NCERT Solutions for Class 9th Science Chapter 9 End-of-chapter questions — Revise, Reflect, Refine

First fix the configurations. A has one electron in the third shell, so the first two shells must be full: A = 2, 8, 1 (Z = 11, sodium). B has six electrons in the second shell: B = 2, 6 (Z = 8, oxygen).

(i) A gives away 1 electron. Removing its single M-shell electron leaves 2, 8 — a complete octet. Taking seven electrons instead would be far harder.

(ii) A forms a monovalent cation, A⁺.

(iii) B takes 2 electrons. With six valence electrons it is two short of an octet; gaining two is much easier than losing six.

(iv) B forms a divalent anion, B²⁻ (configuration 2, 8).

(v) An ionic bond. One element hands electrons over and the other accepts them, so this is a transfer, not a sharing. The oppositely charged ions that result are then held together by electrostatic attraction.

(vi) A₂B.

X has six electrons in its outermost shell, so it is two electrons short of an octet. Taking the smallest such element, X = oxygen (2, 6).

(i) Why a diatomic molecule? Because X cannot complete its octet alone, and the atom nearest to hand is another X atom, which is short by exactly the same amount. Neither can take two electrons from the other, since that would leave the donor four short. The only arrangement that satisfies both is for each atom to contribute two electrons to a shared region — and once both octets are complete there is no reason to add a third atom. So the molecule stops at two, X₂.

(ii) A covalent bond — specifically a double bond, since two pairs of electrons are shared.

(iii) Structure of X₂

(iv) The compound of X with Y. Y has two electrons in its second shell, so its configuration is 2, 2 (Z = 4, beryllium). Y has 2 valence electrons to dispose of and X needs exactly 2, so they combine in a 1 : 1 ratio.

Option (iii) — 2 Fe³⁺ and 3 O²⁻.

Only (iii) has both totals equal to 6, so only (iii) gives a neutral compound. The formula is Fe₂O₃ — ferric oxide, the red-brown compound in rust and in haematite ore.

Only statement (iv) is correct. The other three are false; corrected versions are given below.

(i) False. The two halves have been swapped and the words used loosely.

Corrected: Elements are made up of atoms of only one kind — those atoms may exist singly (He, Ne) or joined as molecules of the element (H₂, O₂, N₂, Cl₂). Compounds are made up of atoms of two or more different elements combined in a fixed ratio, forming either molecules (H₂O, CO₂) or formula units (NaCl, CaCO₃).

(ii) False. Atoms of the same kind give a molecule of an element, not of a compound.

Corrected: The molecule of a compound is always made up of two or more atoms of different kinds. For example, HCl has one hydrogen and one chlorine atom; H₂SO₄ has three different elements.

(iii) False. Nitrogen has five valence electrons (2, 5), needs three more, and gets them by sharing three pairs with one other nitrogen atom.

Corrected: One molecule of nitrogen gas contains two nitrogen atoms, joined by a triple bond — N ≡ N, formula N₂.

(iv) True.

Write the cation first, put the charge numbers beneath, criss-cross them as subscripts, and divide by any common factor.

Note the brackets in Al(NO₃)₃: there are three nitrate ions, and the bracket keeps each NO₃ group intact. Without it, AlNO₃₃ would be meaningless.

When are brackets used? Only when two or more identical polyatomic ions are present. So Al₂(CO₃)₃ needs them (three carbonate ions), K₂SO₄ does not (one sulfate ion), and NH₄Cl does not (one ammonium ion).

Option (ii).

A chloride ion is a chlorine atom that has gained one electron, so the diagram must show:

Count the dots on each of the four diagrams printed in the book:

Atomic masses used (from Chapter 8): H = 1 u, C = 12 u, N = 14 u, O = 16 u, Na = 23 u, P = 31 u.

(i) Ammonium nitrate, NH₄NO₃ → 80 u

(ii) Phosphoric acid, H₃PO₄ → 98 u

(iii) Sodium hydrogencarbonate, NaHCO₃ → 84 u

Work out the ion each element forms, then balance the charges.

Names: (i) magnesium nitride, (ii) lithium nitride, (iii) sodium sulfide, (iv) aluminium oxide.

Criss-cross the charge numbers in every cell, then simplify. Brackets go round a polyatomic ion whenever its subscript is 2 or more.

Two cells deserve a second look:

• AlPO₄ — criss-crossing 3 and 3 gives Al₃(PO₄)₃, which must be divided by the common factor 3 to reach the simplest ratio 1 : 1.
• CuSO₄ — criss-crossing 2 and 2 gives Cu₂(SO₄)₂, which divides by 2 to give CuSO₄.

The law is obeyed — both sides come to 11.3 g.

Hence the Law of Conservation of Mass is valid for this reaction.

(i) Atomic number Z = 11, mass number A = 23.

Note that electrons play no part in either — their mass is negligible, and it is the proton count that fixes the identity of the element.

(ii) It is a cation, carrying a charge of +1.

There is one electron fewer than the number of protons, so one positive charge is left unbalanced. An atom that has lost electrons is a cation.

(iii) Electronic configuration: 2, 8. The species has only 10 electrons, so the K shell takes 2 and the L shell the remaining 8 — the same arrangement as neon. (The neutral atom would have been 2, 8, 1.)

(iv) It is the sodium ion, Na⁺ — more precisely ²³₁₁Na⁺.

A is 2, 8, 5 — five valence electrons, three short of an octet (Z = 15, phosphorus). B is 2, 8, 7 — seven valence electrons, one short (Z = 17, chlorine).

(i) B is more reactive.

B is only one electron away from stability, so it reacts readily and with almost anything. A must gather three electrons, which is a much harder demand, so it reacts less eagerly.

(ii) They form covalent bonds. Both A and B have more than four valence electrons, so neither will hand electrons over — giving away five or seven electrons would need an enormous amount of energy. And neither can simply take electrons from the other, because both are trying to gain. The only workable arrangement is sharing.

So A shares one electron with each of three B atoms. Each shared pair completes one gap: A gains three shared pairs and reaches eight, while each B gains one shared pair and reaches eight.

(iii) Formula: AB₃ (in real terms PCl₃, phosphorus trichloride).

Option (iii) — A is true, but R is false.

A is true. Copper sulfate is an ionic compound. In the solid its Cu²⁺ and SO₄²⁻ ions are locked in fixed lattice positions, so no charge can move and it does not conduct. Melt it and the lattice breaks down; the ions become mobile and the melt conducts.

R is false — it has the two states exactly the wrong way round.

Since R is a false statement, options (i) and (ii) are ruled out; and since A is true, (iv) is ruled out. The answer is (iii).

Use two rules: neutrons = mass number − protons, and electrons = protons − charge (a positive charge means electrons are missing, a negative charge means extra electrons are present).