [ O \right ]$


$\left [ SO_{2}+\left [ O \right ]+H_{2}O\rightarrow H_{2}SO_{4} \right ]\times 5$


$2KMnO_{4}+3H_{2}SO_{4}\rightarrow K_{2}SO_{4}+2MnSO_{4}+2H_{2}SO_{4}$

$SO_{2}+2H_{2}O\rightarrow H_{2}SO_{4}+2\left [ H \right ]$


$Fe_{2}\left ( SO_{4} \right )_{3}+2\left [ H \right ]\rightarrow 2FeSO_{4}+H_{2}SO_{4}$


$Fe_{2}\left ( SO_{4} \right )_{3}+SO_{2}+2H_{2}O\rightarrow 2FeSO_{4}+2H_{2}SO_{4}$

Question 71. On heating lead (II) nitrate gives a brown gas “A”. The gas “A” on cooling changes to colourless solid “B”. Solid “B” on heating with NO changes to a blue solid C. Identify A, B and C and also write reactions involved and draw the structures of B and C.
Answer:

The gas A is $NO_{2}$ . The reactions are given as under:
$2Pb\left ( NO_{3} \right )_{2}\overset{Heat}{\rightarrow}2PbO+4NO_{2}+O_{2}$


$2NO_{2}\overset{Cooling}{\rightarrow}N_{2}O_{4}$


$2NO+N_{2}O_{4}\overset{Heat}{\rightarrow}2N_{2}O_{3}$


Question 72. On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treated with 3 mol of hydrogen $(H_{2})$ in the presence of a catalyst gives another gas (C) which is basic in nature. Gas $C$ on further oxidation in moist condition gives a compound (D) which is a part of acid rain. Identify compounds (A) to (D) and also give necessary equations of all the steps involved.
Answer:

$(A)=NH_{4}NO_{2},\; \; (B)=N_{2},\; \; \; \; (C)=NH_{3},\; \; \; \; \; (D)=HNO_{3}$
$NH_{4}NO_{2}\overset{Heat}{\rightarrow}N_{2}+2H_{2}O$


$N_{2}+3H_{2}\overset{Catalyst}{\rightarrow}2NH_{3}$


$4NH_{3}+5O_{2}\rightarrow 4NO+6H_{2}O$


$4NO+O_{2}\rightarrow 2NO_{2}$


$3NO_{2}+H_{2}O\rightarrow 2HNO_{3}+NO$

NCERT Exemplar Class 12 Chemistry Chapter 7: Higher Order Thinking Skills (HOTS) Questions

The class 12 chemistry chapter 7 questions and answers are given below that will help you tackle complex problems. Students can follow p-block elements notes to learn the concepts in detail. The questions below will help you evaluate your understanding of the concepts.

Question 1. When a salt is treated with a sodium hydroxide solution, it gives gas X. On passing gas X through reagent Y, a brown-colored precipitate is formed. X and Y, respectively, are

1) $\mathrm{X}=\mathrm{NH}_3$ and $\mathrm{Y}=\mathrm{HgO}$

2) $\mathrm{X}=\mathrm{NH}_3$ and $\mathrm{Y}=\mathrm{K}_2 \mathrm{HgI}_4+\mathrm{KOH}$

3) $\mathrm{X}=\mathrm{NH}_4 \mathrm{Cl}$ and $\mathrm{Y}=\mathrm{KOH}$

4) $\mathrm{X}=\mathrm{HCl}$ and $\mathrm{Y}=\mathrm{NH}_4 \mathrm{Cl}$

Answer:

$\mathrm{NH}_4^{+}+\mathrm{NaOH} \longrightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{NH}_3 \uparrow$

$\mathrm{NH}_3$ is identify by $\mathrm{K}_2\left[\mathrm{HgI}_4\right]+\mathrm{KOH}$

Hence, the correct answer is option (2).

Question 2. Evaluate the following statements related to group 14 elements for their correctness.
(A)Covalent radius decreases down the group from $\mathrm{C}$ to $\mathrm{Pb}$ in a regular manner.
(B) Electronegativity decreases from $\mathrm{C}$ to $\mathrm{Pb}$ down the group gradually.
(C) Maximum covalence of $\mathrm{C}$ is 4 whereas other elements can expand their covalence due to the presence of d orbitals.
(D) Heavier elements do not form $\mathrm{p} \pi-\mathrm{p} \pi$ bonds.
(E) Carbon can exhibit negative oxidation states.

Choose the correct answer from the options given below:

1) (C), (D) and (E) Only

2) (A) and (B) Only

3) (A), (B) and (C) Only

4) (C) and (D) Only

Answer:

(A) Down the group; the radius increases
(B) Electronegativity does not decrease gradually from $\mathrm{C}$ to $\mathrm{Pb}$.
(C) Correct.
(D) Correct.
(E) Range of oxidation state of carbon: -4 to +4

Hence, the correct answer is option (1).

Question 3. The species that do not contain perodxie ions is:

(1) PbO2

(2) H2O2

(3) SrO2

(4) BaO2

Answer:

All members, when heated with oxygen, form oxides. There are mainly two types of oxides, i.e., monoxide and dioxide of formula MO and MO2, respectively. SiO only exists at high temperatures. Oxides in higher oxidation states of elements are generally more acidic than those in lower oxidation states. The dioxides - CO2, SiO2, and GeO2 are acidic, whereas SnO2 and PbO2 are amphoteric in nature. Among monoxides, CO is neutral, GeO is distinctly acidic, whereas SnO and PbO are amphoteric.

In PbO2 , Pb is in +4 oxidation state and oxygen is in the -2 oxidation State. In all other cases, the peroxide ion ($O_{2}^{2-}$) is present.

Hence, the correct answer is option (1).

Approach to Solve Questions of Class 12 Chemistry Chapter 7

To effectively solve questions of the chapter "P-Block Element", a systematic and structured approach that helps understand the concepts and the theories is given below:

1. Focus on Key Concepts

Each group has similar themes. Focus on:

  • Electronic configuration

  • Oxidation states and variable valency

  • Trends in physical and chemical properties (electronegativity, metallic character, ionization enthalpy)

  • Important compounds (e.g., NH3, HNO3, SO2, H2SO4, PCl5, etc.)

  • Anomalous behaviour of the first element (like N or O)

  • Trends down the group (acidic/basic nature, thermal stability, etc.)

The questions are directly asked in the exams from these topics.

2. Use a Comparative Approach

When solving theory questions compare elements down the group (e.g., N vs P vs As…). Also, compare group-wise behaviours (Group 15 vs Group 16). Learn more from p-block elements class 12 notes.

3. Focus on structures and hybridisation

Practice how to identify molecular structures (e.g., NH3, SO2, PCl5, ClF3). Also, try to understand types of hybridization (e.g., sp3d for PCl5). The NCERT Exemplar Class 12 Chemistry Solutions Chapter 7 will help you understand these concepts better.

4. Practice Conceptual and Memory-Based Questions

Refer to the NCERT in-text questions and exercises. You can also follow previous year questions as to get a clear idea of the exam pattern. You can get these type of questions in the NCERT Class 12 solutions available on our website. Practice previous year questions and solve mock tests. Some examples are given below

  • Why is N2 less reactive than P4?

  • Write one use of H2SO4.

  • Compare the thermal stability of H2O, H2S, H2Se.

  • What is the oxidation state of Cl in $ClO_{3}^{-}$?

5. Revise with Tables and Charts

Make group-wise comparison tables for similar trends. You can also prepare short flashcards for compounds, oxidation states, and trends. Make suitable short notes and revise them timely.

Topics and Subtopics Covered in the NCERT Exemplar Class 12 Chemistry Chapter 7

The following are the important topics covered in chapter 7. Practice more at NCERT Class 12 chemistry chapter 7 exemplar solutions.

NCERT Exemplar Class 12 Chemistry Chapter 7: Important Trends

The following are the important trends covered in the Class 12 NCERT Chemistry chapter 7.

  • Electronegativity: Increases from left to right across a period and generally decreases down a group.
  • Ionization Enthalpy: Generally increases from left to right across a period and decreases down a group, with some exceptions.
  • Metallic Character: Decreases from left to right and increases down a group.
  • Atomic Radius: Generally increases down a group and decreases from left to right across a period.
  • Oxidation States: Elements in the p-block exhibit a variety of oxidation states, generally related to their group number such as, group 15 elements can show +3 and +5 oxidation states.

NCERT Exemplar Solutions Class 12 Chemistry Chapter-Wise

Follow the links below to get chapter-wise NCERT Class 12 solutions for your exam preparation.

NCERT Exemplar Class 12 Chemistry Chapter 1 Solid State

NCERT Exemplar Class 12 Chemistry Chapter 2 Solutions

NCERT Exemplar Class 12 Chemistry Chapter 3 Electrochemistry

NCERT Exemplar Class 12 Chemistry Chapter 4 Chemical Kinetics
NCERT Exemplar Class 12 Chemistry Chapter 5 Surface Chemistry

NCERT Exemplar Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements

NCERT Exemplar Class 12 Chemistry Chapter 7 The p-Block Elements

NCERT Exemplar Class 12 Chemistry Chapter 8 The d & f Block Elements

NCERT Exemplar Class 12 Chemistry Chapter 9 Coordination Compounds

NCERT Exemplar Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes

NCERT Exemplar Class 12 Chemistry Chapter 11 Alcohols, Phenols, and Ethers

NCERT Exemplar Class 12 Chemistry Chapter 12 Aldehydes, Ketones, and Carboxylic Acids

NCERT Exemplar Class 12 Chemistry Chapter 13 Amines

NCERT Exemplar Class 12 Chemistry Chapter 14 Biomolecules

NCERT Exemplar Class 12 Chemistry Chapter 15 Polymers

NCERT Exemplar Class 12 Chemistry Chapter 16 Chemistry in Everyday Life

NCERT Solutions for Class 12 Chemistry Chapter-Wise

Click on the links below to access the NCERT Class 12 solutions for all the chapters to make your preparations better.

NCERT solutions for class 12 Chemistry chapter 1 Solutions

NCERT Solutions for class 12 Chemistry chapter 2 Electrochemistry

NCERT solutions for class 12 Chemistry chapter 3 Chemical Kinetics

NCERT solutions for class 12 Chemistry chapter 4 The d and f block elements

NCERT solutions for class 12 Chemistry chapter 5 Coordination compounds

NCERT Solutions for class 12 Chemistry chapter 6 Haloalkanes and Haloarenes

NCERT solutions for class 12 Chemistry chapter 7 Alcohols, Phenols, and Ethers

NCERT Solutions for class 12 Chemistry chapter 8 Aldehydes, Ketones and Carboxylic Acids

NCERT solutions for class 12 Chemistry chapter 9 Amines

NCERT solutions for class 12 Chemistry chapter 10 Biomolecules

NCERT Solutions subject-wise

The NCERT subject-wise solutions will help you broaden your concepts and will also help in revision. Learn more from Class 12 NCERT notes.

NCERT Solution for Class 12 Physics

NCERT Solution for Class 12 Chemistry
NCERT Solution for Class 12 Maths
NCERT Solution for Class 12 Biology

NCERT Exemplar Class 12 Solutions subject-wise

Excel your preparation with NCERT exemplar solutions. Click on the link below

NCERT Exemplar Class 12 Mathematics Solutions

NCERT Exemplar Class 12 Physics Solutions

NCERT Exemplar Class 12 Biology Solutions

NCERT Class 12 subject-wise notes

You can follow the links given in the table below to get access to the Class 12 NCERT notes.

NCERT Notes for Class 12 Physics

NCERT Notes for Class 12 Chemistry
NCERT Notes for Class 12 Maths
NCERT Notes for Class 12 Biology

NCERT Books and NCERT Syllabus

You can find links to the Class 12 NCERT chemistry book and syllabus for the respective subjects.

NCERT Books Class 12 Chemistry

NCERT Syllabus Class 12 Chemistry
NCERT Books Class 12
NCERT Syllabus Class 12

Frequently Asked Questions (FAQs)

Q: What are p-block elements, and where are they located in the periodic table?
A:

P-block elements are the elements found in groups 13 to 18 of the periodic table. They include metals, non-metals, and metalloids, and are characterized by the presence of their outermost electrons in the p-orbitals.

Q: How do the properties of p-block elements vary across a period?
A:

As we move from left to right across a period in the p-block, the elements typically become less metallic and more non-metallic. This shift is accompanied by an increase in electronegativity, ionization energy, and a decrease in atomic size.

Q: How do the oxidation states of elements in Group 15 vary as discussed in class 12 NCERT chemistry book? ?
A:

Group 15 elements are Nitrogen, Phosphorus, etc. they can exhibit a range of oxidation states from -3 to +5.

Q: What is the significance of the "staircase line" in the p-block?
A:

The staircase line or diagonal line separating metals and nonmetals in the p-block. It is important because the elements immediately bordering this line are metalloids.

Q: What are the general trends of electronegativity and ionization enthalpy across the p-block?
A:

The general trends of electronegativity and ionization enthalpy across the p-block are:

  • Electronegativity: Increases from left to right across a period and generally decreases down a group.
  • Ionization Enthalpy: Generally increases from left to right across a period and decreases down a group (with some exceptions).
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Questions related to CBSE Class 12th

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Have a question related to CBSE Class 12th ?
how may marks should i get in my 12th board exams to get fee concesion in sastra university

Hello Aspirant,

SASTRA University commonly provides concessions and scholarships based on merit in class 12 board exams and JEE Main purposes with regard to board merit you need above 95% in PCM (or on aggregate) to get bigger concessions, usually if you scored 90% and above you may get partial concessions. I suppose the exact cut offs may change yearly on application rates too.

Read Complete Answer
G
gopi
4 Sep'25
Which course is best for computer science students after 12th standard

Hello,

After 12th, if you are interested in computer science, the best courses are:

  • B.Tech in Computer Science Engineering (CSE) – most popular choice.

  • BCA (Bachelor of Computer Applications) – good for software and IT jobs.

  • B.Sc. Computer Science / IT – good for higher studies and research.

  • B.Tech in Information Technology (IT) – focuses on IT and networking.

All these courses have good career scope. Choose based on your interest in coding, software, hardware, or IT field.

Hope it helps !

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P
Prachi Kumari
4 Sep'25
I am a CBSE class 12 private candidate will my maksheet of CBSE supplement exam will be delivered on main exam form address for supplementary exam form address ,my both addresses are different please tell

Hello Vanshika,

CBSE generally forwards the marksheet for the supplementary exam to the correspondence address as identified in the supplementary exam application form. It is not sent to the address indicated in the main exam form. Addresses that differ will use the supplementary exam address.

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G
gopi
3 Sep'25
12th science gseb ncert 2025 me fail hua to muje 2026 me exam deni h abhi tak kis school me admission nhi Lia to abhi muje kisi school me admission milega ? Private candidate me exam de sakta hu? Private candidate process kya hota h kon form bharata h ? 12th science gseb 2026 me Private candidate me bina school ke exam de sakate h ??

Hii ansh!

Yes, the Gujarat board does allow students to register as private candidates for class 12th examinations that means that even if you are not enrolled in any school you can sit for the examination. For this you will have to fill out a special private candidate application form and submit it along with relevant documents like marksheet, ID card, photos, etc to the Gujarat state education board office. You can submit it either offline or online.

However, the 2026 application windows has not opened yet and when it will typically schools will handle the registration through their board login so as of now you will have to reach out any GSEB affiliated school or board office itself in order to seek guidance on how to apply as a private candidate for 2026.

Hope it helps !

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S
Shivani Chaurasia
7 Sep'25
I gave my 12th exam this year in which I got less marks. Then I gave improvement exam in which also I could not score full marks. So can I become a private candidate and appear for the exam again next year or not?

Hello

Yes, if you’re not satisfied with your marks even after the improvement exam, many education boards allow you to reappear as a private candidate next year to improve your scores. This means you can register independently, study at your own pace, and take the exams without attending regular classes. It’s a good option to improve your results and open up more opportunities for higher studies or careers. Just make sure to check the specific rules and deadlines of your education board so you don’t miss the registration window. Keep your focus, and you will do better next time.

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S
SEJAL JAIN
20 Aug'25
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Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
Read Complete Answer

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

 Option 4)

12.89×10−3 kg

 
Read Complete Answer

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
Read Complete Answer

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
Read Complete Answer

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
Read Complete Answer

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
Read Complete Answer

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

 Option 4)

be a function of the molecular mass of the substance.
Read Complete Answer

With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
Read Complete Answer

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
Read Complete Answer

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
Read Complete Answer

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