CBSE Class 12 Chemistry NCERT Solutions: p-Block Elements

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

NCERT Solutions PDF Class 12 PDF

CBSE Class 12 Chemistry Chapter 7: p-Block Elements NCERT Solutions offer comprehensive explanations for this crucial topic. Students will find detailed solutions covering reactions of chlorides with sulfuric acid, qualitative analysis of copper ions, and the structure of cyclotrimetaphosphoric acid. The chapter also explores the concept of p-pi-d-pi bonding and the identification of isoelectronic and isostructural species. These solutions are crafted to provide clear, step-by-step guidance for all exercises, ensuring a thorough understanding of the p-block elements. This resource is an excellent tool for students aiming to strengthen their grasp of the chapter's concepts and excel in their examinations.

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Board	CBSE
Class	Class 12
Subject	Chemistry Exemplar
Session	2026
Language	English
Type	NCERT Solutions
Chapter	Chapter 7

Chapter summary

Chapter 7 of the Class 12 Chemistry syllabus, 'p-Block Elements', is explored through these NCERT Solutions. The solutions address multiple-choice questions related to the reactivity of halides with sulfuric acid, the identification of metal ions through qualitative analysis involving hydrogen sulfide and nitric acid, the bonding and structure of inorganic compounds like cyclotrimetaphosphoric acid, and the principles of p-pi-d-pi bonding. It also covers the identification of isoelectronic and isostructural species, crucial for understanding chemical periodicity and bonding.

Learning outcomes

Understand the reactions of halide salts with concentrated sulfuric acid.
Analyze the steps involved in qualitative analysis of metal ions using H2S and HNO3.
Determine the number of single and double bonds in inorganic molecules like cyclotrimetaphosphoric acid.
Identify elements capable of forming p-pi-d-pi bonds.
Compare and identify isoelectronic and isostructural species.
Explain the formation of complex ions in solution.

Topics covered

Paper topics

Reactions of concentrated H₂SO₄ with halides
Qualitative analysis of copper ions
Formation of metal sulfides
Oxidation by nitric acid
Ammonia complex formation
Structure of cyclotrimetaphosphoric acid
Sigma and pi bonds
pπ - dπ bonding
Isoelectronic species
Isostructural species
Hybridization
p-Block elements

Important topics

Qualitative analysis procedures and observations
pπ - dπ bonding capabilities
Identifying isoelectronic and isostructural species
Structure and bonding in oxyacids of phosphorus
Redox reactions involving sulfuric acid and halides

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Questions and Solutions

Multiple Choice Questions (MCQs) - Question 1

Q. 1 On addition of conc.

\mathrm{H_2SO_4}

to a chloride salt, colourless fumes are evolved but in case of iodide salt, violet fumes come out. This is because

$\mathrm{H_2SO_4}$ reduces HI to $\mathrm{I_2}$
HI is of violet colour
HI gets oxidised to $\mathrm{I_2}$
HI changes to $\mathrm{HIO_3}$

Solution: The correct option is (c). Concentrated sulfuric acid is a strong oxidizing agent. While it reacts with chloride salts to produce HCl gas (colorless fumes), its reaction with iodide salts is different. Hydrogen iodide (HI) is a stronger reducing agent than sulfuric acid. Therefore, HI reduces sulfuric acid to sulfur dioxide (

\mathrm{SO_2}

) and is itself oxidized to iodine (

\mathrm{I_2}

). Iodine exists as violet fumes. The reaction is represented as:

$\mathrm{H_2SO_4} + 2\mathrm{HI} \longrightarrow \mathrm{SO_2} + \mathrm{I_2} + 2\mathrm{H_2O}$

(Violet fumes)

Multiple Choice Questions (MCQs) - Question 2

Q. 2 In qualitative analysis when H₂S is passed through an aqueous solution of salt acidified with dil. HCl, a black precipitate is obtained. On boiling the precipitate with dil. HNO₃, it forms a solution of blue colour. Addition of excess of aqueous solution of ammonia to this solution gives ........

deep blue precipitate of Cu (OH)₂
deep blue solution of $[Cu(NH_3)_4]^{2+}$
deep blue solution of $Cu(NO_3)_2$
deep blue solution of $Cu (OH)_2 \cdot Cu (NO_3)_2$

Solution: The correct option is (b). In qualitative analysis, when hydrogen sulfide (H₂S) is passed through an aqueous solution of a salt acidified with dilute hydrochloric acid (HCl), a black precipitate of copper(II) sulfide (CuS) is formed if copper ions are present. The reaction is:

$CuSO_4 + H_2S \xrightarrow{\text{dil. HCl}} CuS \downarrow + H_2SO_4$

(Black precipitate) On boiling this black precipitate (CuS) with dilute nitric acid (HNO₃), it dissolves to form a blue-colored solution of copper(II) nitrate (Cu(NO₃)₂). The reactions involved are:

$3CuS + 8HNO_3 \longrightarrow 3Cu(NO_3)_2 + 2NO + 3S + 4H_2O$

$S + 2HNO_3 \longrightarrow H_2SO_4 + NO$

When excess aqueous ammonia solution is added to this blue solution containing copper(II) ions, a deep blue solution is formed due to the formation of the tetraamminecopper(II) complex ion,

[Cu(NH_3)_4]^{2+}

. The reaction is:

$Cu^{2+} + 4NH_3 \longrightarrow [Cu(NH_3)_4]^{2+}$

(Deep blue solution)

Multiple Choice Questions (MCQs) - Question 3

Q. 3 In a cyclotrimetaphosphoric acid molecule, how many single and double bonds are present?

3 double bonds; 9 single bonds
6 double bonds; 6 single bonds
3 double bonds; 12 single bonds
Zero double bond; 12 single bonds

Solution: The correct option is (c). A molecule of cyclotrimetaphosphoric acid has the formula

(HPO_3)_3

H_3P_3O_9

. It consists of a six-membered ring with alternating phosphorus and oxygen atoms. Each phosphorus atom is bonded to one oxygen atom via a double bond (P=O) and to two other phosphorus atoms and one hydroxyl group (-OH) via single bonds. Therefore, in one molecule, there are 3 P=O double bonds and 9 P-O single bonds (including those within the ring and the P-OH bonds). The structure can be visualized with three

\pi

bonds (double bonds) and twelve

\sigma

bonds (single bonds).

Multiple Choice Questions (MCQs) - Question 4

Q. 4 Which of the following elements can be involved in

p\pi - d\pi

bonding?

Carbon
Nitrogen
Phosphorus
Boron

Solution: The correct option is (c).

p\pi - d\pi

bonding occurs when there is an overlap between a filled p-orbital of one atom and a vacant d-orbital of another atom. Among the given elements, phosphorus (P) has vacant 3d orbitals in its valence shell. This allows phosphorus to form

p\pi - d\pi

bonds, for example, in compounds like

PO_4^{3-}

SO_2

. Carbon (C) and Nitrogen (N) do not have vacant d-orbitals in their valence shells, so they cannot participate in

p\pi - d\pi

bonding. Boron (B) can form

p\pi - p\pi

bonds but not typically

p\pi - d\pi

bonds in the same way as phosphorus.

Multiple Choice Questions (MCQs) - Question 5

Q. 5 Which of the following pairs of ions are isoelectronic and isostructural?

$CO_3^{2-}$ , $NO_3^{-}$
$CIO_3^{-}$ , $CO_3^{2-}$
$SO_3^{2-}$ , $NO_3^{-}$
$CIO_3^{-}$ , $SO_3^{2-}$

Solution: The correct option is (a). Isoelectronic species have the same total number of electrons, while isostructural species have the same arrangement of atoms and similar bonding. Let's analyze the options: For

CO_3^{2-}

: Total electrons = (Number of electrons in C) + 3 * (Number of electrons in O) + (Charge) = 6 + 3 * 8 + 2 = 6 + 24 + 2 = 32 electrons. For

NO_3^{-}

: Total electrons = (Number of electrons in N) + 3 * (Number of electrons in O) + (Charge) = 7 + 3 * 8 + 1 = 7 + 24 + 1 = 32 electrons. Since both

CO_3^{2-}

and

NO_3^{-}

have 32 electrons, they are isoelectronic. Structurally, both the carbonate ion (

CO_3^{2-}

) and the nitrate ion (

NO_3^{-}

) have a central atom (C or N) that is

sp^2

hybridized, resulting in a trigonal planar geometry. Both ions have resonance structures with one double bond and two single bonds (or delocalized pi system), making them isostructural. Let's check other options: (b)

CIO_3^{-}

has 26 + 3*8 + 1 = 51 electrons.

CO_3^{2-}

has 32 electrons. Not isoelectronic. (c)

SO_3^{2-}

has 16 + 3*8 + 2 = 42 electrons.

NO_3^{-}

has 32 electrons. Not isoelectronic. (d)

CIO_3^{-}

has 51 electrons.

SO_3^{2-}

has 42 electrons. Not isoelectronic. Therefore,

CO_3^{2-}

and

NO_3^{-}

are the only pair that is both isoelectronic and isostructural.

Common mistakes

Confusing oxidizing and reducing properties of acids and their reaction products.
Incorrectly identifying the products of precipitation and complex formation in qualitative analysis.
Miscounting sigma and pi bonds in complex inorganic structures.
Difficulty in recognizing the conditions required for p-pi-d-pi bonding.
Errors in calculating the total number of electrons for isoelectronic species.

Revision tips

Review the redox reactions between concentrated sulfuric acid and halides.
Memorize the characteristic reactions and observations in qualitative analysis for common ions.
Practice drawing Lewis structures to determine bond types and hybridization.
Focus on understanding the electronic configurations that enable p-pi-d-pi bonding.
Work through examples of isoelectronic and isostructural species to solidify the concept.

Practice MCQs

Q1. On adding concentrated H₂SO₄ to a chloride salt, colorless fumes are evolved, but on adding it to an iodide salt, violet fumes are observed. What is the reason for this difference?

Q2. In qualitative analysis, a black precipitate is formed when H₂S is passed through an aqueous solution of a salt acidified with dil. HCl. This precipitate dissolves in dil. HNO₃ to form a blue solution. What is the final observation upon adding excess aqueous ammonia to this blue solution?

Q3. How many single and double bonds are present in one molecule of cyclotrimetaphosphoric acid (H₃P₃O₉)?

Q4. Which of the following elements is capable of forming pπ - dπ bonding?

Q5. Which pair of ions are both isoelectronic and isostructural?

Frequently asked questions

What is the significance of p-pi-d-pi bonding in p-block elements?

p-pi-d-pi bonding occurs when an element with a p-orbital containing lone pairs overlaps with an empty d-orbital of another atom. This type of bonding is important for elements like phosphorus, enabling them to form stable compounds with specific structures and reactivities.

How can we determine if two ions are isoelectronic and isostructural?

Isoelectronic species have the same total number of electrons. Isostructural species have the same arrangement of atoms and the same number of bonds and lone pairs. Both conditions must be met for ions to be considered isoelectronic and isostructural.

What is the role of concentrated H₂SO₄ in the reaction with iodide salts?

Concentrated sulfuric acid acts as an oxidizing agent. When reacting with iodide salts, it oxidizes the iodide ion (I⁻) to iodine (I₂), which is observed as violet fumes. Sulfuric acid itself is reduced, typically to SO₂.

Explain the formation of the deep blue color in the qualitative analysis of copper ions.

The deep blue color observed is due to the formation of the tetraamminecopper(II) complex ion, [Cu(NH₃)₄]²⁺. This complex forms when copper(II) ions react with excess aqueous ammonia.

How are the bonds in cyclotrimetaphosphoric acid determined?

The structure of cyclotrimetaphosphoric acid (H₃P₃O₉) is determined by drawing its Lewis structure, which reveals three P=O double bonds and nine P-O single bonds, along with the P-OH bonds.

Are these NCERT solutions suitable for exam revision?

Yes, these solutions provide clear, step-by-step explanations for complex problems, helping students understand the underlying concepts and methods, which is ideal for revising the p-Block Elements chapter for exams.

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