NCERT Solutions Class 9 Science Chapter 9 — Atomic Foundations of Matter

NCERT Class 9 Science — Atomic Foundations of Matter, Chapter 9 Revise, Reflect, Refine. All 15 questions solved.

Also here. Solutions to the in-text questions for this chapter and the activities for this chapter are on their own pages.

The exercise divides into three. Questions 1 to 4, 13 and 14 are about why atoms combine — which electrons are given, taken or shared, and whether the bond that results is ionic or covalent. Questions 5, 6, 9 and 10 are writing formulae from ions. Questions 7, 8, 11, 12 and 15 are counting — electrons, neutrons, formula unit mass and mass in a reaction. Four things carry all of it:

• Atoms combine to complete their outermost shell. An atom with few valence electrons loses them and becomes a cation; one with nearly eight gains electrons and becomes an anion.
• Metal + non-metal gives an ionic bond (electrons transferred); non-metal + non-metal gives a covalent bond (electrons shared).
• A formula balances the charges. Take as many of each ion as makes the total positive charge equal the total negative charge.
• Mass is conserved in every chemical reaction, and formula unit mass is just the sum of the atomic masses in the formula.

Key insight. Write the electronic configuration before answering anything. “One electron in the third shell” is 2, 8, 1 — and the moment that is on paper, you know the element loses one electron, forms a singly-charged cation, and combines with an oxide ion in a two-to-one ratio. Questions 1, 12 and 13 are all the same move, and the configuration is what makes each of them a single step.

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.

Solution. Write both configurations out first; everything follows from them.

A has one electron in the third shell, so its shells fill 2, 8, then 1 — that is 2, 8, 1, which is 11 electrons, the element sodium.

B has six electrons in its second shell, so 2, 6 — 8 electrons, the element oxygen.

(i) A has just one electron in its outermost shell. Losing it leaves the full shell of 8 beneath as the outermost, which is a stable arrangement — far easier than gaining seven. So A tends to give 1 electron.

(ii) Losing an electron leaves A with 11 protons and 10 electrons, a net charge of $+1$. It forms a cation, $\text{A}^+$.

(iii) B has six electrons in its outermost shell and needs two more for an octet. Gaining two is far easier than losing six, so B tends to take 2 electrons.

(iv) Gaining two electrons gives B a net charge of $-2$: an anion, $\text{B}^{2-}$.

(v) One element gives electrons and the other takes them, so the electrons are transferred, not shared — the bond is ionic.

(vi) Balance the charges. Each A gives one electron and each B needs two, so two A atoms are needed for every one B:

$$2 \times (+1) + 1 \times (-2) = 0 \quad\Longrightarrow\quad \mathbf{A_2B}$$

(With A as sodium and B as oxygen, that compound is $\text{Na}_2\text{O}$.)

Question 2

An element X has six electrons in its outer shell and forms a diatomic molecule.

Solution. Six electrons in the outermost shell means the configuration 2, 6 — eight electrons in all, so X is oxygen.

(i) Why is it diatomic? X needs two more electrons to complete its octet, and there is no other kind of atom available — only more atoms of X. Two atoms of X solve the problem between them: each contributes two electrons to be shared, so two pairs are shared between them, and each atom then counts eight. Neither atom can be satisfied alone, and a third would have nothing left to share, so the molecule is exactly two atoms, $\text{X}_2$.

(ii) What kind of bond? Both atoms are non-metals with the same attraction for electrons, so neither can take electrons from the other. They share, so the bond is covalent — and since two pairs are shared, it is a double bond.

(iii) The structure of $\text{X}_2$. Each oxygen brings six valence electrons. Two pairs are shared between them, and each atom keeps two lone pairs:

$$\text{O} = \text{O}$$

(iv) The molecule X forms with Y. Y has two electrons in its second shell, so its configuration is 2, 2 — four electrons, the element beryllium. Y has two electrons to lose and X needs two, so they combine in a one-to-one ratio, giving XY (beryllium oxide, BeO): Y transfers both its valence electrons to X, leaving $\text{Y}^{2+}$ and $\text{X}^{2-}$.

Worth noting: although this is written as a transfer, the beryllium–oxygen bond has a good deal of covalent character, because the small, doubly-charged beryllium ion holds the shared electrons strongly. At this level the transfer picture is the one to draw.

Question 3

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?

Solution. Multiply the number of each ion by its charge and check both totals against 6.

(i) $2 \times (+3) = +6$ ✓, but $3 \times (-1) = -3$ ✗.

(ii) $3 \times (+2) = +6$ ✓, but $1 \times (-3) = -3$ ✗.

(iii) $2 \times (+3) = +6$ ✓ and $3 \times (-2) = -6$ ✓ — both correct.

(iv) $3 \times (+2) = +6$ ✓, but $2 \times (-2) = -4$ ✗.

Only (iii) gives 6+ and 6−, and it is also the only one that is electrically neutral overall — which is what a real compound must be. It is iron(III) oxide, $\text{Fe}_2\text{O}_3$.

Question 4

Choose the correct statement(s) and correct the false statement(s).

Solution.

(i) “Elements are made up of molecules and compounds are made up of atoms.” — FALSE, and it has the two the wrong way round.

Corrected: Elements are made up of atoms of only one kind; those atoms may exist singly or joined as molecules of the element ($\text{O}_2$, $\text{N}_2$). Compounds are made up of atoms of two or more different elements, combined as molecules or as ions.

(ii) “The molecule of a compound is always made up of two or more atoms of the same kind.” — FALSE.

Corrected: A molecule of a compound is made of two or more atoms of different kinds — water is two hydrogen and one oxygen. Atoms of the same kind joined together make a molecule of an element, not of a compound.

(iii) “One molecule of nitrogen gas contains three nitrogen atoms.” — FALSE.

Corrected: One molecule of nitrogen gas contains two nitrogen atoms, $\text{N}_2$. Nitrogen has five valence electrons and needs three more, which two atoms achieve by sharing three pairs — a triple bond.

(iv) “Water is made of two hydrogen atoms, covalently bonded with one oxygen atom.” — TRUE. Oxygen needs two electrons and each hydrogen one, so oxygen shares a pair with each of two hydrogen atoms: $\text{H}_2\text{O}$.

Question 5

Write the chemical formulae for the following compounds.

Solution. Find the charge on each ion, then take as many of each as balances the charges.

(i) Aluminium nitrate. $\text{Al}^{3+}$ and $\text{NO}_3^-$. One aluminium needs three nitrates; since nitrate is a group of atoms, it is bracketed before the subscript:

$$\text{Al(NO}_3)_3$$

(ii) Calcium oxide. $\text{Ca}^{2+}$ and $\text{O}^{2-}$. The charges are equal and opposite, so one of each:

$$\text{CaO}$$

(iii) Ferric oxide. “Ferric” is iron in its $+3$ state, so $\text{Fe}^{3+}$ and $\text{O}^{2-}$. Two irons give $+6$ and three oxides $-6$:

$$\text{Fe}_2\text{O}_3$$

Question 6

Write the formulae of the compounds formed from the following pairs of ions.

Solution. The same balancing, each time.

(i) $\text{Ca}^{2+}$ and $\text{Br}^-$ — one $+2$ needs two $-1$: $\text{CaBr}_2$

(ii) $\text{Al}^{3+}$ and $\text{CO}_3^{2-}$ — the lowest common multiple of 3 and 2 is 6, so two aluminium ($+6$) and three carbonate ($-6$): $\text{Al}_2(\text{CO}_3)_3$

(iii) $\text{K}^+$ and $\text{SO}_4^{2-}$ — two potassium for one sulphate: $\text{K}_2\text{SO}_4$

(iv) $\text{NH}_4^+$ and $\text{Cl}^-$ — equal and opposite, one of each: $\text{NH}_4\text{Cl}$

Note the brackets in (ii): they are needed because the subscript 3 applies to the whole carbonate group. In (iii) no bracket is needed, since only one sulphate is present.

Question 7

Which of the diagrams in Fig. 9.18 correctly represents the Cl⁻ ion? (Atomic number of chlorine = 17.)

Solution. Work out what the ion must look like before examining the diagrams.

A chlorine atom has atomic number 17, so 17 electrons, arranged 2, 8, 7. A chloride ion is what it becomes after gaining one electron to complete its octet — so it has 18 electrons, arranged

$$2,\ 8,\ 8$$

So the correct diagram must show three shells carrying 2, 8 and 8 electrons, eighteen in all.

Counting the dots shell by shell in the four diagrams, only (ii) has this arrangement: two electrons in the innermost shell, eight in the second and eight in the outermost.

Question 8

Determine the formula unit mass of the following substances.

Solution. The formula unit mass is the sum of the atomic masses of every atom in the formula, counted with its subscript. Use H = 1, C = 12, N = 14, O = 16, Na = 23, P = 31.

(i) Ammonium nitrate, $\text{NH}_4\text{NO}_3$. Count the atoms carefully — there are two nitrogens, one in the ammonium and one in the nitrate:

$$2(14) + 4(1) + 3(16) = 28 + 4 + 48 = 80\ \text{u}$$

(ii) Phosphoric acid, $\text{H}_3\text{PO}_4$:

$$3(1) + 31 + 4(16) = 3 + 31 + 64 = 98\ \text{u}$$

(iii) Sodium hydrogencarbonate, $\text{NaHCO}_3$:

$$23 + 1 + 12 + 3(16) = 23 + 1 + 12 + 48 = 84\ \text{u}$$

Question 9

Write the formulae for the compounds formed by the reaction of:

Solution. Work out the ion each element forms from its position, then balance.

(i) Magnesium and nitrogen. Magnesium is 2, 8, 2 and loses two electrons: $\text{Mg}^{2+}$. Nitrogen has five valence electrons and gains three: $\text{N}^{3-}$. Balancing $+2$ against $-3$ needs three magnesium and two nitrogen:

$$\text{Mg}_3\text{N}_2$$

(ii) Lithium and nitrogen. $\text{Li}^+$ and $\text{N}^{3-}$ — three lithium for each nitrogen:

$$\text{Li}_3\text{N}$$

(iii) Sodium and sulfur. $\text{Na}^+$ and $\text{S}^{2-}$ — two sodium for one sulfur:

$$\text{Na}_2\text{S}$$

(iv) Aluminium and oxygen. $\text{Al}^{3+}$ and $\text{O}^{2-}$ — two aluminium ($+6$) and three oxygen ($-6$):

$$\text{Al}_2\text{O}_3$$

Question 10

Complete Table 9.3 by writing the formulae of the compounds formed by the cations on the left and the anions at the top.

Solution. Twelve formulae, each by the same balancing. Bracket a group of atoms whenever more than one of it is needed.

	$\text{NO}_3^-$	$\text{SO}_4^{2-}$	$\text{PO}_4^{3-}$
$\text{NH}_4^+$	$\text{NH}_4\text{NO}_3$	$(\text{NH}_4)_2\text{SO}_4$	$(\text{NH}_4)_3\text{PO}_4$
$\text{Li}^+$	$\text{LiNO}_3$	$\text{Li}_2\text{SO}_4$	$\text{Li}_3\text{PO}_4$
$\text{Al}^{3+}$	$\text{Al(NO}_3)_3$	$\text{Al}_2(\text{SO}_4)_3$	$\text{AlPO}_4$
$\text{Cu}^{2+}$	$\text{Cu(NO}_3)_2$	$\text{CuSO}_4$	$\text{Cu}_3(\text{PO}_4)_2$

Two patterns are worth seeing. Along the $\text{Al}^{3+}$ row, the $+3$ and $-3$ of the phosphate cancel exactly, so $\text{AlPO}_4$ needs no subscripts at all. And along the $\text{Cu}^{2+}$ row, the sulphate’s $-2$ matches the copper’s $+2$, giving the familiar $\text{CuSO}_4$.

Question 11

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.

Solution. The law says that in a chemical reaction the total mass of the reactants equals the total mass of the products, since atoms are only rearranged. So add each side and compare.

Reactants:

$$5.3 + 6.0 = 11.3\ \text{g}$$

Products:

$$2.2 + 0.9 + 8.2 = 11.3\ \text{g}$$

The two totals are equal, so the law of conservation of mass is verified for this reaction.

It is worth noting why the gas has to be counted. Carbon dioxide escapes if the reaction is done in an open vessel, and the mass then appears to fall — which is exactly why such experiments are carried out in a closed flask.

Question 12

If a species has 11 protons, 12 neutrons and 10 electrons then —

Solution. (i) The atomic number is the number of protons, and the mass number is protons plus neutrons:

$$Z = 11, \qquad A = 11 + 12 = 23$$

(ii) Compare protons with electrons. There are 11 positive charges and only 10 negative, so the species carries a net charge of $+1$. A positively charged species is a cation — this one has lost an electron from the neutral atom.

(iii) Ten electrons arrange themselves as 2, 8 — the K shell full with 2 and the L shell full with 8. Note that this is the configuration of the ion, not of the atom, which would be 2, 8, 1.

(iv) The element is fixed by the atomic number, and $Z = 11$ is sodium. With a charge of $+1$, the species is the sodium ion, $\text{Na}^+$.

Question 13

Two elements, A and B, have the configurations A: 2, 8, 5 and B: 2, 8, 7.

Solution. Read off the valence electrons: A has 5, B has 7. Both are non-metals, and both need to gain electrons to reach an octet — A needs three, B needs one.

(i) Which is more reactive? B. It is only one electron short of a complete octet, so it attracts that electron very strongly and reacts readily. A, needing three, has a much harder task. (B is chlorine, $Z = 17$; A is phosphorus, $Z = 15$.)

(ii) Ionic or covalent? Both are non-metals, and both want to gain electrons — neither is willing to give any up. With no atom prepared to lose electrons, no transfer can take place, so they share instead. The bond is covalent.

(iii) The formula. A needs three electrons and each B can share only one, so one A combines with three B:

$$\mathbf{AB_3}$$

Each of the three B atoms shares a pair with A: A then counts $5 + 3 = 8$ electrons, and each B counts $7 + 1 = 8$. (The compound is $\text{PCl}_3$.)

Question 14

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.

Solution. A is true. An ionic compound conducts only when its ions can move. In the solid state they are locked in a rigid lattice and cannot, so the solid does not conduct; melting breaks that lattice apart and frees the ions, so the molten compound does.

R is false — and it states exactly the reverse. It claims the ions are fixed in the molten state and move freely in the solid, which is the opposite of the truth and would predict the opposite of the observation in A.

So the assertion is true and the reason is false: option (iii). The correct reason would read: in the solid state the ions are held fixed in the lattice, while in the molten state they are free to move.

Question 15

The species ²⁷Al, ⁸⁰Br⁻ and ²⁰¹Hg²⁺ have 13, 35 and 80 protons respectively. How many electrons and neutrons do they have?

Solution. Two rules do all three. Neutrons $= A – Z$, using the superscript as the mass number. Electrons equal the protons for a neutral atom, with one more for each negative charge and one fewer for each positive charge.

$^{27}\text{Al}$ — no charge shown, so it is a neutral atom:

$$\text{electrons} = 13, \qquad \text{neutrons} = 27 – 13 = 14$$

$^{80}\text{Br}^-$ — a charge of $-1$ means one electron more than the 35 protons:

$$\text{electrons} = 35 + 1 = 36, \qquad \text{neutrons} = 80 – 35 = 45$$

$^{201}\text{Hg}^{2+}$ — a charge of $+2$ means two electrons fewer than the 80 protons:

$$\text{electrons} = 80 – 2 = 78, \qquad \text{neutrons} = 201 – 80 = 121$$

Notice that the neutron count is unaffected by the charge in every case: gaining or losing electrons changes nothing in the nucleus.

Common mistakes

• Giving the ion’s charge the wrong sign in Questions 1 and 12. Losing a negative electron leaves a positive ion; gaining one gives a negative ion. Counting protons against electrons settles it every time.
• Writing the atom’s configuration instead of the ion’s in Question 12(iii). The species has 10 electrons, so it is 2, 8 — the sodium atom would be 2, 8, 1.
• Forgetting the second nitrogen in $\text{NH}_4\text{NO}_3$ in Question 8. There are two, one in each part of the formula, so the mass is 80 u and not 66 u.
• Leaving out the brackets in Question 6(ii) and Question 10. $\text{Al}_2(\text{CO}_3)_3$ means three carbonate groups; written $\text{Al}_2\text{CO}_{33}$ it means nothing at all.
• Choosing the diagram with 17 electrons in Question 7. The question asks for the chloride ion, which has gained an electron — 18 electrons, 2, 8, 8.
• Accepting R in Question 14 because A is true. R describes the ions as fixed when molten and mobile when solid, which is the reverse of the truth; a reason can be false even when the assertion it accompanies is correct.

Practise next

• The in-text Pause and Ponder questions of this chapter, twenty-four across ten boxes, which drill the same ion charges, formulae and molecular masses in shorter form.
• Chapter 8, Journey Inside the Atom, whose electronic configurations are what every answer on this page is built on.