NCERT Solutions for Class 9th Science Chapter 8 End-of-chapter exercise — Revise, Reflect, Refine

Book page 158 – 160 Updated on2026-09-08

Q1.

Choose the correct options and explain the reason for the correct and incorrect options in the context of Ernest Rutherford’s gold foil experiment: (i) The experiment clearly showed the existence of neutrons in the nucleus. (ii) The results disproved the plum pudding model and led to the idea of a nucleus at the centre of the atom. (iii) The large deflection of a few alpha particles indicated that most of the mass of the atom and positive charge are packed into a tiny centre. (iv) The way alpha particles were deflected showed that electrons move around the nucleus.

Answer

Correct options: (ii) and (iii). Incorrect: (i) and (iv).

Option	Verdict	Reason
(i) showed neutrons exist	Incorrect	Neutrons are uncharged, so they exert no electrostatic force on an α-particle and cannot show up in a scattering experiment. The neutron was discovered 21 years later, in 1932, by James Chadwick.
(ii) disproved plum pudding, led to the nucleus	Correct	Thomson's uniformly spread positive charge could produce only tiny deflections. Large-angle deflection and backscattering are impossible in that model, so the model had to go — and a small, dense, positive centre took its place.
(iii) mass and positive charge packed into a tiny centre	Correct	Only a target that is both highly charged and much heavier than the α-particle can turn it through a large angle. Since most particles passed straight through, that target must occupy a minute fraction of the atom's volume.
(iv) showed how electrons move	Incorrect	The experiment says nothing about electrons. An α-particle is about 7300 times heavier than an electron, so passing electrons barely alter its path. Electron arrangement came later, from Bohr's work on spectra.

Why it happens: an experiment can only tell you about what it interacts with. α-particles are charged and heavy, so they probe concentrated charge and concentrated mass — nothing else.

Q2.

Which of the following statements are correct or incorrect according to the Bohr’s atomic model? Give a reason for each statement. (i) Electrons lose energy while moving in fixed orbits and slowly fall into the nucleus. (ii) Electrons can exist anywhere around the nucleus with no fixed energy. (iii) Electrons revolve around the nucleus in orbits of fixed energy without losing energy. (iv) Electrons can be found between energy levels as they move around the nucleus.

Answer

Only (iii) is correct. (i), (ii) and (iv) are incorrect.

Statement	Verdict	Reason
(i) lose energy and fall into the nucleus	Incorrect	That is the classical prediction that Bohr rejected. It is what would happen in Rutherford's model, and it is the failure Bohr set out to fix. In a stationary state the energy stays constant.
(ii) anywhere, with no fixed energy	Incorrect	Bohr's central idea is the opposite — electrons occupy only certain allowed shells, each with a definite energy. Energy is quantised, not arbitrary.
(iii) revolve in orbits of fixed energy without losing energy	Correct	This is exactly the postulate of stationary states. The shells K, L, M, N (n = 1, 2, 3, 4) each hold a definite energy, and an electron moving in one does not radiate.
(iv) found between energy levels	Incorrect	The space between two shells is not allowed. An electron changes shell by absorbing or releasing exactly the energy gap between them — it is in one level or the other, never in between.

Why it happens: the four statements are a checklist of the two ideas Bohr added — that only certain energies are allowed, and that an electron in an allowed state does not radiate. Statement (i) denies the second; statements (ii) and (iv) deny the first.

Q3.

The composition of the nuclei of three atomic species X, Y, and Z are given as follows. Number of protons — X: 18, Y: 17, Z: 17. Number of neutrons — X: 19, Y: 18, Z: 20. Explain the relation between the following: (i) Y and Z (ii) Z and X

Answer

(i) Y and Z are isotopes. (ii) Z and X are isobars.

First work out Z (atomic number) and A (mass number) for each species, using A = p⁺ + n⁰.

Species	Protons = Z	Neutrons	Mass number A = p + n	Element
X	18	19	18 + 19 = 37	³⁷₁₈Ar (argon)
Y	17	18	17 + 18 = 35	³⁵₁₇Cl (chlorine)
Z	17	20	17 + 20 = 37	³⁷₁₇Cl (chlorine)

(i) Y and Z — isotopes. Both have 17 protons, so both are chlorine, but their mass numbers are 35 and 37. Same atomic number, different mass number = isotopes. They have the same electronic configuration (2, 8, 7), so identical chemical properties; only their masses differ. These are in fact the two natural isotopes of chlorine, present in the ratio 3 : 1.

(ii) Z and X — isobars. Both have mass number 37, so both nuclei contain 37 nucleons, but Z has 17 protons and X has 18. Same mass number, different atomic number = isobars. They are different elements — chlorine and argon — with completely different chemistry: chlorine (2, 8, 7) needs one electron and is highly reactive, argon (2, 8, 8) has a full octet and is inert.

Why it happens: the two ideas are easy to keep apart if you ask which number is being held fixed. Iso-topes hold the proton number fixed, so they are the same element. Iso-bars hold the total nucleon number fixed, so they weigh the same but are different elements.

Q4.

What conclusion did Rutherford draw about the position and characteristics of the atom’s positively charged part based on the few alpha particles that bounced back or were deflected at large angles in the gold foil experiment?

Answer

Rutherford concluded that the positive charge is not spread through the atom but is concentrated in an extremely small region at the centre, which he called the nucleus.

Position. At the centre of the atom, occupying only a minute fraction of its volume. The atom's diameter is about 10^–10 m; the nucleus is about 10^–15 m across — some 10⁵ times smaller. Everything outside is essentially empty space through which the electrons move.

Characteristics.

It carries all the positive charge of the atom.
It contains almost all the mass — the electrons outside are so light that their mass can be ignored.
It is therefore extremely dense.

How each observation forced each conclusion:

Observation	Conclusion
Most α-particles passed straight through, undeflected	The atom is mostly empty space
A few were deflected through large angles	There is a concentrated positive charge that repels them strongly
A very few bounced almost straight back	That charge sits on something far heavier than the α-particle, and occupies a very small volume

Why it happens: the rarity of the large deflections is as important as the deflections themselves. If backscattering had been common, the target would have to be large. Because it was very rare, the target must be tiny — and to still turn the particle right around, everything heavy and positive must be packed into it.

Q5.

Explain and arrange the following statements in the correct chronological order to show how atomic models have evolved over time. (i) Bohr’s model proposed that electrons move in fixed orbits around the nucleus, each with a definite energy. (ii) Thomson’s model depicted the atom as a ʻplum puddingʼ with electrons embedded in a sphere of positive charge. (iii) Rutherford’s model proposed that atoms have a dense central nucleus. (iv) Dalton’s model described atoms as indivisible particles.

Answer

Correct chronological order: (iv) → (ii) → (iii) → (i), that is Dalton → Thomson → Rutherford → Bohr.

Order	Model	Year	What it proposed	Why the next one was needed
1st	(iv) Dalton	1808	All matter is made of indivisible atoms — the first scientific description of matter	Radioactivity and cathode rays showed atoms give out smaller particles
2nd	(ii) Thomson	after 1897	Electrons embedded in a sphere of positive charge — the plum pudding	Gold foil experiment: a few α-particles bounced back
3rd	(iii) Rutherford	1911	Dense central nucleus, mostly empty space, electrons revolving around it	Could not explain why atoms do not collapse
4th	(i) Bohr	1913	Electrons in fixed shells, each of definite energy; no energy loss while in a shell	Later work showed electrons are clouds, not sharp orbits — the quantum mechanical model

The logic of the order. Each model had to wait for the discovery that made it possible. Thomson's model could not come before the electron was found in 1897. Rutherford's could not come before the gold foil experiment. Bohr's could not come before there was a nucleus to orbit — his model is Rutherford's with one new rule added.

Why it happens: notice that no model was thrown away entirely. Dalton's atom survives as the unit of an element; Thomson's electron survives; Rutherford's nucleus survives inside Bohr's model and inside the modern one. Science advances by keeping what works and repairing what fails.

Q6.

Electrons move around the nucleus in orbits. Why do they not fly away from the atom? Explain what keeps them attracted to the nucleus.

Answer

They do not fly away because of the electrostatic force of attraction between the negatively charged electrons and the positively charged nucleus.

The nucleus contains protons, each carrying charge +1. Each electron carries –1. Unlike charges attract.
This attraction is directed continuously towards the centre, so it keeps pulling the electron back in as the electron moves. It is the force that bends the electron's path into a closed shell instead of a straight line.
The role it plays is the same as that of gravity for a planet going round the Sun: an inward force that keeps a moving body in a closed path.

The other half of the answer is why the electron does not fall in, which is the opposite risk. Bohr's postulate settles that: while an electron stays in an allowed shell its energy is constant, so it neither radiates away its energy nor spirals inward. It escapes only if it is given enough energy from outside to reach beyond the outermost level — which is exactly what happens when an atom is ionised.

Why it happens: the strength of the attraction also explains why electrons in different shells behave so differently. An electron in the K-shell is close to the nucleus, strongly held, and hard to remove. A valence electron in an outer shell is far away and weakly held — which is why sodium (2, 8, 1) gives up its single outer electron so readily and reacts vigorously with water.

Q7.

Assertion (A): The discovery of subatomic particles helped in understanding the atomic structure. Reason (R): The number of electrons is equal to the number of protons in an atom. 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.

Answer

The correct option is (ii) Both A and R are true, but R is not the correct explanation of A.

A is true. Every step forward in atomic structure came from finding a subatomic particle. The electron (Thomson, 1897) proved the atom was divisible; the proton gave the nucleus its charge; the neutron (Chadwick, 1932) finally explained atomic masses. Without these discoveries there would be no model of the atom at all.

R is true. In a neutral atom the number of electrons does equal the number of protons — that is why the atom carries no net charge, and it is what makes the atomic number Z serve as both the proton count and the electron count.

But R does not explain A. R is one particular fact that we learnt because the particles were discovered — it is a consequence, not a cause. It says nothing about why discovering particles advanced our understanding. A is a statement about the whole history of the subject; R is a single detail inside it.

Why it happens: test the direction of the arrow. Ask “does R lead to A?” Knowing that electrons = protons does not by itself tell you anything about how atomic structure came to be understood. Now ask “does A lead to R?” Yes — it was the discovery of electrons and protons that let us state the equality at all. The arrow runs the wrong way, so R cannot be the explanation of A.

Q8.

Magnesium is essential for many biological processes, including muscle contraction. For an atom of magnesium with a mass number of 24 and atomic number 12, determine the number of (i) protons, (ii) neutrons, (iii) electrons, and also illustrate the arrangement of electrons in a magnesium atom.

Answer

(i) 12 protons (ii) 12 neutrons (iii) 12 electrons. The electronic configuration is 2, 8, 2.

Formulae: Z = number of protons = number of electrons in a neutral atom; n⁰ = A – Z.

Given: A = 24, Z = 12

(i) Number of protons = Z = 12

(ii) Number of neutrons = A – Z
= 24 – 12
= 12

(iii) Number of electrons = Z = 12 (the atom is electrically neutral)

Arrangement of the 12 electrons. Fill K first (maximum 2 × 1² = 2), then L (maximum 2 × 2² = 8), then the rest into M.

K-shell: 2 electrons → 12 – 2 = 10 remain
L-shell: 8 electrons → 10 – 8 = 2 remain
M-shell: 2 electrons
Electronic configuration of Mg = 2, 8, 2

Electron arrangement in a magnesium atom (Z = 12): 12 electrons fill K → L → M as 2, 8, 2. The two outermost electrons are the valence electrons.

Why it happens: the two electrons in the outermost M-shell are the valence electrons. Magnesium loses them to reach the stable configuration 2, 8, which is why its valency is 2 and why it forms Mg²⁺ ions — the form in which magnesium works in your muscles and in chlorophyll.

Q9.

Find the following information for the elements shown in Fig. 8.17: (i) Name of the element (ii) Symbol (iii) Total number of electrons (iv) Number of valence electrons (v) Valency of the element (vi) Number of protons (vii) Atomic number

Answer

Counting the electrons drawn on each shell in Fig. 8.17 gives the configurations 2,1 · 2,5 · 2,8,3 · 2,7, which identify the four elements.

	(a)	(b)	(c)	(d)
Electrons drawn (K, L, M)	2, 1	2, 5	2, 8, 3	2, 7
(i) Name of the element	Lithium	Nitrogen	Aluminium	Fluorine
(ii) Symbol	Li	N	Al	F
(iii) Total number of electrons	3	7	13	9
(iv) Number of valence electrons	1	5	3	7
(v) Valency	1 (loses 1)	3 (gains 3)	3 (loses 3)	1 (gains 1)
(vi) Number of protons	3	7	13	9
(vii) Atomic number Z	3	7	13	9

The four atoms of Fig. 8.17, redrawn with the electrons counted off the printed figure: lithium (2,1), nitrogen (2,5), aluminium (2,8,3) and fluorine (2,7).

How each column was obtained. Add up the electrons on all the shells — that gives the total number of electrons. The atom is neutral, so the number of protons and the atomic number are the same number. The electrons on the outermost shell are the valence electrons. For valency, apply the rule: fewer than 4 valence electrons → lose them; more than 4 → gain enough to reach 8.

(a) 2 + 1 = 3 electrons → Z = 3 → Li; 1 valence electron, so it loses 1 → valency 1
(b) 2 + 5 = 7 electrons → Z = 7 → N; 5 valence electrons, needs 8 – 5 = 3 → valency 3
(c) 2 + 8 + 3 = 13 electrons → Z = 13 → Al; 3 valence electrons, so it loses 3 → valency 3
(d) 2 + 7 = 9 electrons → Z = 9 → F; 7 valence electrons, needs 8 – 7 = 1 → valency 1

Check it yourself: (b) and (c) both come out with valency 3, but for opposite reasons — nitrogen gains three electrons, aluminium loses three. That is why they combine with each other in a 1 : 1 ratio, as AlN.

Q10.

Both Rutherford’s and Bohr’s models have electrons orbiting the nucleus. Why did Rutherford’s model fail to explain atomic stability, while Bohr’s model succeeded?

Answer

Because Rutherford allowed the electron any orbit and let classical physics take its course, while Bohr allowed only certain fixed orbits and postulated that an electron in one of them does not radiate energy.

Why Rutherford's model fails.

An electron moving in a circle is constantly changing direction, so it is accelerating.
Classical physics says an accelerating charge must radiate energy.
Losing energy, the electron's orbit would shrink; it would spiral inward and fall into the nucleus.
The atom would collapse in a tiny fraction of a second — but matter around us is stable. So the model, as it stood, was incomplete.

Why Bohr's model succeeds.

Electrons may occupy only certain allowed shells — stationary states K, L, M, N (n = 1, 2, 3, 4) — and nothing in between.
In a stationary state the electron's energy remains constant even though it is moving, so no energy is radiated.
Energy is exchanged only in jumps between two levels, equal to the difference in their energies. Since there are no allowed states between shells, and none below the K-shell, there is no continuous route inward.

	Rutherford (1911)	Bohr (1913)
Allowed orbits	any radius	only fixed shells
Energy while orbiting	radiated continuously	constant
Prediction for the atom	collapses	stable

Why it happens: Bohr did not change the picture of the atom — he kept Rutherford's nucleus and orbits exactly. What he changed was the rule governing them, by adding a restriction that classical physics does not contain. That is why the models look alike in a drawing but predict opposite fates. Bohr's rule was an assumption, justified by the fact that it worked — it also explained the sharp lines in atomic spectra, which continuous radiation could never do.

Q11.

An atom ⁷⁰X has 31 electrons. How many neutrons are there in its nucleus?

Answer

39 neutrons.

Reading the symbol: the superscript in ⁷⁰X is the mass number, so A = 70. The atom is neutral, so protons = electrons.

Number of electrons = 31
⇒ Number of protons, Z = 31
Mass number, A = 70

Number of neutrons = A – Z
= 70 – 31
= 39 neutrons

Z = 31 identifies the element as gallium, so ⁷⁰X is ⁷⁰₃₁Ga.

Check it yourself: 31 protons + 39 neutrons = 70 nucleons ✓. Notice that neutrons already outnumber protons here (ratio 39 : 31 ≈ 1.26) — the trend the chapter describes, that heavier nuclei need proportionally more neutrons to stay bound.

Q12.

An atom has 79 protons and a mass number of 197. Calculate (i) the number of neutrons, and (ii) the number of electrons.

Answer

(i) 118 neutrons (ii) 79 electrons. The element is gold, ¹⁹⁷₇₉Au.

Given: number of protons = 79, mass number A = 197

(i) Number of neutrons = A – number of protons
= 197 – 79
= 118

(ii) The atom is electrically neutral, so
Number of electrons = number of protons
= 79

Z = 79 is gold — the very metal Geiger and Marsden beat into foil for the scattering experiment.

Why it happens: with 79 protons crammed together, the electrostatic repulsion inside a gold nucleus is enormous. It takes 118 neutrons — a neutron-to-proton ratio of about 1.49 — to add enough short-range nuclear force, and enough spacing between protons, to hold it together. Compare carbon, which manages with 6 and 6.

Q13.

Complete the Table 8.5: (row 1) Atomic number 5, number of neutrons 6; (row 2) Mass number 14, number of electrons 7, Nitrogen; (row 3) Mass number 24, number of protons 12; (row 4) Atomic number 15, number of neutrons 16; (row 5) Mass number 1, number of neutrons 0.

Answer

Every blank follows from two relations: Z = number of protons = number of electrons (neutral atom) and A = number of protons + number of neutrons.

Atomic number	Mass number	Number of neutrons	Number of protons	Number of electrons	Name of the element
5	11	6	5	5	Boron
7	14	7	7	7	Nitrogen
12	24	12	12	12	Magnesium
15	31	16	15	15	Phosphorus
1	1	0	1	1	Hydrogen

Row by row:

Row 1: Z = 5 ⇒ protons = 5, electrons = 5. A = 5 + 6 = 11. Z = 5 → boron, ¹¹₅B

Row 2: electrons = 7 ⇒ Z = 7, protons = 7. Neutrons = 14 – 7 = 7. Given: nitrogen, ¹⁴₇N ✓

Row 3: protons = 12 ⇒ Z = 12, electrons = 12. Neutrons = 24 – 12 = 12. Z = 12 → magnesium, ²⁴₁₂Mg

Row 4: Z = 15 ⇒ protons = 15, electrons = 15. A = 15 + 16 = 31. Z = 15 → phosphorus, ³¹₁₅P

Row 5: A = 1 and neutrons = 0 ⇒ protons = 1 – 0 = 1, so Z = 1 and electrons = 1. Z = 1 → hydrogen, ¹₁H

Did you know? Row 5 is protium, the only atom in the whole periodic table whose nucleus contains no neutron at all — just a lone proton. It makes up about 99.98% of all natural hydrogen.

Q14.

Aman was discussing the structure of atom with his classmates. During the discussion, he learnt that an element X has a mass number of 35 and contains 18 neutrons. Based on this information, answer the following questions: (i) How many electrons and protons does element X have? (ii) What is its atomic number? (iii) Identify the element X. (iv) Write its electronic configuration. (v) How many valence electrons does it have? (vi) What will be the mass number if two neutrons are added to its nucleus? (vii) What will be the relation of X with the new atom?

Answer

Start from A = 35 and n⁰ = 18.

(i) 17 protons and 17 electrons.

Number of protons = A – number of neutrons
= 35 – 18
= 17
The atom is neutral, so number of electrons = number of protons = 17

(ii) Atomic number Z = 17, since Z is defined as the number of protons.

(iii) The element is chlorine (Cl), written ³⁵₁₇Cl.

(iv) Electronic configuration = 2, 8, 7.

K-shell (max 2 × 1² = 2): 2 electrons → 17 – 2 = 15 remain
L-shell (max 2 × 2² = 8): 8 electrons → 15 – 8 = 7 remain
M-shell: 7 electrons
Configuration = 2, 8, 7

(v) 7 valence electrons — the electrons in the outermost (M) shell. Chlorine needs 8 – 7 = 1 more to complete its octet, so its valency is 1 and it is highly reactive.

(vi) New mass number = 37.

New number of neutrons = 18 + 2 = 20
Protons are unchanged at 17
New mass number A' = 17 + 20
= 37, i.e. ³⁷₁₇Cl

(vii) X and the new atom are isotopes of each other. Both have Z = 17, so both are chlorine, but their mass numbers are 35 and 37 — same atomic number, different mass number.

Why it happens: adding neutrons cannot change the element, because the element is defined by the proton count alone. Both atoms still have 17 electrons in the arrangement 2, 8, 7, so both behave identically in every chemical reaction. Only the mass differs. These two are in fact the real isotopes of chlorine, mixed in nature in the ratio 3 : 1, which is why chlorine's average atomic mass is 35.5 u.

Q15.

In an atom, there are 12 protons and 12 neutrons in the nucleus. Now, imagine that all the electrons are replaced with some hypothetical particles that have the same charge as electrons but are 500 times heavier. What effect will this replacement have on the atom’s: (i) Atomic number (ii) Atomic mass (iii) Mass number (iv) Overall charge

Answer

The atom is magnesium: 12 protons, 12 neutrons, and therefore 12 electrons.

(i) Atomic number — no change. It stays 12. The atomic number is defined as the number of protons, and the nucleus has not been touched.

(ii) Atomic mass — it increases, by about 3.3 u. This is the only quantity that changes.

Mass of one electron ≈ 1/1836 of a proton's mass ≈ 5.5 × 10^–4 u

Mass of one heavy particle = 500 × 5.5 × 10^–4 u
= 0.275 u

Total for 12 such particles = 12 × 0.275 u
= ≈ 3.3 u

Original atomic mass ≈ 24 u (12 protons + 12 neutrons; electrons negligible)
New atomic mass ≈ 24 u + 3.3 u = ≈ 27.3 u

The point is not just the number but what it means: ordinarily the 12 electrons contribute only 12 × 5.5 × 10^–4 u ≈ 0.0066 u, which is why we ignore them. At 500 times the mass they add about 14% to the atom, and they can no longer be ignored.

(iii) Mass number — no change. It stays 24.

Mass number A = number of protons + number of neutrons
= 12 + 12
= 24 (electrons never enter this count, whatever their mass)

(iv) Overall charge — no change. The atom remains neutral. The replacement particles have the same charge as electrons, and there are still 12 of them.

Total positive charge = 12 × (+1) = +12
Total negative charge = 12 × (–1) = –12
Net charge = +12 – 12 = 0

Why it happens: this question separates three ideas that students often merge. Charge depends only on how many positive and negative units there are, not on their mass. Mass number is a count of nucleons — a definition, so it cannot be affected by anything outside the nucleus. Atomic mass is an actual mass, so every particle contributes; we drop the electrons only because their contribution is far too small to matter, and that approximation breaks the moment they become heavy.

Did you know? A particle exactly like this really exists — the muon, which has the electron's charge but is about 207 times heavier. Physicists make “muonic atoms” with them. The heavier particle orbits much closer to the nucleus, which makes such atoms a sensitive probe of nuclear size.

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