CBSE Class 12 Chemistry Chapter 13: Organic Compounds Containing Nitrogen - NCERT Solutions

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

NCERT Solutions PDF Class 12 PDF

This resource provides detailed NCERT Solutions for CBSE Class 12 Chemistry, Chapter 13, focusing on Organic Compounds Containing Nitrogen. It covers essential topics such as the properties and reactivity of p-block elements, particularly nitrogen and its compounds. The solutions explain why pentahalides are more covalent than trihalides, the strong reducing nature of BiH₃, and the low reactivity of N₂ due to its triple bond. It also details the conditions for maximizing ammonia yield via the Haber process, the reaction of ammonia with Cu²⁺ ions, the covalence of nitrogen in N₂O₅, and the reasons for the difference in bond angles between PH₄⁺ and PH₃. These solutions are designed to help students understand complex chemical concepts and prepare effectively for their board examinations.

Home

Quick info

Board	CBSE
Class	Class 12
Subject	Chemistry
Session	2026
Language	English
Type	NCERT Solutions
Chapter	Chapter 13

Chapter summary

Chapter 13 of the CBSE Class 12 Chemistry syllabus delves into Organic Compounds Containing Nitrogen. This section offers NCERT Solutions that clarify key concepts like the comparative covalent nature of halides, the reducing properties of group 15 hydrides, and the stability of the nitrogen molecule. It also covers industrial processes like ammonia synthesis and chemical reactions involving ammonia and nitrogen oxides, providing step-by-step explanations for students.

Learning outcomes

Understand the factors affecting the covalent character of halides.
Explain the trend in reducing strength among group 15 hydrides.
Identify the conditions required for the synthesis of ammonia.
Describe the reaction of ammonia with metal ions.
Determine the covalence of nitrogen in its oxides.
Compare bond angles in related phosphorus compounds.

Topics covered

Paper topics

P-Block Elements
Group 15 Elements
Nitrogen and its Compounds
Ammonia (NH₃)
Haber's Process
Nitrogen Oxides
Covalence
Polarizing Power
Reducing Agents
Bond Angles
Hydrides of Group 15
Halides

Important topics

Reactivity and bonding in N₂
Haber's process for ammonia synthesis
Properties of Group 15 hydrides (reducing nature)
Covalence of nitrogen in oxides
Factors affecting covalent character (polarizing power)

PDF preview

Read page by page below. PDF is streamed from the official NCERT website — no download button on this page.

Loading document …

Page of

Loading page …

Questions and Solutions

Question 7.1

Why are pentahalides more covalent than trihalides?

Solution: Pentahalides involve metal ions with a higher oxidation state (e.g., +5) compared to trihalides (e.g., +3). According to Fajan's rules, a higher positive charge on the cation increases its polarizing power. A greater polarizing power leads to a stronger distortion of the electron cloud of the anion, resulting in a more covalent character in the bond. Therefore, pentahalides exhibit greater covalent nature 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 strength of the bond between the central atom and hydrogen. Consequently, the stability of the hydrides decreases from NH₃ to BiH₃. BiH₃, being the least stable, readily loses hydrogen atoms, making it 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 very strong triple covalent bond (N≡N). This triple bond has a very high bond dissociation energy. A significant amount of energy is required to break this bond. Due to this high bond strength, N₂ is relatively unreactive at room temperature.

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). To maximize the yield of ammonia, the following conditions are employed:

High Pressure: A pressure of approximately 200 atm is used. According to Le Chatelier's principle, increasing pressure shifts the equilibrium towards the side with fewer moles of gas, which is the product side (ammonia).
$N_2(g) + 3H_2(g) \leftrightarrow 2NH_3(g)$
Moderate Temperature: A temperature of around 700 K is maintained. While lower temperatures favor equilibrium yield (exothermic reaction), they result in slow reaction rates. Thus, a compromise temperature is used to achieve a reasonable rate and yield.
$\Delta H = -93 \text{ kJ/mol}$
Catalyst: An iron catalyst, promoted with small amounts of K₂O and Al₂O₃, is used. 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 complex ion. This reaction results in the formation of a deep blue precipitate or solution of the tetraamminecopper(II) complex.

The reaction is represented as:

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

The initial solution might be light blue, but the formation of the complex [Cu(NH₃)₄]²⁺ gives it a characteristic deep blue color.

Question 7.6

What is the covalence of nitrogen in N₂O₅?

Solution: The structure of dinitrogen pentoxide (N₂O₅) reveals how nitrogen atoms form bonds. In N₂O₅, each nitrogen atom is bonded to three oxygen atoms. Typically, one nitrogen atom forms a double bond with an oxygen atom, and single bonds with two other oxygen atoms. One of these single bonds is a coordinate covalent bond where nitrogen donates its lone pair to oxygen. Counting all the shared electron pairs around each nitrogen atom, we find that each nitrogen atom forms a total of four covalent bonds (one double bond counts as two, and two single bonds count as one each). Therefore, the covalence of nitrogen in N₂O₅ is 4.

Question 7.7

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

Solution: In the phosphine molecule (PH₃), the phosphorus atom is sp³ hybridized. Three of the sp³ hybrid orbitals are used to form sigma bonds with three hydrogen atoms, and the fourth sp³ hybrid orbital contains a lone pair of electrons. The presence of this lone pair leads to stronger repulsion (lone pair-bond pair repulsion) compared to the repulsion between bond pairs (bond pair-bond pair repulsion). This increased 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 angle, resulting in a pyramidal shape.

In the phosphonium ion (PH₄⁺), the phosphorus atom is also sp³ hybridized, but all four hybrid orbitals are involved in bonding with four hydrogen atoms. There is no lone pair of electrons on the phosphorus atom in PH₄⁺. Consequently, only bond pair-bond pair repulsions are present. These repulsions are weaker than lone pair-bond pair repulsions. As a result, the bond angles in PH₄⁺ are closer to the ideal tetrahedral angle, making the bond angle in PH₄⁺ significantly higher than that in PH₃.

Common mistakes

Confusing polarizing power with ionic character.
Misinterpreting the relationship between stability and reducing strength.
Incorrectly recalling conditions for Haber's process.
Overlooking the role of lone pairs in determining bond angles.

Revision tips

Focus on the trend of properties down Group 15 for hydrides and halides.
Memorize the conditions for Haber's process and the reaction of ammonia with Cu²⁺.
Understand the concept of covalence and its relation to bonding in N₂O₅.
Review VSEPR theory to explain bond angle differences in PH₃ and PH₄⁺.

Practice MCQs

Q1. Why are pentahalides generally more covalent than trihalides?

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

Q3. The low reactivity of N₂ at room temperature is primarily due to:

Q4. Which of the following is NOT a condition to maximize ammonia yield in Haber's process?

Q5. When ammonia reacts with Cu²⁺ solution, a deep blue complex is formed. What is the formula of this complex?

Q6. What is the covalence of nitrogen in N₂O₅?

Frequently asked questions

What is the main focus of CBSE Class 12 Chemistry Chapter 13 NCERT Solutions?

These solutions focus on Organic Compounds Containing Nitrogen, covering topics like p-block elements, ammonia, and nitrogen oxides, with detailed explanations of chemical properties and reactions.

Why is N₂ less reactive at room temperature according to the solutions?

The solutions explain that N₂ is less reactive due to the very strong triple covalent bond between the two nitrogen atoms, which has a high bond dissociation energy.

What are the key conditions mentioned for maximizing ammonia yield?

The solutions state that high pressure (around 200 atm), a moderate temperature (around 700 K), and the use of an iron catalyst are crucial for maximizing ammonia yield in the Haber process.

How do the solutions explain the difference in bond angles between PH₄⁺ and PH₃?

The solutions clarify that PH₃ has a lone pair on phosphorus, causing lone pair-bond pair repulsion that reduces the bond angle. PH₄⁺ lacks this lone pair, resulting in a higher bond angle due to only bond pair-bond pair repulsion.

Are these solutions helpful for exam preparation?

Yes, these NCERT Solutions provide clear, step-by-step explanations for complex concepts, helping students understand the underlying chemistry and prepare effectively for board exams.

Content reviewed by the NCERT Help team. Editorial Team and update policy

NCERT Solutions PDF PDF on NCERT Help. URL unchanged for search indexing.