CBSE Class 12 Chemistry Chapter 7: The p-Block Elements - NCERT Solutions

Reviewed by NCERT Help Editorial Team · Updated on 29 Jul 2026

NCERT Solutions PDF Class 12 PDF

CBSE Class 12 Chemistry, Chapter 7: The p-Block Elements, delves into the fascinating world of elements in groups 15 to 18. This chapter explores key concepts such as the covalent character of halides, the varying reducing strengths of hydrides in Group 15, and the unique inertness of nitrogen gas. It also covers the essential conditions for synthesizing ammonia through the Haber process and the reaction of ammonia with copper(II) ions. Furthermore, the solutions explain the covalence of nitrogen in N₂O₅ and clarify the difference in bond angles between PH₄⁺ and PH₃. Understanding these topics is vital for grasping periodic trends and the chemical properties of p-block elements, providing a strong foundation for students preparing for their board examinations.

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Board	CBSE
Class	Class 12
Subject	Chemiry
Session	2026
Language	English
Type	NCERT Solutions
Chapter	Chapter 7: The p – Block Elements - Intext Questions Solutions

Chapter summary

This chapter's NCERT Solutions focus on the intext questions related to the p-block elements, specifically group 15. It explains the factors influencing the covalent character of halides, the trend in reducing properties of hydrides, and the reasons for nitrogen's low reactivity. Key concepts like the Haber process for ammonia synthesis and the coordination chemistry of ammonia are also addressed. The solutions clarify bonding and structural aspects, such as the covalence of nitrogen and bond angle variations in related species.

Learning outcomes

Understand the factors affecting covalent character in halides.
Explain the trend in reducing strength of group 15 hydrides.
Identify conditions for maximizing ammonia yield in the Haber process.
Describe the reaction of ammonia with metal ions.
Determine the covalence of an element from its structure.
Explain variations in bond angles based on hybridization and lone pairs.

Topics covered

Paper topics

Covalent character of halides
Reducing strength of hydrides
Reactivity of nitrogen
Haber process for ammonia
Reaction of ammonia with metal ions
Covalence of nitrogen
Bond angles in hydrides and their cations
Hybridization and lone pair effects

Important topics

Trends in reactivity of p-block elements
Factors affecting covalent character
Reducing properties of group 15 hydrides
Ammonia synthesis and properties
Bond angle variations

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Questions and Solutions

Question 7.1

Why are pentahalides more covalent than trihalides?

Solution: Pentahalides exhibit a higher oxidation state (+5) compared to trihalides (+3) for the central atom. According to Fajan's rules, a higher positive charge on the cation increases its polarizing power. This increased polarization of the anion (halide ion) leads to a greater extent of electron cloud distortion and sharing, resulting in a more covalent bond. Therefore, pentahalides are more covalent than trihalides.

Question 7.2

Why is BiH₃ the strongest reducing agent amongst all the hydrides of Group 15 elements?

Solution: As we move down Group 15 from Nitrogen (N) to Bismuth (Bi), the atomic size of the central atom increases. This leads to a decrease in the bond strength between the central atom and hydrogen (M-H bond strength decreases). Consequently, the stability of the hydrides decreases from NH₃ to BiH₃. A less stable hydride can more easily donate hydrogen atoms, making it a stronger reducing agent. Therefore, BiH₃, being the least stable hydride, acts as the strongest reducing agent among the Group 15 hydrides.

Question 7.3

Why is N₂ less reactive at room temperature?

Solution: The nitrogen molecule (N₂) consists of two nitrogen atoms joined by a triple covalent bond (N≡N). This triple bond is extremely strong and possesses a very high bond dissociation energy. A significant amount of energy is required to break this bond. Due to this high bond strength, the N₂ molecule is very stable and unreactive under normal conditions (room temperature and pressure).

Question 7.4

Mention the conditions required to maximise the yield of ammonia.

Solution: Ammonia is synthesized industrially via the Haber's process, represented by the reversible reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) + Heat. To maximize the yield of ammonia, the following conditions are employed, based on Le Chatelier's principle:

High Pressure: The forward reaction involves a decrease in the number of moles (4 moles of reactants form 2 moles of product). Therefore, high pressure (typically around 200 atm) favors the formation of ammonia.
Moderate Temperature: The reaction is exothermic. While low temperatures favor higher equilibrium yield, they result in a very slow reaction rate. A compromise temperature of about 700 K is used to achieve a reasonable rate of reaction along with a satisfactory yield.
Catalyst: Iron (Fe) in the form of finely divided particles, promoted with small amounts of K₂O and Al₂O₃, is used as a catalyst. The catalyst increases the rate of both forward and backward reactions, helping the system reach equilibrium faster without significantly affecting the equilibrium yield.

Question 7.5

How does ammonia react with a solution of Cu²⁺?

Solution: Ammonia (NH₃) acts as a Lewis base because the nitrogen atom has a lone pair of electrons. When ammonia reacts with a solution containing Cu²⁺ ions, it donates its lone pair of electrons to the Cu²⁺ ion, forming a coordination complex. The reaction is:

$\operatorname{Cu}^{2+}_{(aq)} + 4\operatorname{NH}_{3(aq)} \leftrightarrow \left[\operatorname{Cu}(\operatorname{NH}_3)_4\right]^{2+}_{(aq)}$

The initial solution containing Cu²⁺ ions is typically blue. Upon addition of excess ammonia, a deep blue colored complex, tetraamminecopper(II) ion, is formed.

Question 7.6

What is the covalence of nitrogen in N₂O₅?

Solution: The structure of dinitrogen pentoxide (N₂O₅) can be visualized as two NO₂ groups linked by an oxygen atom, or more accurately, it exists as two resonance structures, one involving a direct N-N bond and the other involving an N-O-N linkage. In the structure where nitrogen atoms are linked via oxygen, each nitrogen atom is bonded to three oxygen atoms (two double bonds and one single bond to the bridging oxygen, or vice versa depending on resonance). Considering the bonds formed by each nitrogen atom, including potential resonance structures and the bridging oxygen, the covalence of nitrogen in N₂O₅ is determined to be 4. This means each nitrogen atom forms four covalent bonds.

Question 7.7

Bond angle in PH₄⁺ is higher than that in PH₃. Why?

Solution: In phosphine (PH₃), the phosphorus atom is sp³ hybridized. Three of the sp³ hybrid orbitals form sigma bonds with the hydrogen atoms, and the fourth sp³ hybrid orbital contains a lone pair of electrons. According to VSEPR theory, the repulsion between a lone pair and bond pairs (lone pair-bond pair repulsion) is greater than the repulsion between two bond pairs (bond pair-bond pair repulsion). This stronger lone pair-bond pair repulsion pushes the P-H bonds closer together, reducing the H-P-H bond angle from the ideal tetrahedral angle of 109.5° to a smaller value (around 93.6°). In the phosphonium ion (PH₄⁺), the phosphorus atom is also sp³ hybridized, but all four sp³ hybrid orbitals are involved in bonding with four hydrogen atoms. There is no lone pair of electrons on the phosphorus atom in PH₄⁺. Consequently, the repulsions are only bond pair-bond pair repulsions. These repulsions are weaker than lone pair-bond pair repulsions. This allows the H-P-H bond angles to be closer to the ideal tetrahedral angle, resulting in a larger bond angle in PH₄⁺ compared to PH₃.

Common mistakes

Confusing polarizing power with ionic character.
Incorrectly applying VSEPR theory to explain bond angles without considering lone pairs.
Not recalling the specific conditions for the Haber process.
Misinterpreting the role of lone pairs in chemical bonding and reactivity.

Revision tips

Focus on understanding the trends in properties down the group for p-block elements.
Draw structures to visualize bonding and explain covalence and bond angles.
Memorize the conditions for the Haber process and the reaction of ammonia with metal ions.
Relate concepts like polarizing power and bond dissociation energy to reactivity.

Practice MCQs

Q1. Why are pentahalides generally more covalent than trihalides?

Q2. Which hydride of Group 15 is the strongest reducing agent?

Q3. What is the primary reason for the low reactivity of N₂ at room temperature?

Q4. In the Haber process for ammonia synthesis, which condition helps maximize the yield?

Q5. What is the observed color change when ammonia reacts with Cu²⁺ solution?

Q6. What is the covalence of nitrogen in N₂O₅?

Q7. Why is the bond angle in PH₄⁺ higher than in PH₃?

Frequently asked questions

What is the main focus of the NCERT Solutions for Chapter 7, The p-Block Elements?

These solutions focus on the intext questions of Chapter 7 for Class 12 Chemistry, explaining concepts like the covalent nature of halides, reducing properties of hydrides, nitrogen's reactivity, ammonia synthesis, and bonding characteristics of p-block elements.

How do these solutions help in understanding the reactivity of p-block elements?

The solutions explain why pentahalides are more covalent than trihalides and why N₂ is less reactive, linking these properties to oxidation states, polarizing power, and bond strengths.

What is explained regarding the hydrides of Group 15 elements?

The solutions clarify why BiH₃ is the strongest reducing agent among the hydrides of Group 15 elements, relating it to the decreasing stability of hydrides down the group.

Are the conditions for ammonia synthesis covered?

Yes, the solutions detail the conditions required to maximize the yield of ammonia in the Haber process, including high pressure, specific temperature, and the use of a catalyst.

How are bonding and structure explained in these solutions?

Concepts like the covalence of nitrogen in N₂O₅ and the difference in bond angles between PH₄⁺ and PH₃ are explained by considering hybridization and the effect of lone pairs on bond angles.

What is the significance of the reaction of ammonia with Cu²⁺?

The reaction illustrates ammonia acting as a Lewis base, donating an electron pair to form a complex ion with Cu²⁺, leading to a distinct color change from blue to deep blue.

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