Question 1

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) If A and B were to combine, what kind of bond would be formed? (vi) What would be the formula for the compound thus formed?

Accepted Answer

(i) tends to give 1 electron (ii) Cation ( A^+ ) (iii) tends to take 2 electrons (iv) Anion ( B^2- ) (v) Ionic (vi) A 2 B. - (i) A tends to give 1 electron. - (ii) It forms a cation, A^+. - (iii) B tends to take 2 electrons. - (iv) It forms an anion, B^2-. - (v) An ionic bond — electrons are transferred from A to B, not shared. - (vi) A 2B. - Working: "one electron in the third shell" fixes A's configuration as 2, 8, 1. Its valence shell has fewer than 4 electrons, so it donates its single electron and is left with the stable 2, 8. - "Six electrons in the second shell" fixes B as 2, 6. With more than 4 valence electrons it accepts 2 to complete its octet. - Two A^+ ions are needed for each B^2- so the charges cancel: 2 × (1+) + (2-) = 0, giving A 2B.

Question 2

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

Accepted Answer

- No — the Law of Conservation of Mass is not violated. - Burning is a chemical reaction: ethanol + oxygen (from the air) → carbon dioxide + water vapour. - Both products are gases, and the beaker is open, so they escape into the air — that is why no residue is left behind. - Nothing was destroyed; the mass simply left the beaker. Mass of ethanol + mass of oxygen used = mass of carbon dioxide + mass of water vapour formed. - This is Experimental set-up 1 of Activity 9.2 again: the balance read less only because the CO 2 escaped. Sealing the system (set-up 2) made the initial and final readings match. - To test it properly, burn the ethanol in a closed vessel and weigh the vessel with everything inside it, before and after.

Question 3

When 20 g of hydrogen reacts completely with 160 g of oxygen, how much water is formed according to the Law of Conservation Mass? 166 Exploration Grade 9

Accepted Answer

180 g. - Water formed = 180 g. - Mass is neither created nor destroyed in a chemical reaction, so mass of product = total mass of reactants. - 20 + 160 = 180 g of water. - The ratio agrees with the chapter too: 20: 160 = 1: 8, which is the fixed hydrogen: oxygen mass ratio in water.

Question 4

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.

Accepted Answer

- (i) Because X has six valence electrons and needs two more to complete its octet, and no single X atom can supply them to itself. Two X atoms each share two of their electrons, so both reach 8 — hence the element exists as X 2. - (ii) A covalent bond, and because two pairs are shared it is a double bond. - (iii) The drawing must show: two atoms labelled X side by side, each with its 6 valence electrons marked (dots on one atom, crosses on the other); two shared pairs — 4 electrons — in the overlap between them; two lone pairs left on each X; each X counting 8 electrons in all; and the line formula X=X written alongside, the two lines standing for the two shared pairs. - (iv) Y is 2, 2 — two valence electrons, fewer than four, so it donates both rather than sharing. Y becomes Y^2+ (left with a complete duplet, 2) and X becomes X^2- (2, 8), joined by an ionic bond in the ratio 1: 1 → YX. - The drawing for (iv) must show: Y with shells 2, 2 on the left and X with shells 2, 6 on the right; two curved arrows labelled 2e^- running from Y's outer shell to X's outer shell; the products drawn as [2]^2+ labelled Y^2+ and [2,8]^2- labelled X^2-, each in square brackets with its charge outside; the label ionic bond between them; and the formula YX written below.

Question 5

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?

Accepted Answer

30 g. - Oxygen present = 30 g. - A compound has a fixed composition, so sulfur: oxygen = 40: 60 = 2: 3 by mass in every sample of it. - Mass of oxygen = 60/40 × 20 = 3/2 × 20 = 30 g. - Check: total sample = 20 + 30 = 50 g, so sulfur is 20/50 × 100 = 40 % and oxygen 30/50 × 100 = 60 % — as stated.

Question 6

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 (iii) 2 Fe^3+ and 3 O^2- (iv) 3 Ca^2+ and 2 SO 4

Accepted Answer

(iii). - (iii) 2 Fe^3+ and 3 O^2-. - Positive charge: 2 × (3+) = 6+. Negative charge: 3 × (2-) = 6-. They balance exactly. - (i) fails: 2 Al^3+ gives 6+, but 3 Cl^- gives only 3-. - (ii) fails: 3 Mg^2+ gives 6+, but 1 PO 4^3- gives only 3-. - (iv) fails: 3 Ca^2+ gives 6+, but 2 SO 4^2- gives only 4-. - The compound in (iii) is ferric oxide, Fe 2O 3 — the simplest whole-number ratio of the same two ions.

Question 7

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.

Accepted Answer

(iv). - Only (iv) is correct; (i), (ii) and (iii) are false. - (i) False. Corrected: elements are made up of atoms of one kind (which may join together as molecules of that element, such as O 2 ), and compounds are made up of atoms of two or more different kinds, combined as molecules or as ions. - (ii) False. Corrected: the molecule of a compound is made up of two or more atoms of different kinds — it is a molecule of an element that contains atoms of the same kind. - (iii) False. Corrected: one molecule of nitrogen gas contains two nitrogen atoms, N 2, joined by a triple bond. - (iv) True. Oxygen needs 2 electrons and each hydrogen needs 1, so two hydrogen atoms each share one electron with one oxygen atom — two covalent bonds, giving H 2O.

Question 8

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?

Accepted Answer

12 g. - Oxygen required = 12 g. - In carbon monoxide the mass ratio carbon: oxygen is fixed at 3: 4. - Mass of oxygen = 4/3 × 9 = 12 g. - So 9 + 12 = 21 g of carbon monoxide is formed.

Question 9

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

Accepted Answer

- Because a compound has a fixed composition, while a mixture does not. - In a compound the elements are chemically combined in a definite ratio by mass — 9 g of pure water always gives 1 g hydrogen and 8 g oxygen, whether it came from a river, a borewell or the ocean. - In a mixture the substances are only mixed, not chemically combined, so they can be taken in any proportion — any amount of salt can be stirred into any amount of water (Activity 9.1). - A mixture also keeps the separate properties of its components, so there is no fixed combining ratio for the law to describe.

Question 10

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

Accepted Answer

(i) Al( NO 3 )3 (ii) CaO (iii) Fe 2 O 3. - (i) Aluminium nitrate — Al(NO 3) 3: Al^3+ with NO 3^-; criss-cross 3 and 1 to get one Al to three nitrate ions, and use brackets because more than one polyatomic ion is present. - (ii) Calcium oxide — CaO: Ca^2+ with O^2-; criss-crossing gives Ca 2O 2, which is divided by the common factor 2. - (iii) Ferric oxide — Fe 2O 3: ferric is Fe^3+ and oxide is O^2-; criss-cross 3 and 2, with no common factor to remove.

Question 11

Write the formulae of the compounds formed from the following pairs of ions. (i) Ca^2+ and Br (ii) Al^3+ and CO 3 (iii) K^+ and SO 4 (iv) NH 4 + and Cl

Accepted Answer

(i) CaBr 2 (ii) Al 2 ( CO 3 )3 (iii) K 2 SO 4 (iv) NH 4 Cl. - (i) Ca^2+ and Br^- → CaBr 2 — criss-cross 2 and 1; (2+) + 2 × (1-) = 0. - (ii) Al^3+ and CO 3^2- → Al 2(CO 3) 3 — criss-cross 3 and 2; brackets because three carbonate ions are present. - (iii) K^+ and SO 4^2- → K 2SO 4 — criss-cross 1 and 2; only one sulfate ion, so no brackets. - (iv) NH 4^+ and Cl^- → NH 4Cl — both valency 1, so one of each and no brackets.

Question 12

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.

Accepted Answer

- Yes — the results do justify the Law of Constant Proportions. - Student X obtained copper: oxygen = 4: 1. - Student Y obtained 8: 2, which reduces to 8/2: 2/2 = 4: 1. - The two ratios are identical, so both students made the same oxide of copper — Y simply used twice as much of each element. - That is exactly what the law says: however it is prepared, a compound contains its elements in the same fixed ratio by mass.

Question 13

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. Exploration Grade 9

Accepted Answer

(iii) -; one; two. - (iii) A is true, but R is false. - A is true. 2 + 16 = 18 g, and 2: 16 = 1: 8, the fixed hydrogen: oxygen mass ratio of water given in the chapter. - R is false as worded. Dalton's postulate is that atoms combine in the ratio of simple whole numbers — that is a ratio of the numbers of atoms, not a ratio by mass. - The two ratios are different quantities. In water the atoms combine as H: O = 2: 1 (the formula H 2O means two hydrogen atoms to one oxygen atom), while the masses combine as 1: 8. Dalton's whole-number rule speaks about the first, and R attaches it to the second. - Nor is a mass ratio obliged to be whole numbers at all — sodium chloride combines in the mass ratio 23: 35.5. - So R states the postulate wrongly and cannot explain A.

Question 14

Which of the following, in Fig. 9.18, correctly represents Cl number of chlorine = 17). (i)

Accepted Answer

(ii). - (ii) is the correct representation of the Cl^- ion. - Chlorine has atomic number 17, so a neutral chlorine atom has 17 protons and 17 electrons, filling as 2, 8, 7. - A Cl^- ion is chlorine that has gained one electron, so it still has 17 protons but now 18 electrons: 2 + 8 + 8 = 18, i.e. the shells fill as 2, 8, 8. - Counting the electrons drawn on each of the four diagrams in Fig. 9.18: - (i) 2, 7, 8 — seventeen electrons in total, which is the right number for a neutral atom, but the shells are filled out of order: the second shell holds up to eight and must be completed before the third begins. - (ii) 2, 8, 8 — eighteen electrons, one more than the seventeen protons, which is exactly a chloride ion. This is the answer. - (iii) 2, 8, 9 — nineteen electrons, two more than the protons; the outermost shell also exceeds the eight that this shell can hold. - (iv) 2, 8, 7 — the neutral chlorine atom, not the ion, and the commonest wrong choice here. - The distinction being tested is that gaining an electron changes the electron count and the outermost shell, never the number of protons — the nucleus is untouched, which is why the ion is still chlorine.

Question 15

Nitrogen has five valence electrons. Draw the structure of the nitrogen molecule ( N 2 ).

Accepted Answer

- Two nitrogen atoms share three pairs of electrons, giving a triple bond: N ≡ N. - Nitrogen has 5 valence electrons and needs 3 more for an octet; neither atom can take 3 from the other, so each contributes 3 electrons to be shared. - After sharing, each nitrogen counts 5 + 3 = 8 valence electrons — a complete octet, so N 2 is stable. - The drawing must contain: - two atoms labelled N, side by side, each starting with its 5 valence electrons marked round it; - dots for one atom's electrons and crosses for the other's, so the reader can see which electron came from where; - three shared pairs (6 electrons) drawn in the overlap between the two symbols; - one lone pair left on each nitrogen, drawn on its outer side; - the line formula N≡N written alongside, the three lines standing for the three shared pairs.

Question 16

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 hydrogencarbonate ( NaHCO 3 ), used to relieve acidity and helps in digestion.

Accepted Answer

(i) 80 u (ii) 98 u (iii) 84 u. - (i) Ammonium nitrate, NH 4NO 3 = 80 u - 2 N, 4 H, 3 O: (14 × 2) + (1 × 4) + (16 × 3) = 28 + 4 + 48 = 80 u. - (ii) Phosphoric acid, H 3PO 4 = 98 u - 3 H, 1 P, 4 O: (1 × 3) + (31 × 1) + (16 × 4) = 3 + 31 + 64 = 98 u. - (iii) Sodium hydrogencarbonate, NaHCO 3 = 84 u - 1 Na, 1 H, 1 C, 3 O: 23 + 1 + 12 + (16 × 3) = 23 + 1 + 12 + 48 = 84 u. - Atomic masses used: H = 1 u, C = 12 u, N = 14 u, O = 16 u, Na = 23 u, P = 31 u.

Question 17

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

Accepted Answer

(i) Mg 3 N 2 (ii) Li 3 N (iii) Na 2 S (iv) Al 2 O 3. - (i) Magnesium and nitrogen → Mg 3N 2 — Mg^2+ with N^3-; criss-cross 2 and 3, and 3 × (2+) + 2 × (3-) = 0. - (ii) Lithium and nitrogen → Li 3N — Li^+ with N^3-; three lithium ions balance one nitride ion. - (iii) Sodium and sulfur → Na 2S — Na^+ with S^2-; two sodium ions balance one sulfide ion. - (iv) Aluminium and oxygen → Al 2O 3 — Al^3+ with O^2-; criss-cross 3 and 2. - Nitrogen is not in Table 9.1: it has 5 valence electrons and, having more than 4, gains 3 to complete its octet, giving the nitride ion N^3-.

Question 18

The atomic number of fluorine is 9. Explain the formation of the fluorine molecule ( F 2 ). Just as atoms share electrons to form covalent bonds, we too, can share and care to build strong relationships with people around us. This sharing brings unity and stability, laying the foundation for a stronger community, and ultimately, a strong nation. 170 Exploration Grade 9

Accepted Answer

- Two fluorine atoms share one pair of electrons, forming a single covalent bond: F—F. - Fluorine ( Z = 9 ) has the configuration 2, 7 — seven valence electrons, one short of an octet. - Both atoms need to gain an electron, so neither will donate one; the only way for both to become stable is to share. - Each atom puts in one electron, and the shared pair is counted by both atoms, so each fluorine now has 8 valence electrons. - The shared pair attracts both nuclei and holds the atoms together; each atom also keeps three lone pairs. - The molecule is written F 2, or F—F, one line for the one shared pair.

Question 19

Show the formation of the following molecules: (i) Carbon dioxide ( CO 2 ) (ii) Hydrogen sulfide ( H 2 S) (iii) Ammonia ( NH 3 )

Accepted Answer

- (i) CO 2 — carbon (2, 4) has 4 valence electrons and needs 4 more; each oxygen (2, 6) needs 2. Carbon shares two electrons with each oxygen, making two double bonds: O=C=O. - (ii) H 2S — sulfur (2, 8, 6) needs 2 electrons and each hydrogen needs 1, so two hydrogen atoms each share one electron with sulfur: two single bonds, H—S—H. - (iii) NH 3 — nitrogen (2, 5) needs 3 electrons, so three hydrogen atoms each share one with it: three single bonds, with one lone pair left on nitrogen. - Each diagram must show: - CO 2: C in the middle, an O on each side; 4 electrons (2 shared pairs) between C and each O; 2 lone pairs left on each O; C ends with 8 electrons and each O with 8; the line form O=C=O beside it. - H 2S: S in the middle with an H on each side; 2 electrons (1 shared pair) between S and each H; 2 lone pairs left on S; each H reaches its duplet of 2 and S its octet of 8; the line form H—S—H. - NH 3: N in the middle with three H atoms round it; 1 shared pair between N and each H; 1 lone pair left on N; N reaches 8 and each H reaches 2. - In all three, use dots for one element's electrons and crosses for the other's, and label every atom with its symbol.

Question 20

Complete the Table 9.3 by writing the formulae of the compounds formed by the cations on the left and the anions at the top. LiNO 3 is given as an example. NH 4 + Li^+ Al^3+ Cu^2+

Accepted Answer

- NO 3 SO 4 PO 4 2− 3− + NH 4 NH 4 NO 3 ( NH 4 ) 2SO 4 ( NH 4 ) 3PO 4 Li^+ LiNO 3 Li 2 SO 4 Li 3 PO 4 Al^3+ Al( NO 3 )3 Al 2 ( SO 4 )3 AlPO 4 Cu^2+ Cu( NO 3 )2 CuSO 4 Cu 3 ( PO 4 )2 272 Textbook of Science for Grade 9. - The completed table: Cation NO 3^- SO 4^2- PO 4^3- --- --- --- --- NH 4^+ NH 4NO 3 (NH 4) 2SO 4 (NH 4) 3PO 4 Li^+ LiNO 3 (given) Li 2SO 4 Li 3PO 4 Al^3+ Al(NO 3) 3 Al 2(SO 4) 3 AlPO 4 Cu^2+ Cu(NO 3) 2 CuSO 4 Cu 3(PO 4) 2 - Method for every cell: criss-cross the charge numbers, then divide the subscripts by a common factor if there is one. - Al^3+ with PO 4^3- gives Al 3(PO 4) 3, which reduces to AlPO 4; Cu^2+ with SO 4^2- gives Cu 2(SO 4) 2, which reduces to CuSO 4. - Brackets are written only where two or more of the same polyatomic ion appear — so Al(NO 3) 3 and (NH 4) 2SO 4 take them, while NH 4NO 3 and AlPO 4 do not.

Question 21

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

Accepted Answer

- Because neon's valence shell is already complete — its configuration is 2, 8, a full octet. - Atoms lose, gain or share electrons only in order to complete their valence shell and become stable; neon has nothing left to complete. - Its energy is already low, so combining with another atom would not lower it further and would bring no gain in stability. - Neon therefore stays as single atoms, forms no molecules and is chemically unreactive.

Question 22

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.

Accepted Answer

- The Law of Conservation of Mass is valid — the total mass is 11.3 g both before and after. - Total mass of reactants = 5.3 + 6.0 = 11.3 g. - Total mass of products = 2.2 + 0.9 + 8.2 = 11.3 g. - Mass of reactants = mass of products, so no matter was created or destroyed — the atoms have only rearranged. - Note that carbon dioxide gas is released here, so the reaction must be carried out in a closed container; in an open one the escaping gas would make the final reading 2.2 g short, exactly as in Activity 9.2, set-up 1.

Question 23

What kind of ion will oxygen (O) form?

Accepted Answer

Anion ( O^2- ). - Oxygen forms an anion — the oxide ion, O^2-. - Oxygen ( Z = 8 ) has the configuration 2, 6: six valence electrons, two short of an octet. - An atom with more than 4 valence electrons gains electrons rather than losing them, so oxygen accepts 2. - It then has 8 protons but 10 electrons, a net charge of 2−, and the stable arrangement 2, 8.

Question 24

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.

Accepted Answer

(i) Z = 11; A = 23 (ii) Cation (iii) Electronic configuration 2, 8 (iv) Sodium cation ( Na^+ ). - (i) Atomic number Z = 11 (the number of protons); mass number A = 11 + 12 = 23. - (ii) A cation. There are 11 positive protons against only 10 electrons, so one negative charge is missing and the species carries a net charge of 1+ — it has lost one electron. - (iii) Electronic configuration 2, 8 — the 10 electrons fill the K shell (2) and the L shell (8), which is why losing that electron made it stable. - (iv) The sodium cation, Na^+: Z = 11 identifies the element as sodium, and the 1+ charge makes it the cation.

Question 25

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.

Accepted Answer

Cl. - Blank 1 = Cl^- (the chloride ion); blank 2 = one; blank 3 = two. - Chlorine is 2, 8, 7 — one electron short of an octet — so it accepts exactly one electron and becomes Cl^-. - Magnesium gives away 2 electrons, but each chlorine atom can take only 1, so two chlorine atoms are needed for every magnesium atom. - The charges must cancel: (2+) + 2 × (1-) = 0. - One Mg^2+ ion and two Cl^- ions therefore combine to give magnesium chloride, MgCl 2.

Question 26

Two elements, A and B, have the following configurations — A: 2, 8, 5 (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.

Accepted Answer

(i) Element B (ii) Covalent (iii) AB 3. - (i) Element B is the more reactive. It needs only one electron to complete its octet (7 valence electrons), while A needs three (5 valence electrons) — the smaller the gap, the more readily the atom reacts. - (ii) A covalent bond, formed by sharing. Both A and B have more than 4 valence electrons, so both need to gain and neither will donate; the only way both can reach an octet is to share electron pairs. - (iii) AB 3. A must gain a share in 3 electrons and each B can offer a share in only 1, so one A atom shares one electron pair with each of three B atoms. - After bonding, A counts 5 + 3 = 8 electrons and each B counts 7 + 1 = 8 — every atom has its octet.

Question 27

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

Accepted Answer

- Potassium forms K^+ and calcium forms Ca^2+; their chlorides are KCl and CaCl 2. - Potassium ( Z = 19 ) is 2, 8, 8, 1. Losing its single valence electron leaves 19 protons and 18 electrons → K^+, with the stable arrangement 2, 8, 8. - Calcium ( Z = 20 ) is 2, 8, 8, 2. Losing both valence electrons leaves 20 protons and 18 electrons → Ca^2+, also 2, 8, 8. - KCl: the one electron potassium loses is taken by one chlorine atom (2, 8, 7 → 2, 8, 8); K^+ and Cl^- are then held by electrostatic attraction → KCl. - CaCl 2: calcium gives one electron to each of two chlorine atoms, since each chlorine can take only one → Ca^2+ and two Cl^- → CaCl 2. - The four diagrams must show: - K → K^+: concentric shells holding 2, 8, 8, 1 electrons round a nucleus labelled 19 p; a curved arrow out of the outermost shell labelled e^-; the product drawn as shells 2, 8, 8 inside square brackets with + outside, labelled "Potassium cation K^+ ". - Ca → Ca^2+: shells 2, 8, 8, 2 round a nucleus labelled 20 p; two arrows labelled 2e^-; the product as 2, 8, 8 in square brackets with 2+ outside, labelled "Calcium cation Ca^2+ ". - KCl: K (2, 8, 8, 1) beside Cl (2, 8, 7), one arrow from K's valence shell to Cl's valence shell; the products [2,8,8]^+ and [2,8,8]^- labelled K^+ and Cl^-; the gap between them labelled ionic bond; the formula KCl written below. - CaCl 2: Ca in the middle with a Cl on either side; one arrow from Ca to each Cl; the products Ca^2+ and two Cl^-, each in square brackets with its charge; the formula CaCl 2 written below.

Question 28

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.

Accepted Answer

(iii). - (iii) A is true, but R is false. - A is true. Copper sulfate is ionic; it does not conduct as a solid but does conduct once molten. - R is false because it has the two states the wrong way round. The ions are held in fixed positions in the solid lattice, and they become free to move in the molten state. - The chapter's rule is that ions must be free to move for a compound to conduct — which is why solid CuSO 4 does not conduct, while its melt (and its solution in water) does. - Since R is false, it cannot explain A, and options (i), (ii) and (iv) are all ruled out.

Question 29

The species 27Al, 80Br How many electrons and neutrons do they have? y Design and perform an experiment to show and compare that water always contains hydrogen and oxygen in the same ratio, regardless of its source. y Compare atoms and ions of any three elements. Show the number of electrons before and after ion formation using bar graphs. y Make a card game with cations and anions. Possible ideas may include —  Pick cards from the pile or from open cards and match them with cards in your hand to form compounds. You may discard any unwanted cards.  Ask other players for cards and form compounds using the cards they already have. y To learn more about molecules you can explore the link given below —  https://phet.colorado.edu/sims/html/build-a-molecule/latest/build-a-molecule all.html The Quest Continues … Are there any chemical changes that do not obey the Law of Conservation of Mass?

Accepted Answer

Electrons 13, Neutrons 14; Electrons 36, Neutrons 45; Electrons 78, Neutrons 121. - ^27Al: 13 electrons and 14 neutrons. - It is a neutral atom, so electrons = protons = 13; neutrons = 27 - 13 = 14. - ^80Br^-: 36 electrons and 45 neutrons. - The 1− charge means one extra electron: 35 + 1 = 36; neutrons = 80 - 35 = 45. - ^201Hg^2+: 78 electrons and 121 neutrons. - The 2+ charge means two electrons lost: 80 - 2 = 78; neutrons = 201 - 80 = 121. - In every case neutrons = mass number − number of protons, and the charge changes only the electron count — the nucleus is untouched when an ion forms.

Question 30

Illustrate how sodium sulfide ( Na 2 S) is formed. Name of ion Sodium Lithium Potassium Silver Calcium Barium Iron (Ferrous) Iron (Ferric) Copper (Cuprous) Copper (Cupric) Magnesium Zinc Aluminium Fluoride Chloride Bromide Iodide Oxide Sulfide

Accepted Answer

- Na 2S is formed when each of two sodium atoms transfers one electron to a single sulfur atom. - Sodium is 2, 8, 1 — one valence electron to give away; sulfur is 2, 8, 6 — two electrons short of an octet. - Each Na loses 1 electron → Na^+ (2, 8); sulfur accepts both → S^2- (2, 8, 8). - Two sodium ions are needed for one sulfide ion so that the charges cancel: 2 × (1+) + (2-) = 0. - The oppositely charged ions are held together by the electrostatic force of attraction — an ionic bond — giving Na 2S, sodium sulfide. - The illustration must show: two sodium atoms (shells 2, 8, 1) on the left and one sulfur atom (2, 8, 6) on the right; one curved arrow labelled e^- from each sodium's valence shell to the sulfur's valence shell; the products drawn as [2,8]^+ twice, labelled Na^+, and [2,8,8]^2-, labelled S^2-; the label ionic bond between them; and the formula Na 2S written underneath.

Question 31

Name the following: (i) CO 2 (ii) NO 2 (iii) SF 6 (iv) PCl3

Accepted Answer

(i) Carbon dioxide (ii) Nitrogen dioxide (iii) Sulfur hexafluoride (iv) Phosphorous trichloride. - (i) CO 2 — carbon dioxide - (ii) NO 2 — nitrogen dioxide - (iii) SF 6 — sulfur hexafluoride - (iv) PCl 3 — phosphorus trichloride - Rule used: the first element keeps its ordinary name, the second element ends in -ide, and the prefixes mono- (1), di- (2), tri- (3), tetra- (4), penta- (5), hexa- (6) count the atoms. - mono- is dropped before the first element, which is why CO 2 is carbon dioxide and not "monocarbon dioxide", and why NO 2 is nitrogen dioxide.

Question 32

Write the formula for the following: (i) Sodium hydrogencarbonate (ii) Sulfur dioxide (iii) Ferric chloride (iv) Cuprous oxide

Accepted Answer

(i) NaHCO 3 (ii) SO 2 (iii) FeCl 3 (iv) Cu 2 O. - (i) Sodium hydrogencarbonate — NaHCO 3: Na^+ with HCO 3^-, both of valency 1, so one of each. - (ii) Sulfur dioxide — SO 2: the prefix di- fixes two oxygen atoms to one sulfur. - (iii) Ferric chloride — FeCl 3: ferric is Fe^3+ and chloride is Cl^-; criss-crossing 3 and 1 gives one Fe to three Cl. - (iv) Cuprous oxide — Cu 2O: cuprous is Cu^+ and oxide is O^2-; criss-crossing 1 and 2 gives two Cu to one O.

Question 33

Write the formulae for the compounds formed from the following pairs of ions: (i) Fe^3+ and OH^‒ 9.6 Properties of the Ionic and the Covalent Compounds Activity 9.4: Let us experiment 1. Collect the samples of some compounds, such as camphor, sodium chloride, copper sulfate, sugar and naphthalene. (A) Solubility in (i) water, (ii) kerosene, and (iii) petrol 2. Try dissolving each sample separately in the water, kerosene and petrol. 3. Record your observations in Table 9.2. (B) Electrical conductivity in the water Safety first: Do not touch the electrodes when they are connected to the battery but use a low-voltage battery to avoid the risk of shock. Petrol and kerosene are flammable liquids, so be careful while working with them.

Accepted Answer

(i) Fe(OH)3 (ii) K 2 CO 3. - (i) Fe^3+ and OH^- → Fe(OH) 3 - Criss-cross the charge numbers 3 and 1: one Fe^3+ to three OH^-, and 3+ + 3 × (1-) = 0. - Brackets are needed because three of the same polyatomic ion are present — it is Fe(OH) 3, not FeOH 3. - (ii) K^+ and CO 3^2- → K 2CO 3 - Criss-cross 1 and 2: two K^+ to one CO 3^2-, and 2 × (1+) + (2-) = 0. - Only one carbonate ion is present, so no brackets are used.

Question 34

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?

Accepted Answer

- An ionic bond. - In the solid state the ions are locked in fixed positions in the crystal lattice by strong forces, so no charge can move and the compound cannot conduct. - On dissolving in water the ions become free to move, and moving ions carry the current — so the solution conducts. - A covalent compound would fail one test or the other: sugar dissolves but releases no ions, and camphor and naphthalene do not conduct at all.

Question 35

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. You have learnt in section 9.4.2 that in ionic compounds, the ions form

Accepted Answer

(i) MO (ii) Ionic (iii) Conducts electricity in aqueous solution. - (i) Formula: MO - (ii) Type of bond: ionic - (iii) Its aqueous solution conducts electricity - M's valence shell is the M shell holding 2 electrons, so M is 2, 8, 2 — a metal with fewer than 4 valence electrons, which donates them → M^2+. - Oxygen has 6 valence electrons and gains 2 → O^2-. - Criss-crossing the charges gives M 2O 2, and dividing both subscripts by the common factor 2 gives MO. - Metal + non-metal joined by the transfer of electrons is an ionic bond. - The compound is only slightly soluble, but whatever does dissolve releases free M^2+ and O^2- ions into the water, and free ions conduct — so the bulb would glow, if weakly.

Question 36

Find the molecular mass of nitric acid ( HNO 3 ). Atomic mass — H = 1 u; N = 14 u; O = 16 u.

Accepted Answer

63 u. - Molecular mass of HNO 3 = 63 u. - Add the atomic mass of every atom in the formula: 1 H, 1 N and 3 O. - (1 × 1) + (14 × 1) + (16 × 3) = 1 + 14 + 48 = 63 u.

Question 37

Find the molecular mass of methane ( CH 4 ). Atomic mass — C = 12 u; H = 1 u.

Accepted Answer

16 u. - Molecular mass of CH 4 = 16 u. - The formula holds 1 carbon atom and 4 hydrogen atoms. - (12 × 1) + (1 × 4) = 12 + 4 = 16 u.

Question 38

Find the formula unit mass of potassium chloride (KCl). Atomic mass — K = 39 u; Cl = 35.5 u.

Accepted Answer

74.5 u. - Formula unit mass of KCl = 74.5 u. - (39 × 1) + (35.5 × 1) = 74.5 u. - It is called a formula unit mass, not a molecular mass, because KCl is ionic — its ions form a 3-D crystal, so it does not exist as molecules.

Question 39

Find the formula unit mass of magnesium hydroxide, Mg(OH)2. Atomic mass — Mg = 24 u; O = 16 u; H = 1 u. Understanding atoms, molecules and chemical bonding reveals the At a Glance y Mass can neither be created nor destroyed in a chemical reaction. This is known as the Law of Conservation of Mass. y A compound always contains the same elements combined in a fixed ratio by mass, no matter how it is formed or from where it is obtained. This is called the Law of Definite Proportions. y A molecule is defined as an electrically neutral entity consisting of more than one atom that can exist independently and shows all its chemical properties. y Atoms combine to form molecules of elements or compounds to become stable. Atoms are held together by a force called a chemical bond. y A covalent bond is formed by the sharing of electrons between atoms. y An ionic bond is formed by the transfer of electrons between atoms, where one atom loses electrons and the other gains electrons to form cations and anions, respectively. y The chemical formula of a covalent compound represents the elements and number of atoms of each element present in it. y The chemical formula of an ionic compound represents the simplest whole number ratio of atoms of different elements present in it. y Molecular mass is the total mass of a molecule, calculated by adding the atomic masses of all the atoms constituting it. y Formula unit mass of an ionic compound is the sum of the atomic masses of all the atoms present in a formula unit (simplest whole number ratio of ions in an ionic compound).

Accepted Answer

58 u. - Formula unit mass of Mg(OH) 2 = 58 u. - The subscript 2 outside the bracket multiplies both the O and the H inside it, so the unit contains 1 Mg, 2 O and 2 H. - (24 × 1) + (16 × 1) + (1 × 1) × 2 = 24 + (17 × 2) = 24 + 34 = 58 u.

Cation	\(\displaystyle NO_3^-\)	\(\displaystyle SO_4^{2-}\)	\(\displaystyle PO_4^{3-}\)
\(\displaystyle NH_4^+\)	\(\displaystyle NH_4NO_3\)	\(\displaystyle (NH_4)_2SO_4\)	\(\displaystyle (NH_4)_3PO_4\)
\(\displaystyle Li^+\)	\(\displaystyle LiNO_3\) (given)	\(\displaystyle Li_2SO_4\)	\(\displaystyle Li_3PO_4\)
\(\displaystyle Al^{3+}\)	\(\displaystyle Al(NO_3)_3\)	\(\displaystyle Al_2(SO_4)_3\)	\(\displaystyle AlPO_4\)
\(\displaystyle Cu^{2+}\)	\(\displaystyle Cu(NO_3)_2\)	\(\displaystyle CuSO_4\)	\(\displaystyle Cu_3(PO_4)_2\)

Revise, Reflect, Refine 9.1 (part 1 of 22)

Exercise 9.1