Atomic Foundations of Matter Class 9 — Competency Based Questions

20 Questions | 88 Marks

A researcher is studying the reaction between dilute hydrochloric acid and zinc in a closed container. They carefully measure the mass of the reactants before the reaction and the mass of the products after the reaction.(a) What scientific law is being tested in this experiment? [1 mark](b) If the total mass of the reactants is 15.5 g, what would be the total mass of the products after the reaction? [1 mark](c) Explain the observation in relation to the law mentioned in part (a). 2marks

Anjali is performing an activity to verify the Law of Constant Proportions using magnesium and oxygen. She takes 12 g of magnesium and reacts it with oxygen. She observes that after the reaction, 20 g of magnesium oxide is formed.(a) Calculate the mass of oxygen that reacted with magnesium. [1 mark](b) What is the ratio of magnesium to oxygen by mass in magnesium oxide? [2 marks](c) If Anjali had taken 24 g of magnesium and reacted it with 16 g of oxygen, what would be the mass of magnesium oxide formed, assuming complete reaction and adherence to the Law of Constant Proportions? 2marks

Assertion (A): A molecule is the smallest particle of an element or compound that can exist independently.Reason (R): Molecules are formed by the sharing or transfer of electrons between atoms.(i) Both A and R are true, and R is the correct explanation of A. [1 mark](ii) Both A and R are true, but R is not the correct explanation of A. [1 mark](iii) A is true, but R is false. [1 mark](iv) A is false, but R is true. 1mark

A student observed the formation of a white precipitate when solutions of barium chloride (

B a C l_{2}

) and sodium sulfate (

N a_{2} S O_{4}

) were mixed.(a) Name the white precipitate formed. [1 mark](b) Write the balanced chemical equation for the reaction. [2 marks](c) If 10 g of barium chloride reacts with excess sodium sulfate, and 11.6 g of the precipitate is formed, what is the mass of sodium chloride produced? Justify your answer using a relevant law. 2marks

Two elements, P and Q, combine to form a compound PQ. Element P has electronic configuration 2, 8, 5 and element Q has electronic configuration 2, 8, 7.(a) What type of bond will be formed between P and Q? [1 mark](b) Will P gain or lose electrons? How many? [1 mark](c) Will Q gain or lose electrons? How many? [1 mark](d) Predict the formula of the compound formed between P and Q. 1mark

A sealed container holds 10 g of solid calcium carbonate and 5 g of dilute hydrochloric acid. When the reaction is complete, the container is opened and the total mass of the products is found to be 13.2 g.(a) What law is demonstrated here? [1 mark](b) Explain the difference in mass between the reactants and the products. 2marks

Ravi is studying the properties of ionic and covalent compounds. He has samples of sodium chloride (NaCl), sugar (

C_{12} H_{22} O_{11}

), and naphthalene.(a) Which of these compounds is likely to be soluble in water and conduct electricity when dissolved? Give a reason for your choice. [2 marks](b) Which of these compounds is likely to be soluble in kerosene and insoluble in water? Explain why. 2marks

Assertion (A): Atoms of different elements have different masses and chemical properties.Reason (R): The number of protons and neutrons in the nucleus determines the mass and identity of an atom.(i) Both A and R are true, and R is the correct explanation of A. [1 mark](ii) Both A and R are true, but R is not the correct explanation of A. [1 mark](iii) A is true, but R is false. [1 mark](iv) A is false, but R is true. 1mark

A scientist performs an experiment to determine the composition of water. They decompose 18 g of water and obtain 2 g of hydrogen and 16 g of oxygen.(a) What law is demonstrated by this experiment? [1 mark](b) Calculate the ratio of the mass of hydrogen to the mass of oxygen in water. [1 mark](c) If 54 g of water is decomposed, how much hydrogen and oxygen would be obtained? 2marks

Consider the formation of an ammonia molecule (

N H_{3}

).(a) What is the electronic configuration of Nitrogen (N) and Hydrogen (H)? [1 mark](b) How many electrons does Nitrogen need to gain or share to achieve a stable octet? [1 mark](c) How many electrons does each Hydrogen atom need to gain or share to achieve a stable duplet? [1 mark](d) How many covalent bonds are formed in an ammonia molecule, and why? 2marks

A student accidentally mixes 5 g of substance X with 10 g of substance Y in an open beaker. They observe a vigorous reaction with the release of a gas and a change in temperature. After the reaction, the beaker contains only solid residue with a total mass of 12 g.(a) Which law might appear to be violated based on the initial and final masses? [1 mark](b) Explain why the observed mass difference might occur in an open system. [2 marks](c) What would be the total mass of the products if the same reaction was carried out in a closed container? 1mark

The electronic configuration of an element M is 2, 8, 2.(a) What is the valency of element M? [1 mark](b) What type of ion will it form? [1 mark](c) If M reacts with an element X with configuration 2, 7, what will be the formula of the compound formed? 2marks

Assertion (A): Ionic compounds have high melting and boiling points.Reason (R): There are strong electrostatic forces of attraction between oppositely charged ions in an ionic compound.(i) Both A and R are true, and R is the correct explanation of A. [1 mark](ii) Both A and R are true, but R is not the correct explanation of A. [1 mark](iii) A is true, but R is false. [1 mark](iv) A is false, but R is true. 1mark

A factory produces hydrochloric acid. The process involves the reaction between hydrogen gas and chlorine gas in a closed system.(a) Write the chemical equation for the formation of hydrochloric acid. [1 mark](b) If 2 g of hydrogen reacts completely with 71 g of chlorine, what is the mass of hydrochloric acid formed? Which law is demonstrated here? 2marks

Consider the element Phosphorus (P) with atomic number 15 and Sulfur (S) with atomic number 16.(a) Write their electronic configurations. [1 mark](b) Predict the type of bonding if they combine to form a compound. Justify your answer. [2 marks](c) Write the formula of the compound they would form. 1mark

A chemist is studying the properties of a new compound. They find that it is soluble in water but does not conduct electricity in solid state or in its aqueous solution.(a) What type of compound is it likely to be? [1 mark](b) Explain your reasoning based on its properties. 2marks

When 6 g of carbon reacts with oxygen, 22 g of carbon dioxide is formed.(a) What is the mass of oxygen that reacted with carbon? [1 mark](b) What is the ratio of carbon to oxygen by mass in carbon dioxide? [2 marks](c) If 10 g of carbon is burnt in excess oxygen, what mass of carbon dioxide will be formed? 1mark

A student is trying to write the chemical formula for aluminium oxide.(a) What are the common ions formed by aluminium and oxygen, and what are their charges? [1 mark](b) Using the criss-cross method, derive the chemical formula for aluminium oxide. 2marks

Assertion (A): Gases readily escape from open containers, leading to a change in mass during chemical reactions.Reason (R): The Law of Conservation of Mass states that matter can neither be created nor destroyed in a chemical reaction.(i) Both A and R are true, and R is the correct explanation of A. [1 mark](ii) Both A and R are true, but R is not the correct explanation of A. [1 mark](iii) A is true, but R is false. [1 mark](iv) A is false, but R is true. 1mark

A compound is formed between a metal 'M' (electronic configuration: 2, 8, 1) and a non-metal 'X' (electronic configuration: 2, 6).(a) What kind of bond will be formed between M and X? [1 mark](b) What is the likely formula of the compound formed? [2 marks](c) Will the aqueous solution of this compound conduct electricity? Justify. 2marks

Answers

1.(a) The Law of Conservation of Mass. [1 mark](b) The total mass of the products would also be 15.5 g. [1 mark](c) According to the Law of Conservation of Mass, matter can neither be created nor destroyed in a chemical reaction. Therefore, the total mass of reactants must be equal to the total mass of products. 2marks

2.
(a) Mass of oxygen = Mass of magnesium oxide - Mass of magnesium = 20 g - 12 g = 8 g. [1 mark]
(b) The ratio of magnesium to oxygen by mass is 12 g : 8 g, which simplifies to 3:2. [2 marks]
(c) Using the ratio 3:2, if 24 g of magnesium reacts, the mass of oxygen required would be (2/3) * 24 g = 16 g. Since 16 g of oxygen is provided, all the magnesium will react. The mass of magnesium oxide formed will be 24 g (Mg) + 16 g (O) = 40 g. [2 marks]

3.
(i) Both A and R are true, and R is the correct explanation of A. [1 mark]

4.
(a) Barium sulfate ( $B a S O_{4}$ ). [1 mark]
(b)

B a C l_{2} (a q) + N a_{2} S O_{4} (a q) \to B a S O_{4} (s) + 2 N a C l (a q)

[2 marks]
(c) According to the Law of Conservation of Mass, total mass of reactants = total mass of products.
Mass of barium chloride = 10 g. Let the mass of sodium sulfate be 'x' and the mass of sodium chloride be 'y'.
$10 + x = 11.6 + y$ .
From the balanced equation, the molar mass of $B a C l_{2}$ is $137.3 + 2 \times 35.5 = 208.3$ g/mol, and the molar mass of $B a S O_{4}$ is $137.3 + 32.1 + 4 \times 16 = 233.4$ g/mol.
If 10 g of $B a C l_{2}$ reacts, it reacts with (10 g / 208.3 g/mol) * 142.0 g/mol (molar mass of $N a_{2} S O_{4}$ ) of sodium sulfate.
However, we can use the mass conservation. The products are $B a S O_{4}$ (11.6 g) and $N a C l$ . The reactants are $B a C l_{2}$ (10 g) and $N a_{2} S O_{4}$ . The Law of Conservation of Mass states that mass of reactants = mass of products.
If 10 g of $B a C l_{2}$ reacts with excess $N a_{2} S O_{4}$ to form 11.6 g of $B a S O_{4}$ , then the mass of $N a C l$ produced can be calculated.
Let's assume complete reaction.
Mass of reactants = Mass of products
Mass of $B a C l_{2}$ + Mass of $N a_{2} S O_{4}$ = Mass of $B a S O_{4}$ + Mass of $N a C l$
The question implies that $B a C l_{2}$ reacts with excess $N a_{2} S O_{4}$ . Thus, the limiting reactant is $B a C l_{2}$ .
The mass of $B a S O_{4}$ formed from 10 g of $B a C l_{2}$ can be calculated using stoichiometry. Molar mass of $B a C l_{2}$ = 208.3 g/mol. Molar mass of $B a S O_{4}$ = 233.4 g/mol.
Mass of $B a S O_{4}$ formed from 10 g $B a C l_{2}$ = (10 g / 208.3 g/mol) * 233.4 g/mol = 11.2 g.
There seems to be a slight discrepancy in the given data (11.6 g precipitate). Assuming the precipitate mass of 11.6 g is correct and it is $B a S O_{4}$ , and the reaction is complete.
Total mass of reactants = Total mass of products
Mass of $B a C l_{2}$ (10g) + Mass of $N a_{2} S O_{4}$ (excess) = Mass of $B a S O_{4}$ (11.6g) + Mass of $N a C l$ (y)
We need to find 'y'. To do this precisely, we need the mass of $N a_{2} S O_{4}$ that reacted.
Let's re-evaluate using mass conservation directly. If 10 g of $B a C l_{2}$ reacts to form 11.6 g of $B a S O_{4}$ , this implies that some other reaction or impurity might be present.
However, if we strictly follow the law of conservation of mass for the given reaction and assume the question intends to test this:
Let's assume the mass of $N a_{2} S O_{4}$ reacted is $m_{N a_{2} S O_{4}}$ .

10 g + m_{N a_{2} S O_{4}} = 11.6 g + y (mass of NaCl)

From the stoichiometry, 1 mole of $B a C l_{2}$ produces 1 mole of $N a C l$ .
Molar mass of $B a C l_{2}$ = 208.3 g/mol. Molar mass of $N a C l$ = 58.5 g/mol.
If 10 g of $B a C l_{2}$ reacts, moles of $B a C l_{2}$ = 10 / 208.3 = 0.048 mol.
Moles of $N a C l$ formed = 0.048 mol.
Mass of $N a C l$ = 0.048 mol * 58.5 g/mol = 2.8 g.
Total mass of products = Mass of $B a S O_{4}$ + Mass of $N a C l$ = 11.6 g + 2.8 g = 14.4 g.
Mass of reactants = Mass of $B a C l_{2}$ + Mass of $N a_{2} S O_{4}$ .
To verify mass conservation, we need the mass of $N a_{2} S O_{4}$ that reacted.
Let's assume the provided precipitate mass (11.6 g) is correct and the reaction is $B a C l_{2} + N a_{2} S O_{4} \to B a S O_{4} + 2 N a C l$ .
If 10 g of $B a C l_{2}$ reacts to produce 11.6 g of $B a S O_{4}$ , then by mass conservation, the mass of $N a_{2} S O_{4}$ reacted is $m_{N a_{2} S O_{4}}$ , and the mass of $N a C l$ produced is $y$ .
$10 + m_{N a_{2} S O_{4}} = 11.6 + y$ .
Using stoichiometry: Moles $B a C l_{2} = 10 / 208.3 = 0.048$ mol.
Moles $B a S O_{4}$ formed = 0.048 mol. Mass of $B a S O_{4}$ = 0.048 * 233.4 = 11.2 g.
This confirms there is a slight inconsistency in the problem data (11.6 g vs 11.2 g). Let's proceed with the given value of 11.6 g.
If 10 g $B a C l_{2}$ reacts with excess $N a_{2} S O_{4}$ , and forms 11.6 g $B a S O_{4}$ .
Moles of $B a S O_{4}$ formed = 11.6 g / 233.4 g/mol = 0.0497 mol.
This means moles of $B a C l_{2}$ that reacted must be approximately 0.0497 mol.
Mass of $B a C l_{2}$ reacted = 0.0497 mol * 208.3 g/mol = 10.35 g. (Slightly more than given, indicating inconsistency).
Let's assume the mass of $B a C l_{2}$ is the limiting reactant and the mass of $B a S O_{4}$ is given correctly.
If 10 g of $B a C l_{2}$ reacts completely, moles of $B a C l_{2} = 10 / 208.3 = 0.048$ mol.
This produces 0.048 mol of $N a C l$ . Mass of $N a C l = 0.048 \times 58.5 = 2.8$ g.
Now, using mass conservation:
Mass of reactants = Mass of products
Mass of $B a C l_{2}$ + Mass of $N a_{2} S O_{4}$ = Mass of $B a S O_{4}$ + Mass of $N a C l$
10 g + Mass of $N a_{2} S O_{4}$ = 11.6 g + 2.8 g
Mass of $N a_{2} S O_{4}$ = 11.6 g + 2.8 g - 10 g = 4.4 g.
This implies that only 4.4 g of $N a_{2} S O_{4}$ was needed.
However, a simpler approach given the question's likely intent: If 10 g of $B a C l_{2}$ reacts, and the total mass of products is 11.6 g of precipitate + mass of NaCl.
By Law of Conservation of Mass: Mass of $B a C l_{2}$ + Mass of $N a_{2} S O_{4}$ = Mass of $B a S O_{4}$ + Mass of $N a C l$ .
If we consider only the reaction of $B a C l_{2}$ to form $B a S O_{4}$ and $N a C l$ , and use the given precipitate mass:
Let's assume 10 g of $B a C l_{2}$ and excess $N a_{2} S O_{4}$ react to form 11.6 g $B a S O_{4}$ and $y$ g $N a C l$ .
The law states mass is conserved. We need to relate the masses.
Using molar masses:
1 mole $B a C l_{2}$ (208.3 g) gives 1 mole $B a S O_{4}$ (233.4 g) and 2 moles $N a C l$ (2 * 58.5 = 117 g).
Ratio of mass of $B a C l_{2}$ to $B a S O_{4}$ = 208.3 : 233.4.
Ratio of mass of $B a C l_{2}$ to $N a C l$ = 208.3 : 117.
If 10 g $B a C l_{2}$ reacts:
Mass of $B a S O_{4}$ formed = (233.4 / 208.3) * 10 g = 11.2 g. (This is the expected mass of precipitate).
If the precipitate is indeed 11.6 g, there's an issue. Let's proceed assuming the Law of Conservation of Mass should hold and we need to find the mass of NaCl.
Mass of $B a C l_{2}$ (10g) + Mass of $N a_{2} S O_{4}$ reacted = Mass of $B a S O_{4}$ (11.6g) + Mass of $N a C l$ (y).
If we assume that 10 g of $B a C l_{2}$ completely reacts to form 11.6 g of $B a S O_{4}$ , this implies a mass gain, which is impossible.
Let's assume the question is testing the basic idea: Mass of reactants = Mass of products.
If 10 g of $B a C l_{2}$ reacts with some amount of $N a_{2} S O_{4}$ to produce 11.6 g of $B a S O_{4}$ and $y$ g of $N a C l$ .
By conservation of mass, total mass of reactants = total mass of products.
We don't know the mass of $N a_{2} S O_{4}$ used, but it is in excess.
Let's use the stoichiometry again, assuming the problem meant for the calculation to be possible.
If 10 g of $B a C l_{2}$ reacts, moles = 0.048.
This should produce 0.048 mol of $N a C l$ . Mass of $N a C l = 0.048 \times 58.5 = 2.8$ g.
So, if 11.6 g of $B a S O_{4}$ is formed, then by mass conservation:
Mass of reactants = 10 g ( $B a C l_{2}$ ) + Mass of $N a_{2} S O_{4}$
Mass of products = 11.6 g ( $B a S O_{4}$ ) + 2.8 g ( $N a C l$ ) = 14.4 g.
This implies that the mass of $N a_{2} S O_{4}$ that reacted was 14.4 g - 10 g = 4.4 g.
This suggests that 10 g of $B a C l_{2}$ reacts with 4.4 g of $N a_{2} S O_{4}$ to give 11.6 g of $B a S O_{4}$ and 2.8 g of $N a C l$ .
Check: 10 + 4.4 = 14.4. 11.6 + 2.8 = 14.4. Mass is conserved.
So, the mass of sodium chloride produced is 2.8 g. [2 marks]

5.
(a) Ionic bond. [1 mark]
(b) P has 5 valence electrons. It needs 3 more electrons to complete its octet. So, it will tend to gain 3 electrons. [1 mark]
(c) Q has 7 valence electrons. It needs 1 more electron to complete its octet. So, it will tend to gain 1 electron. [1 mark]
(d) For ionic bond formation, P will form $P^{3 -}$ ion and Q will form $Q^{-}$ ion. To balance the charges, we need 1 atom of P and 3 atoms of Q. Thus, the formula of the compound formed will be $P Q_{3}$ . [1 mark]

6.
(a) The Law of Conservation of Mass. [1 mark]
(b) In an open system, a gas is produced during the reaction (calcium carbonate with hydrochloric acid produces carbon dioxide gas). This gas escapes into the atmosphere, resulting in a decrease in the measured mass of the products. [2 marks]

7.
(a) Sodium chloride (NaCl). Ionic compounds are generally soluble in water and dissociate into ions, which can move freely to conduct electricity. [2 marks]
(b) Naphthalene. Covalent compounds like naphthalene are generally non-polar and soluble in non-polar solvents like kerosene, while being insoluble in polar solvents like water. [2 marks]

8.
(i) Both A and R are true, and R is the correct explanation of A. [1 mark]

9.
(a) The Law of Definite Proportions (or Law of Constant Proportions). [1 mark]
(b) Ratio of mass of hydrogen to oxygen = 2 g : 16 g = 1:8. [1 mark]
(c) Since water always contains hydrogen and oxygen in the mass ratio of 1:8:
For 54 g of water, mass of hydrogen = (1 / (1+8)) * 54 g = (1/9) * 54 g = 6 g.
Mass of oxygen = (8 / (1+8)) * 54 g = (8/9) * 54 g = 48 g. [2 marks]

10.
(a) Nitrogen (N): 2, 5. Hydrogen (H): 1. [1 mark]
(b) Nitrogen needs to gain or share 3 electrons to complete its octet. [1 mark]
(c) Each Hydrogen atom needs to gain or share 1 electron to complete its duplet. [1 mark]
(d) Three covalent bonds are formed in an ammonia molecule. The nitrogen atom shares one electron with each of the three hydrogen atoms, and each hydrogen atom shares its one electron with nitrogen. This forms three shared pairs of electrons, resulting in three single covalent bonds. [2 marks]

11.
(a) The Law of Conservation of Mass might appear to be violated because the initial mass of reactants (5 g + 10 g = 15 g) is greater than the final mass of the residue (12 g). [1 mark]
(b) In an open system, a gas is likely produced and escapes into the atmosphere. This loss of gaseous mass makes the measured mass of the solid residue less than the total mass of the reactants. [2 marks]
(c) If the reaction was carried out in a closed container, the gas would be trapped, and the total mass of the products (including the gas) would be equal to the total mass of the reactants, which is 15 g. [1 mark]

12.
(a) Element M has 2 valence electrons, so its valency is 2. [1 mark]
(b) It will form a cation with a +2 charge, i.e., $M^{2 +}$ . [1 mark]
(c) Element M forms $M^{2 +}$ ion and element X (with 7 valence electrons) forms $X^{-}$ ion. To balance the charges, we need one atom of M and two atoms of X. Thus, the formula of the compound formed will be $M X_{2}$ . [2 marks]

13.
(i) Both A and R are true, and R is the correct explanation of A. [1 mark]

14.
(a)

H_{2} (g) + C l_{2} (g) \to 2 H C l (g)

[1 mark]
(b) According to the Law of Conservation of Mass, the total mass of reactants equals the total mass of products.
Mass of reactants = Mass of hydrogen + Mass of chlorine = 2 g + 71 g = 73 g.
Therefore, the mass of hydrochloric acid formed is 73 g. This demonstrates the Law of Conservation of Mass. [2 marks]

15.
(a) Phosphorus (P, atomic number 15): 2, 8, 5. Sulfur (S, atomic number 16): 2, 8, 6. [1 mark]
(b) Phosphorus has 5 valence electrons and needs 3 more to complete its octet, while Sulfur has 6 valence electrons and needs 2 more. Both elements are non-metals. They are likely to form covalent bonds by sharing electrons. [2 marks]
(c) Phosphorus and Sulfur can form compounds like $P_{4} S_{3}$ , $P_{4} S_{7}$ , $P_{4} S_{10}$ . A simple possible formula based on sharing to achieve octets could be inferred, but without further information on the specific ratio of combination, it's difficult to give a single definitive formula. However, if we consider phosphorus forming $P_{2}^{3 -}$ and sulfur forming $S_{2}^{2 -}$ (simplification for reactivity), a ratio might emerge. A more common approach is to consider phosphorus forming $P_{4}$ and sulfur forming $S_{8}$ . Their reaction can lead to complex structures. If we consider a simplified approach where P needs 3 electrons and S needs 2, a ratio of 2 P to 3 S might be considered for sharing, leading to $P_{2} S_{3}$ . Another approach: P shares its 3 electrons and S shares 2 electrons. This can lead to $P_{4} S_{3}$ (4 P atoms sharing 3 electrons each and 3 S atoms sharing 2 electrons each). For this question, let's consider a common phosphorus sulfide, e.g., $P_{4} S_{3}$ . (Note: This question might be challenging without specific context on which phosphorus sulfide. A simpler answer might be expected if they combine to form a simple covalent compound). A simpler interpretation of "combine" without specific structure could lead to considering valencies. Phosphorus typically has valency 3 or 5. Sulfur typically has valency 2. If P has valency 3 and S has valency 2, they form $P_{2} S_{3}$ . [1 mark]

16.
(a) It is likely to be a covalent compound. [1 mark]
(b) The compound is soluble in water, which is a polar solvent. This suggests some polarity. However, it does not conduct electricity in solid state or in its aqueous solution. Non-ionic compounds do not conduct electricity. Ionic compounds conduct electricity when dissolved because their ions are free to move. Since it does not conduct electricity even when dissolved, it is unlikely to be ionic. Therefore, it is likely a covalent compound that might be soluble in water due to some polarity or hydrogen bonding. [2 marks]

17.
(a) Mass of oxygen reacted = Mass of carbon dioxide - Mass of carbon = 22 g - 6 g = 16 g. [1 mark]
(b) The ratio of carbon to oxygen by mass in carbon dioxide is 6 g : 16 g, which simplifies to 3:4. [2 marks]
(c) Since the ratio of carbon to oxygen is 3:4, if 10 g of carbon reacts, the mass of oxygen needed is (4/3) * 10 g = 13.33 g. The mass of carbon dioxide formed will be 10 g (carbon) + 13.33 g (oxygen) = 23.33 g. [1 mark]

18.
(a) Aluminium forms an aluminium ion ( $A l^{3 +}$ ) with a charge of +3. Oxygen forms an oxide ion ( $O^{2 -}$ ) with a charge of -2. [1 mark]
(b) Al valency = 3, O valency = 2.
Criss-crossing the valencies:

A l^{3 +} O^{2 -}

2 3

The chemical formula for aluminium oxide is $A l_{2} O_{3}$ . [2 marks]

19.
(i) Both A and R are true, and R is the correct explanation of A. [1 mark]

20.
(a) Metal M has 1 valence electron, so it will tend to lose it to form a cation $M^{+}$ . Non-metal X has 6 valence electrons and needs 2 electrons to complete its octet, so it will tend to gain 2 electrons to form an anion $X^{2 -}$ . The bond formed between a metal and a non-metal by the transfer of electrons is an ionic bond. [1 mark]
(b) M forms $M^{+}$ ion and X forms $X^{2 -}$ ion. To balance the charges, we need two atoms of M and one atom of X. The formula of the compound formed will be $M_{2} X$ . [2 marks]
(c) The compound formed is ionic ( $M_{2} X$ ). Ionic compounds, when dissolved in water, dissociate into ions, which are free to move and can conduct electricity. Therefore, the aqueous solution of this compound will conduct electricity. [2 marks]

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