Chapter 9- Atomic Foundations of Matter

NCERT Solutions for Class 9 Science Chapter 9: Atomic Foundations of Matter

Class 9 Science Chapter 9: Atomic Foundations of Matter builds directly on Chapter 8 (Journey Inside the Atom) and takes students into how atoms combine to form the matter around us. Curated by ALLEN Experts and aligned with the new NCERT Exploration textbook (2026–27), these chapter-wise NCERT solutions cover the Law of Conservation of Mass, the Law of Constant Proportions, Dalton's Atomic Theory, ionic and covalent bonding, writing chemical formulae using valency, and calculating molecular mass and formula unit mass — with step-by-step answers to every in-text and end-of-chapter question, plus important questions for focused practice.

1.0Download NCERT Solutions for Class 9 Science Chapter 9 – Atomic Foundations of Matter PDF

Get instant access to the free PDF of NCERT Solutions for this chapter and revise offline at your own pace. Each solution is broken down into clear, exam-ready steps so you can quickly locate the exact concept or numerical you're stuck on, cross-check your working, and revise just before a test without flipping through the entire textbook.

NCERT Solutions Class 9 Science Chapter 9

2.0NCERT Class 9 Science Chapter 9 Solutions: Learning Outcomes

Understand the Law of Conservation of Mass: Verify through activities that mass is neither created nor destroyed during a chemical reaction.
Understand the Law of Constant Proportions: Explain that a pure chemical compound always contains the same elements combined in a fixed proportion by mass.
Learn Dalton's Atomic Theory: State the postulates of Dalton's Atomic Theory and explain how it accounts for the two laws above.
Differentiate Atoms and Molecules: Understand how atoms combine to form molecules of elements and compounds.
Understand Ionic Bonding: Explain how ionic bonds form through the transfer of electrons, using examples like sodium chloride.
Understand Covalent Bonding: Explain how covalent bonds form through the sharing of electrons, using examples like hydrogen, oxygen, and water.
Compare Ionic and Covalent Compounds: Differentiate the properties of ionic and covalent compounds, including solubility and electrical conductivity.
Write Chemical Formulae Using Valency: Apply the valency of elements and radicals to correctly write chemical formulae of compounds.
Calculate Molecular Mass: Compute the molecular mass of covalent compounds by summing the atomic masses of constituent atoms.
Calculate Formula Unit Mass: Compute the formula unit mass of ionic compounds and distinguish it from molecular mass.
Solve NCERT Exercise Problems: Confidently attempt and solve all in-text activities and end-of-chapter numerical and conceptual questions.
Build a Foundation for Advanced Topics: Develop conceptual clarity that supports higher-level chemistry, including mole concept and chemical bonding in later classes.

3.0Detailed NCERT Class 9 Science Chapter 9 Atomic Foundations of Matter Solutions

1. Water can be obtained from various sources. Are all these samples of water chemically identical?

Ans. No. Water from different sources contains different dissolved impurities such as salts, minerals, gases, and microorganisms. However, the chemical formula of pure water is always $H_{2} O$ . Thus, pure water is chemically identical, but natural water samples are not.

2. Oxygen is sometimes represented as O and sometimes as $O_{2}$ . What is the difference between these symbols?

Ans. 0 represents one oxygen atom. $O_{2}$ represents one oxygen molecule containing two oxygen atoms chemically bonded together. Oxygen normally exists in nature as $O_{2}$ gas.

3. Why does dissolved salt in water conduct electricity, but sugar does not?

Ans. Salt ( NaCl ) dissolves in water and breaks into charged ions ( $Na^{+}$ and $Cl^{-}$ ). These ions carry electric current through the solution.

Sugar dissolves in water as neutral molecules and does not form ions. Since there are no charged particles to carry current, sugar solution does not conduct electricity. Example: $NaCl \to Na^{+} + Cl^{-}$ Therefore, salt solution conducts electricity, whereas sugar solution does not.

4. A student burns 10 g of ethanol in an open beaker. After the reaction, no residue is left in the beaker. Does this mean the Law of Conservation of Mass is violated? Explain.

Ans. Burning of ethanol in an open beaker When ethanol burns, it reacts with oxygen from the air to form carbon dioxide and water vapor.

Ethanol + Oxygen → Carbon dioxide + Water vapor"

Although no residue is left in the beaker, the products escape into the air as gases. Therefore, the total mass is still conserved.

Conclusion: No, the Law of Conservation of Mass is not violated. The gaseous products have simply escaped into the surroundings.

5. When 20 g of hydrogen reacts completely with 160 g of oxygen, how much water is formed according to the Law of Conservation of Mass?

Ans. Formation of water Given: Mass of hydrogen $= 20 g$ Mass of oxygen $= 160 g$ According to the Law of Conservation of Mass:

Mass of reactants $=$ Mass of products $20 + 160 = 180 g$ Mass of water formed $= 180 g$

6. A compound consists of $40%$ sulfur and 60% oxygen by mass. In a sample of the same compound containing 20 g of sulfur, what mass of oxygen must be present to satisfy the Law of Constant Proportions?

Ans. Mass of oxygen in the compound Given: Compound contains 40% sulfur and 60% oxygen.

Sulfur present $= 20 g$ Ratio of sulfur : oxygen $40 : 60 = 2 : 3$ If sulfur $= 20 g$ , Oxygen $= \frac{3}{2} \times 20 = 30 g$ Mass of oxygen required $= 30 g$

7. Carbon monoxide ( CO ) contains carbon and oxygen in the mass ratio of 3:4. How much oxygen will combine with 9 g of carbon to form carbon monoxide?

Ans. Formation of carbon monoxide Given: Ratio of Carbon : oxygen $= 3 : 4$ Carbon $= 9 g$ Using proportion: $\frac{3}{4} = \frac{9}{x}$ $x = \frac{9 \times 4}{3}$ $x = 12 g$ Mass of oxygen required $= 12 g$

8. The Law of Definite Proportions holds true for compounds but not for mixtures. Give reason.

Ans. Why the Law of Definite Proportions applies to compounds but not mixtures. In a compound, elements combine in a fixed ratio by mass. For example, water always contains hydrogen and oxygen in the ratio $1 : 8$ by mass. In a mixture, substances can be mixed in any proportion. For example, salt and sand can be mixed in different amounts. Therefore, the Law of Definite Proportions applies only to compounds and not to mixtures.

9. Students X and Y , both prepared an oxide of copper by combining copper and oxygen in the ratios of $4 : 1$ and $8 : 2$ , respectively. Do their results justify the Law of Constant Proportions? Explain. Ans. Copper oxide experiment Given ratios: Student $X \to 4 : 1$ Student Y → 8:2 Simplify Student Y's ratio: $8 : 2 = 4 : 1$ Both students obtained the same ratio of copper to oxygen. Conclusion: Yes, their results justify the Law of Constant Proportions because copper and oxygen combine in the same fixed ratio by mass in copper oxide.

10. Assertion (A): 2 g of hydrogen combines with 16 g of oxygen to form 18 g of water. Reason (R): According to Dalton's Atomic Theory, atoms combine in a simple whole number ratio by mass to form compounds.

Choose the correct option: (i) Both A and R are true, and R is the correct explanation of A. (ii) Both A and R are true, but R is not the correct explanation of A. (iii) A is true, but R is false. (iv) A is false, but R is true.

Ans. 2 g of hydrogen combines with 16 g of oxygen to form 18 g of water. This follows the Law of Conservation of Mass. $2 + 16 = 18 g$ According to Dalton's Atomic Theory, atoms combine in a simple whole number ratio by mass to form compounds. However, the reason explains the formation of compounds in fixed ratios, not specifically why the total mass of water formed is 18 g . Therefore, the reason is not the correct explanation of the assertion.

Option (ii) is correct.

11. Nitrogen has five valence electrons. Draw the structure of the nitrogen molecule $(N_{2})$ . Ans. Structure of Nitrogen Molecule ( $N_{2}$ ) Nitrogen has atomic number 7. Electronic configuration: 2,5 So, each nitrogen atom has 5 valence electrons.

To complete its octet, each nitrogen atom needs 3 more electrons.

Therefore, two nitrogen atoms share three pairs of electrons and form a triple covalent bond.

Structure: : $N \equiv N$ : Each nitrogen atom has one lone pair of electrons.

12. The atomic number of fluorine is 9 . Explain the formation of the fluorine molecule (F2). Ans. Formation of Fluorine Molecule ( $F_{2}$ ) Fluorine has atomic number 9. Electronic configuration: 2, 7 Each fluorine atom has 7 valence electrons and needs 1 electron to complete its octet.

Two fluorine atoms share one pair of electrons to form a single covalent bond.

Structure: : $\ddot{F} - F: \ddot{F}$ :

Flourine atom (F) Flourine atom (F)

Flourine molecule ( $F_{2}$ ) Formation of a Flourine molecule Each fluorine atom attains 8 electrons in its outermost shell.

13. Show the formation of the following molecules: (i) Carbon dioxide $(CO_{2})$ (ii) Hydrogen sulfide $(H_{2} S)$ (iii) Ammonia $(NH_{3})$

Ans. Formation of Molecules (i) Carbon Dioxide $(CO_{2})$

Carbon has 4 valence electrons and oxygen has 6 valence electrons.

Carbon shares two pairs of electrons with each oxygen atom.

Structure: $0 = C = 0$

Two double covalent bonds are formed. (ii) Hydrogen Sulfide $(H_{2} S)$

Sulfur has 6 valence electrons and needs 2 more electrons.

Each hydrogen shares one electron with sulfur. Structure: H- $.. S$ - H Two single covalent bonds are formed. (iii) Ammonia $(NH_{3})$

Nitrogen has 5 valence electrons and needs 3 more electrons.

It shares one electron each with three hydrogen atoms. Structure:

Nitrogen also has one lone pair of electrons.

14. Neon (atomic number 10) neither transfers nor shares its valence electrons. Explain.

Ans. Why Neon neither transfers nor shares electrons

Neon has atomic number 10. Electronic configuration: 2,8 Its outermost shell already contains 8 electrons, which is a stable octet configuration. Therefore, neon: does not need to gain electrons, does not need to lose electrons, does not need to share electrons.

Hence, neon is chemically inert (unreactive).

15. What kind of ion will oxygen ( O ) form?

Ans. Kind of ion formed by oxygen Oxygen has atomic number 8. Electronic configuration: 2,6 It needs 2 more electrons to complete its octet.

Therefore, oxygen gains 2 electrons and forms a negative ion (anion) called the oxide ion. $O + 2 e^{-} \to Oxide (Anion) O^{2 -}$ Ion formed: $O^{2 -}$ So, oxygen forms a divalent negative ion.

16. Fill in the blanks.

Among magnesium and chlorine, magnesium atom can give two electrons to become $Mg^{2 +}$ . However, chlorine can take only one electron to become $____$ . Now, $____$ ion of magnesium and $____$ ions of chlorine combine to give magnesium chloride.

Ans. Among magnesium and chlorine, magnesium atom can give two electrons to become $Mg^{2 +}$ . However, chlorine can take only one electron to become $Cl^{-}$ (chloride ion). Now, one ion of magnesium and two ions of chlorine combine to give magnesium chloride.

17. Show the formation of cations of potassium (K) and calcium (Ca) atoms, and the formation of their corresponding chlorides using diagrams.

Ans. Formation of cations and their chlorides (i) Formation of Potassium Chloride (KCl) Potassium ( K ) has 1 valence electron. $K \to K^{+} + e^{-}$ Chlorine gains that electron: $Cl + e^{-} \to Cl^{-}$ Thus, $K^{+} + Cl^{-} \to KCl$

Diagram representation $K (2, 8, 8, 1) \to K^{+} (2, 8, 8) + e^{-}$ $Cl (2, 8, 7) + e^{-} \to Cl^{-} (2, 8, 8)$ (ii) Formation of Calcium Chloride ( $CaCl_{2}$ )

Calcium has 2 valence electrons. $Ca \to Ca^{2 +} + 2 e^{-}$ Each chlorine atom gains one electron: $2 Cl + 2 e^{-} \to 2 Cl^{-}$ Thus, $Ca^{2 +} + 2 Cl^{-} \to CaCl_{2}$

Diagram representation

$Ca (2, 8, 8, 2) \to Ca^{2 +} (2, 8, 8) + 2 e^{-} 2 Cl (2, 8, 7) + 2 e^{-} \to 2 Cl^{-} (2, 8, 8)$

Illustrate how sodium sulfide $(Na_{2} S)$ is formed.

Ans. Formation of Sodium Sulfide ( $Na_{2} S$ ) Sodium has 1 valence electron and sulfur has 6 valence electrons.

Two sodium atoms each donate one electron to sulfur.

Reactions $2 Na \to 2 Na^{+} + 2 e^{-}$ $S + 2 e^{-} \to S^{2 -}$ Thus, $2 Na^{+} + S^{2 -} \to Na_{2} S$

2 Na (2, 8, 1) \to Na^{+} (2, 8) + 2 e^{-}

S (2, 8, 6) 2 e^{-} S^{2 -} (2, 8, 8)

19. Name the following: (i) $CO_{2}$ $____$ (ii) $NO_{2}$ $____$ (iii) $SF_{6}$ $____$ (iv) $PCl_{3}$ $____$

Ans. Name the following (i) $CO_{2} \to$ Carbon dioxide (ii) $NO_{2} \to$ Nitrogen dioxide (iii) $SF_{6} \to$ Sulfur hexafluoride (iv) $PCl_{3} \to$ Phosphorus trichloride

20. Write the formula for the following: (i) Sodium hydrogen carbonate $____$ (ii) Sulfur dioxide $____$ (iii) Ferric chloride $____$ (iv) Cuprous oxide $____$

Ans. Write the formula (i) Sodium hydrogencarbonate → $NaHCO_{3}$ (ii) Sulfur dioxide → $SO_{2}$ (iii) Ferric chloride $\to FeCl_{3}$ (iv) Cuprous oxide → $Cu_{2} O$

21. Write the formulae for the compounds formed from the following pairs of ions: (i) $Fe^{3 +}$ and $OH^{-}$ (ii) $K^{+}$ and $CO_{3}^{2 -}$ .

Ans. Write the formulae for the compounds (i) $Fe^{3 +}$ and $OH^{-}$

Iron ion has charge +3 and hydroxide ion has charge -1 .

Three hydroxide ions are needed to balance one iron ion. $Fe^{3 +} + 3 OH^{-} \to Fe (OH)_{3}$ Formula: $Fe (OH)_{3}$

(ii)

K^{+}

and

CO_{3}^{2 -}

Carbonate ion has charge -2 and potassium ion has charge +1 .

Two potassium ions are needed to balance one carbonate ion. $2 K^{+} + CO_{3}^{2 -} \to K_{2} CO_{3}$ Formula: $(K_{2} CO_{3})$

22. What type of chemical bond is present in a solid compound that does not conduct electricity in the solid state but conducts electricity when dissolved in water? Ans. Type of chemical bond A compound that: does not conduct electricity in solid state, but conducts electricity in aqueous solution, contains an ionic bond.

This is because ions are fixed in solid state but become free to move in water.

Ionic bond.

23. Metal M, with two electrons in its valence shell (M shell), reacts with oxygen to form a compound that is slightly soluble in water.

Predict its: (i) Formula (ii) Type of bond (iii) Electrical conductivity of its aqueous solution.

Ans. Metal M reacting with oxygen The metal has 2 valence electrons, so it forms $M^{2 +}$ . Oxygen forms $O^{2 -}$ . (i) Formula

(ii) Type of bond

Since metal transfers electrons to oxygen, the bond is: Ionic bond. $M \to M^{2 +} + 2 e^{-}$ $O + 2 e^{-} \to O^{2 -}$ (iii) Electrical conductivity of aqueous solution

The compound forms ions in water, so its aqueous solution conducts electricity.

24. Find the molecular mass of nitric acid $(HNO_{3})$ . Atomic mass - H = 1 u ; N = 14 u ; $0 = 16 u$ . Ans. Molecular mass of nitric acid ( $HNO_{3}$ ) $HNO_{3}$ Atomic masses: $H = 1 u$ $N = 14 u$ $0 = 16 u$ Calculation: $1 + 14 + (3 \times 16)$ $= 1 + 14 + 48 = 63 u$ Molecular mass: ( 63 u )

25. Find the molecular mass of methane $(CH_{4})$ .

Atomic mass- $C = 12 u; H = 1 u$ . Ans. Molecular mass of methane ( $CH_{4}$ ) $CH_{4}$ Atomic masses: $C = 12 u$ $H = 1 u$ Calculation: $12 + (4 \times 1) = 16 u$ Molecular mass: ( 16 u )

26. Find the formula unit mass of potassium chloride (KCl). Atomic mass $- K = 39 u; Cl = 35.5 u$ . Ans. Formula unit mass of potassium chloride (KCl) Calculation: $39 + 35.5 = 74.5 u$ Formula unit mass: ( 74.5 u )

27. Find the formula unit mass of magnesium hydroxide, $Mg (OH)_{2}$ . Atomic mass-Mg = $24 u; O = 16 u; H = 1$ u.

Ans. Formula unit mass of magnesium hydroxide $Mg (OH)_{2}$

Atomic masses: $Mg = 24 u$ $0 = 16 u$ $H = 1 u$ Mass of one OH group: $16 + 1 = 17 u$ Since there are two OH groups: $24 + (2 \times 17)$ $= 24 + 34 = 58 u$ Formula unit mass: ( 58 u ).

28. A particular element (A) has one electron in its third shell. There is another element (B) with six electrons in its second shell. (i) How many electrons does A tend to give or take to become stable? (ii) What kind of ion would it form? (iii) How many electrons does B tend to give or take to become stable? (iv) What kind of ion would it form? (v) A and B were to combine, what kind of bond would be formed? (vi) What would be the formula for the compound thus formed?

Ans. Based on the properties described, Element A (typically Sodium, Na) will lose 1 electron to become a +1 cation, while Element B (typically Oxygen, O) will gain 2 electrons to become a -2 anion. They form an ionic bond resulting in a compound with a $2 : 1$ ratio. (i) Electrons A tends to give or take: A tends to lose (give) 1 electron from its third shell to achieve a stable octet in the second shell. (ii) Kind of ion A forms: A forms a positive ion (cation), specifically $A^{+}$ . $2, 8, 1 Na \to 2, 8 Na^{+} + e^{-}$ (iii) Electrons B tends to give or take: B has 6 electrons in its second shell, so it tends to gain (take) 2 electrons to complete its octet $(2, 8)$ . $2, 6 O + 2 e^{-} \to 2, 8 Octate (stable) O^{2 -}$ (iv) Kind of ion B forms: B forms a negative ion (anion), specifically $B^{2 -}$ (v) Kind of bond formed: An ionic bond is formed due to the electrostatic attraction between positive (A) and negative (B) ions. (vi) Formula for the compound: The formula is $A_{2} B$ (e.g., $Na_{2} O$ ).

29. An element X has six electrons in its outer shell and forms a diatomic molecule. (i) Why would that be so? (ii) What kind of bond would it form? (iii) Draw the structure of the molecule it would form. (iv) A certain other element Y has two electrons in its second shell. Draw the structure of the molecule that X would form with Y.

Ans. Element X (likely Oxygen, S, or similar) has 6 valence electrons, requiring 2 more to complete its octet for stability. It forms a diatomic molecule by sharing two pairs of electrons with another identical atom, creating a double covalent bond to satisfy the octet rule. (i) The element has 6 electrons in its valence shell, leaving it 2 electrons short of a stable octet (8 electrons). It forms a diatomic molecule ( $X_{2}$ ) to share 2 electrons with another atom, allowing both to achieve a stable electronic configuration. (ii) A double covalent bond (a covalent bond formed by the sharing of two pairs of electrons). (iii) Each atom shares two valence electrons, creating two shared pairs between them.

$: .. X = .. X :$

(iv) Molecule X would form with Y : Element Y has two electrons in its second shell (assumed valence shell based on common chemical bonding context).

Therefore, Y must lose 2 electrons, while X must gain 2 . They form an ionic bond where X takes 2 electrons from Y.

Formula: YX or XY Structure: $[Y]^{2 +} [: X :]^{2 -}$

30. You want to design a new ionic compound, where the total positive charge is $6 +$ and the total negative charge is 6-. Which of the following combinations gives the correct number of ions? (i) $2 Al^{3 +}$ and $3 Cl^{-}$ (ii) $3 Mg^{2 +}$ and $1 PO_{4}^{3 -}$ (iii) $Fe^{3 +}$ and $3 O^{2 -}$ (iv) $3 Ca^{2 +}$ and $2 SO_{4}^{2 -}$

Ans. The correction combination is (iii) $2 Fe^{3 +}$ and $3 0^{2 -}$ .

Here is the breakdown of why this is the correct choice, along with why the others don't fit the $6 +$ and $6 -$ criteria:

Total Positive Charge: Two iron ions, each with a $3 +$ charge, give a total of: $2 \times (+ 3) = + 6$ Total Negative Charge: Three oxide ions, each with a $2 -$ charge, give a total of: $3 \times (- 2) = - 6$ Why the other options are incorrect: (i) $2 Al^{3 +}$ and $3 Cl^{-}$ * Positive: $2 \times (+ 3)$ $= + 6$ Negative: $3 \times (- 1) = - 3$ (ii) $3 Mg^{2 +}$ and $1 PO_{4}^{3 -} *$ Positive: $3 \times (+ 2)$ $= + 6$ Negative: $1 \times (- 3) = - 3$ (iv) $3 Ca^{2 +}$ and $2 SO_{4}^{2 -}$ * Positive: $3 \times (+ 2)$ $= + 6$ Negative: $2 \times (- 2) = - 4$

31. Choose the correct statement(s) and correct the false statement(s). (i) Elements are made up of molecules and compounds are made up of atoms. (ii) The molecule of a compound is always made up of two or more atoms of the same kind. (iii)One molecule of nitrogen gas contains three nitrogen atoms. (iv)Water is made of two hydrogen atoms, covalently bonded with one oxygen atom.

Ans. (i) False Elements are made up of atoms (or molecules of the same element), while compounds are made up of molecules or ions containing two or more different types of atoms. (ii) False

A molecule of a compound is made up of two or more atoms of different kinds (elements). (iii) False

One molecule of nitrogen gas ( $N_{2}$ ) contains two nitrogen atoms, not three. (iv) True

Water ( $H_{2} O$ )) is a compound made of two hydrogen atoms covalently bonded to one oxygen atom.

32. Write the chemical formulae for the following compounds. (i) Aluminium nitrate (ii) Calcium oxide (iii) Ferric oxide

Ans. The chemical formulae for the requested compounds are: (i) Aluminium nitrate: $Al (NO_{3})_{3}$ (ii) Calcium oxide: CaO (iii) Ferric oxide (Iron(III) oxide): $Fe_{2} O_{3}$

33. Write the formulae of the compounds formed from the following pairs of ions. (i) $Ca^{2 +}$ and $Br^{-}$ (ii) $Al^{3 +}$ and $CO_{3}^{2 -}$ (iii) $K^{+}$ and $SO_{4}^{2 -}$ (iv) $NH_{4}^{+}$ and $Cl^{-}$

To write the formulae for these compounds, we use the "criss-cross" method. The goal is to balance the positive and negative charges so the overall compound is neutral. Ans. To write the formulae for these compounds, we use the "criss-cross" method. The goal is to balance the positive and negative charges so the overall compound is neutral. Here are the formulae for the pairs you listed: (i) $Ca^{2 +}$ and $Br^{-} \to CaBr_{2}$

Calcium has a +2 charge and Bromine has a -1 charge. You need two Bromine ions to balance one Calcium ion. (ii) $Al^{3 +}$ and $CO_{3}^{2 -} \to Al_{2} (CO_{3})_{3}$

Aluminum has a +3 charge and Carbonate has a -2 charge. The lowest common multiple is 6 . You need two Aluminum ions ( +6 ) and three Carbonate ions (-6). (iii) $K^{+}$ and $SO_{4}^{2 -} \to K_{2} SO_{4}$

Potassium has a +1 charge and Sulfate has a -2 charge. You need two Potassium ions to balance one Sulfate ion. (iv) $NH_{4}^{+}$ and $Cl^{-} \to NH_{4} Cl$

Ammonium has a +1 charge and Chloride has a -1 charge. Since the charges are equal and opposite, they combine in a 1:1 ratio.

34. Which of the following, in Figure, correctly represents $Cl^{-}$ ion (Atomic number of chlorine $= 17$ ).

Ans. (ii) correct

Configuration: 2, 8, 8 .

By achieving a 2, 8, 8 configuration, the chloride ion satisfies the Octet Rule (having 8 electrons in its outermost shell). This makes it highly stable, giving it the exact same electronic structure as the noble gas Argon. $Cl + e^{-} \to Cl^{-}$ Cl : Neutral chlorine atom (2, 8, 7 configuration) $e^{-}$ : The gained electron $Cl^{-}$ : Chlorine anion (2, 8, 8 configuration) (i) Configuration: 2, 7, 8

Status: Incorrect / Impossible for a ground-state atom. Explanation: Electrons must completely fill the inner shells before moving to an outer shell (unless the atom is in a highly unstable, temporary "excited state"). The second shell can hold a maximum of 8 electrons. It cannot have 7 electrons while the third shell already has 8. Correction: If this represents 17 total electrons ( $2 + 7 + 8 = 17$ ), the correct ground-state configuration is 2,8,7 (Chlorine). (iii) Configuration: 2, 8, 9 Status: Incorrect / Highly Unstable for main-group elements. Explanation: According to chemical stability rules, the outermost (valence) shell of an atom cannot hold more than 8 electrons if it is the furthest shell from the nucleus (the Octet Rule).

Correction: If an element has 19 total electrons ( $2 + 8 + 9 = 19$ ), the 9th electron in the third shell jumps to the fourth shell to maintain stability. The correct configuration is $2, 8, 8, 1$ , which belongs to Potassium (K). (iv) Configuration: 2, 8, 7

Status: Incorrect

Explanation: This configuration follows all electron-filling rules perfectly. The first shell is full with 2, the second is full with 8 , and the third holds 7 valence electrons.

Element Info: This represents a total of 17 electrons, which belongs to Chlorine (Cl), a highly reactive halogen in Group 17 of the periodic table.

35. Determine the formula unit mass of the following substances. (i) Ammonium nitrate $(NH_{4} NO_{3})$ , used as a nitrogen fertiliser, which is essential for plant growth. (ii) Phosphoric acid $(H_{3} PO_{4})$ , used to make phosphate fertiliser and detergents. (iii) Sodium hydrogen carbonate ( $NaHCO_{3}$ ), used to relieve acidity and helps in digestion.

Ans. To calculate these, we add up the atomic masses of all the atoms in each formula: (Atomic masses: $H = 1 u, C = 12 u, N = 14 u, O = 16 u$ , $Na = 23 u, P = 31 u$ ) (i) Ammonium nitrate $(NH_{4} NO_{3})$ :

$Mass = (2 \times N) + (4 \times H) + (3 \times 0) Mass = (2 \times 14) + (4 \times 1) + (3 \times 16) = 28 + 4 + 48 = 80 u$

(ii) Phosphoric acid $(H_{3} PO_{4})$ :

$Mass = (3 \times H) + (1 \times P) + (4 \times 0) Mass = (3 \times 1) + 31 + (4 \times 16) = 3 + 31 + 64 = 98 u$

(iii)Sodium hydrogen carbonate $(NaHCO_{3})$ :

$Mass = (1 \times Na) + (1 \times H) + (3 \times 0) Mass = 23 + 1 + 12 + (3 \times 16) = 36 + 48 = 84 u$

Write the formulae for the compounds formed by the reaction of: (i) Magnesium and nitrogen (ii) Lithium and nitrogen (iii) Sodium and sulfur (iv) Aluminium and oxygen

Ans.

Reaction pair	Valency of metal	Valency of nonmetal	Chemical formula	Compound name
Magnesium and nitrogen	+2	-3	$Mg_{3} N_{2}$	Magnesium nitride
Lithium and nitrogen	+1	-3	$Li_{3} N$	Lithium nitride
Sodium and sulfur	+1	-2	$Na_{2} S$	Sodium sulfide
Aluminium and oxygen	+3	-2	$Al_{2} O_{3}$	Aluminium oxide

(i) Magnesium $(Mg^{2 +})$ and Nitrogen ( $N^{3 -}$ ): To balance the charges, three Mg atoms (total +6 ) and two N atoms (total - 6) are required, leading to $Mg_{3} N_{2}$ .

(ii) Lithium (

Li^{+}

) and Nitrogen (

N^{3 -}

Three Li atoms are needed to balance the -3 charge of a single nitrogen atom, resulting in $Li_{3} N$ .

(iii) Sodium (

Na^{+}

) and Sulfur (

S^{2 -}

Two Na atoms are required to neutralize the -2 charge of a sulfur atom, giving $Na_{2} S$ .

(iv) Aluminium

(Al^{3 +})

and Oxygen

(O^{2 -})

Two Al atoms (total -6 ) and three 0 atoms (total -6) combine to form the stable compound $Al_{2} O_{3}$ .

37. Complete the Table by writing the formulae of the compounds formed by the cations on the left and the anions at the top.

LiNO_{3}

is given table as an example.

	$NO_{3}^{-}$	$SO_{4}^{2 -}$	$PO_{4}^{3 -}$
$NH_{4}^{+}$
$Li^{+}$	$LiNO_{3}$
$Al^{3 +}$
$Cu^{2 +}$

Ans.

Cation/Anion	$NO_{3}^{-}$ (Nitrate)	$SO_{4}^{2 -}$ (Sulfate)	$PO_{4}^{3 -}$ (Phosphate)
$NH_{4}^{+}$ (Ammonium)	$NH_{4} NO_{3}$	$(NH_{4})_{2} SO_{4}$	$(NH_{4})_{3} PO_{4}$
$Li^{+}$ (Lithium)	$LiNO_{3}$	$Li_{2} SO_{4}$	$Li_{3} PO_{4}$
$Al^{3 +}$ (Aluminium)	$Al (NO_{3})_{3}$	$Al_{2} (SO_{4})_{3}$	$AlPO_{4}$
$Cu^{2 +}$ (Copper II)	$Cu (NO_{3})_{2}$	$CuSO_{4}$	$Cu_{3} (PO_{4})_{2}$

5.3 g of sodium carbonate and 6.0 g of acetic acid react to produce 2.2 g of carbon dioxide, 0.9 g of water, and 8.2 g of sodium acetate. Verify whether the law of conservation of mass is valid.

Ans. 1. Calculate the total mass of the Reactants:

The reactants are sodium carbonate and acetic acid.

Mass of Sodium Carbonate $= 5.3 g$
Mass of Acetic Acid $= 6.0 g$
Total Mass (Reactants) $= 5.3 g + 6.0 g = 11.3 g$

Calculate the total mass of the Products:

The products are carbon dioxide, water, and sodium acetate.

Mass of Carbon Dioxide $= 2.2 g$
Mass of Water $= 0.9 g$
Mass of Sodium Acetate $= 8.2 g$
Total Mass (Products) $= 2.2 g + 0.9 g + 8.2 g = 11.3 g$ Conclusion: Since the Total Mass of Reactants ( 11.3 g ) is exactly equal to the Total Mass of Products $(11.3 g)$ , the Law of Conservation of Mass is verified and valid for this reaction.

39. If a species has 11 protons, 12 neutrons and 10 electrons then (i) What is its atomic number and mass number? (ii) Is it neutral, a cation or an anion? Explain. (iii) Write its electronic configuration. (iv) Name the species.

Ans. Based on the provided information (11 protons, 12 neutrons, 10 electrons), the species is a positively charged sodium cation ( $Na^{+}$ ) with an atomic number of 11 and a mass number of 23, having a 2,8 electronic configuration. (i) Atomic Number & Mass Number:

Atomic Number (Z): Equal to the number of protons, which is 11.

Mass Number (A): Equal to Protons + Neutrons = $11 + 12 = 23$ . (ii) Nature of the Species:

It is a Cation (positively charged ion). The number of protons (positive charges) is 11 , and the number of electrons (negative charges) is 10 . Since there are more protons than electrons ( $11 > 10$ ), the species has a net positive charge of +1 (written as $Na^{+}$ ). (iii) Electronic Configuration:

With 10 electrons to arrange: 2, 8. (iv) Name of the Species:

Sodium ion (or Sodium cation).

40. Two elements, $A$ and $B$ , have the following configurations - A: 2, 8, 5 B: 2, 8, 7 (i) Which element is more reactive? (ii) Will A and B form ionic or covalent bonds when they combine? Explain using electron transfer or sharing. (iii) Predict the formula of the compound they would form. Ans. (i) Element B is more reactive. Element A has 5 valence electrons and needs to gain 3 electrons to reach a stable octet. Element B has 7 valence electrons and only needs to gain one electron to become stable. Since it is easier and requires less energy to gain one electron than three, B is more chemically reactive. (ii) They will form covalent bonds.

Both A and B are non-metals (A is Phosphorus and B is Chlorine). Nonmetals typically react with each other by sharing electrons rather than transferring them. Since neither atom is willing to completely give up its valence electrons, they overlap their orbits to share pairs of electrons so that both can achieve a stable noble gas configuration ( 8 electrons in the outer shell). (iii) Predict the formula of the compound. The formula is $AB_{3}$ . Element A needs 3 electrons to complete its octet. Element B needs 1 electron to complete its octet. To satisfy element A , it must share one electron with three different B atoms. This allows A to get the 3 electrons it needs, while each of the three B atoms gets the 1 electron it needs.

41. Assertion (A): Copper sulfate conducts electricity in the molten state but not in the solid state.

Reason (R): Copper and sulfate ions are fixed in the lattice in molten state, while in solid state they can move freely. Choose the correct option: (i) Both A and R are true, and R is the correct explanation of A. (ii) Both A and R are true, but R is not the correct explanation of A. (iii) A is true, but R is false. (iv) A is false, but R is true.

Ans. The correct option is (iii) A is true, but R is false.

Assertion (A) is true: Copper sulfate (CuSO4) is an ionic compound. In its solid state, the ions are held in a rigid crystal lattice and cannot move. When molten or dissolved in water, the lattice breaks down, allowing the ions ( $Cu^{2 +}$ and $SO_{4}^{2 -}$ ) to move freely and conduct electricity. Reason (R) is false: The reason states that in the solid state ions can move freely and in the molten state they are fixed, which is the exact opposite of reality. In the solid state, ions are fixed; in the molten state, they are free.

42. The species $27 Al, 80 Br^{-}$ and $201 Hg^{2 +}$ have 13,35 and 80 protons, respectively. How many electrons and neutrons do they have? Ans. To determine the number of electrons and neutrons for these species, you need to use the atomic number (Z) and the mass number (A).

Number of Neutrons = Mass Number (A) - Atomic Number (Z)

Number of Electrons $=$ Atomic Number (Z) - Charge.

Calculate for Aluminum ( $^{27}$ Al)

Aluminum ( $^{27} Al$ ) is a neutral atom with an atomic number of 13 and a mass number of 27.

Neutrons: $A - Z = 27 - 13 = 14$ neutrons.

Electrons: Since it is neutral (charge $= 0$ ), the number of electrons equals the number of protons (13 electrons).

2. Calculate for Bromide Ion ( $^{80} Br^{-}$ )

The bromide ion $(80 Br^{-})$ has an atomic number of 35, a mass number of 80 , and a charge of -1 .

Neutrons: $A - Z = 80 - 35 = 45$ neutrons. Electrons: Because it has a - 1 charge, it has gained one electron. Therefore, $35 + 1 = 36$ electrons.

3. Calculate for Mercury Ion $(^{201} Hg^{2 +})$

The mercury ion $(^{201} Hg^{2 +})$ has an atomic number of 80, a mass number of 201, and a charge of +2 .

Neutrons: $A - Z = 201 - 80 = 121$ neutrons. Electrons: Because it has a +2 charge, it has lost two electrons. Therefore, $80 - 2 = 78$ electrons.

4.0Key Topics in NCERT Class 9 Science Chapter 9 Atomic Foundations of Matter

Topic	What Students Learn
Law of Conservation of Mass	Verify that the total mass of reactants equals the total mass of products in a chemical reaction.
Law of Constant Proportions	Understand that a compound always contains its constituent elements in a fixed mass ratio, regardless of source.
Dalton's Atomic Theory	Learn the postulates of Dalton's theory and how it explains the two fundamental laws of chemical combination.
Atoms and Molecules	Differentiate between atoms of elements and molecules formed by the combination of atoms.
Covalent Bonding	Understand bond formation through the sharing of electrons, with examples like H₂, O₂, Cl₂, H₂O, and HCl.
Ionic Bonding	Understand bond formation through the transfer of electrons, with examples like sodium chloride (NaCl).
Properties of Ionic and Covalent Compounds	Compare melting points, solubility, and electrical conductivity of ionic versus covalent compounds.
Chemical Formulae and Valency	Learn to write correct chemical formulae for compounds using the valency of elements and polyatomic ions.
Molecular Mass	Calculate the mass of a molecule by adding the atomic masses of all atoms present in it.
Formula Unit Mass	Calculate the mass of one formula unit of an ionic compound and understand how it differs from molecular mass.

5.0Mind Map / Concept Recap

6.0Related Study Materials for Class 9 Science

Strengthen your Class 9 Science preparation with additional study resources designed according to the latest CBSE and NCERT syllabus. Explore chapter-wise notes, NCERT Solutions, important questions, sample papers, and revision materials to improve conceptual understanding and exam performance.

CBSE Class 9 Science Syllabus

Class 9 Science Revision Notes

NCERT Textbook for Class 9 Science

CBSE Sample Papers for Class 9 Science

7.0Advantages of Chapter 9 Science Class 9 NCERT Solutions

Concept Clarity: Detailed explanations help students understand every topic from basic to advanced levels.
Step-by-Step Solutions: Each answer follows a logical approach, making it easy to learn the correct method of solving questions.
Exam-Oriented Preparation: Covers all in-text and exercise questions frequently asked in school examinations.
Improved Problem-Solving Skills: Helps students develop analytical thinking and apply concepts confidently.
Accurate and Reliable Answers: Solutions are prepared according to the latest NCERT textbook and CBSE guidelines.
Quick Revision Resource: Well-organised answers enable students to revise important concepts before tests and exams.
Strengthens Fundamentals: Builds a strong conceptual foundation that supports higher classes and competitive exam preparation.

NCERT Solutions for Class 9 Science Chapter 9: Atomic Foundations of Matter

1.0Download NCERT Solutions for Class 9 Science Chapter 9 – Atomic Foundations of Matter PDF

NCERT Solutions Class 9 Science Chapter 9

2.0NCERT Class 9 Science Chapter 9 Solutions: Learning Outcomes

Understand the Law of Conservation of Mass: Verify through activities that mass is neither created nor destroyed during a chemical reaction.
Understand the Law of Constant Proportions: Explain that a pure chemical compound always contains the same elements combined in a fixed proportion by mass.
Learn Dalton's Atomic Theory: State the postulates of Dalton's Atomic Theory and explain how it accounts for the two laws above.
Differentiate Atoms and Molecules: Understand how atoms combine to form molecules of elements and compounds.
Understand Ionic Bonding: Explain how ionic bonds form through the transfer of electrons, using examples like sodium chloride.
Understand Covalent Bonding: Explain how covalent bonds form through the sharing of electrons, using examples like hydrogen, oxygen, and water.
Compare Ionic and Covalent Compounds: Differentiate the properties of ionic and covalent compounds, including solubility and electrical conductivity.
Write Chemical Formulae Using Valency: Apply the valency of elements and radicals to correctly write chemical formulae of compounds.
Calculate Molecular Mass: Compute the molecular mass of covalent compounds by summing the atomic masses of constituent atoms.
Calculate Formula Unit Mass: Compute the formula unit mass of ionic compounds and distinguish it from molecular mass.
Solve NCERT Exercise Problems: Confidently attempt and solve all in-text activities and end-of-chapter numerical and conceptual questions.
Build a Foundation for Advanced Topics: Develop conceptual clarity that supports higher-level chemistry, including mole concept and chemical bonding in later classes.