NCERT Solutions for Class 12 Chemistry P Block Elements: Questions with Answers PDF, Topics

Candidates can check and download here NCERT Chemistry class 12 solution for P Block Elements PDF. The chapter's solution is prepared by the subject experts and are easy to understand. P Block Elements is an important chapter and many questions are asked from this chapter in class 12 board exams as well as JEE Mains. 

Download Here: NCERT Solution for Class XII Chemistry P Block Elements PDF

 

Intext Q 7.1 Why are pentahalides more covalent than trihalides?

A 7.1 Pentahalides (like MX5) means the compounds in which metal is bonded with five halogen atoms. Thus, the oxidation state of metal here is +5. Similarly, trihalides (like MX3) means the compounds in which metal is bonded with three halogen atoms. Thus, the oxidation state of metal here is +3. Now, as the polarizing power is directly proportional to the charge, the metals with the higher charge will have higher polarizing power. Hence, Pentahalides are more covalent than trihalides. Note: Polarizing power is the ability of a cation to distort an anion.

Intext Q 7.2 Why is BiH3 the strongest reducing agent amongst all the hydrides of Group 15 elements?

A 7.2 Atomic size increases down the group and thus the stability of the hydrides of Group-15 elements decreases. Now, the least stable hydride is the one which is more reactive and possessing higher reducing strength. The reducing character of hydrides increases on moving from NH3 to BiH3. Hence, BiH3 is the strongest reducing agent among hydrides of Group-15 elements.

Intext Q 7.3 Why is N2 less reactive at room temperature?

A 7.3 N2 (Dinitrogen) is formed by sharing three electron pairs between two nitrogen atoms, which are joined by a triple bond as shown below: Due to a complete octet of the nitrogen atoms and bonded by strong triple bonds, due to which its bond dissociation energy is very high. Due to this reason, N2 is very less reactive at room temperature.

Intext Q 7.4 Mention the conditions required to maximize the yield of ammonia.

A 7.4 Ammonia (NH3) is generally prepared through a many numbers of processes out of which Haber’s process is the most important one. Following the Le Chatlier’s principle and taking into considerations the reaction conditions of the Haber’s process, the yield of ammonia can be maximized by High pressure (~ 200 atm) High temperature (~ 700K) A mixture of Iron oxide with small amounts of K2O and Al2O3 used as a catalyst. (This mixture acts as a positive catalyst in the Haber’s process)

The p-block elements chapter is deleted from the CBSE 2025 syllabus. However, this is one of the most important topics for NEET and JEE exams. You can expect 3 - 4 questions from this chapter. These questions may be directly concept-based or related to chemical reactions. Check important topics below;

Topics Covered in Chemistry P-block Elements Chapter

• General chemical properties of elements such as atomic radius, electronegativity, oxidation states, covalency, and others.
• Chemical properties of specific elements such as Carbon, Nitrogen, Oxygen, inert gases, and others.
• Important compounds of all groups such as SO₂, NH₃, CO, CO₂, H₂SO₄, Boric acid and others
• Other important chemical and physical behaviour of special elements.

Important Chemical Reactions

Boric Acid Formation

$$N{a}_2{B}_4{O}_7\cdot 10H_{2}O+2HCl\to 4H_{3}B{O}_3+2NaCl+5H_{2}O$$

Phosgene Formation:

$$CO+C{l}_2\to COC{l}_2$$

Silicon Tetrachloride Hydrolysis:

$$SiCl₄ + 2H₂O → SiO₂ + 4HCl$$

Hydrolysis of PCl₃:

 PCl₃ + 3H₂O → H₃PO₃ + 3HCl

Bleaching Powder Preparation

$$Ca(OH)₂ + Cl₂ → Ca(OCl)₂ + H₂O$$

Hydrolysis of XeF₆

$$XeF₆ + 3H₂O → XeO₃ + 6HF$$