Question 1

Suppose you made up your own ‘atom’, as Thomson described, using clay for the positive charge and small beads for the electrons spread through it. What will happen if: (i) the positive charge on the clay is lesser than the total negative charge of the beads? (ii) by mistake, the clay itself carries a bit of negative charge? Would your model still represent a neutral atom?

Accepted Answer

In Thomson’s model an atom is electrically neutral only because the positive charge of the sphere is exactly equal to the total negative charge of the electrons embedded in it. Both parts of the question disturb this balance. (i) If the positive charge of the clay is less than the total negative charge of the beads, the negative charge wins. The model as a whole carries a net negative charge . It no longer represents a neutral atom — it behaves like a negatively charged particle (a negative ion). (ii) No, it would not represent a neutral atom. If the clay itself carries some negative charge, the ‘positive sphere’ is not fully positive any more, so the total positive charge becomes smaller than the total negative charge of the beads (plus the clay’s own extra negative charge). The model again ends up with a net negative charge . In a real atom the positive part carries only positive charge, and its amount equals the total charge of the electrons. So a Thomson-style model is neutral only when the positive charge and the total negative charge are exactly equal.

Question 2

Could an orange or a lemon, which also contain seeds inside soft pulp, be a good comparison? In what ways does it match Thomson’s idea and where does it fall short?

Accepted Answer

An orange or a lemon is a partly useful comparison, much like the watermelon in the book. Where it matches Thomson’s idea: • The soft pulp fills the whole fruit, like Thomson’s sphere of positive charge that fills the whole atom. • The seeds are small objects sitting inside the pulp, like the electrons embedded in the positive sphere. • There is no hard central core, just as Thomson’s atom had no nucleus . Where it falls short: • Seeds and pulp carry no electric charge , so the fruit cannot show how positive and negative charges balance to make the atom neutral — the main point of Thomson’s model. • In an orange or lemon the seeds are bunched near the centre and the pulp is divided into segments by skins; Thomson’s electrons were spread throughout a uniform sphere. • The fruit has a peel and a definite number of seeds that has nothing to do with the type of fruit, whereas the number of electrons is fixed for each element. • Like Thomson’s model itself, the picture cannot explain the gold foil results (a few α-particles bouncing back), because it has no small, dense, positive centre.

Question 3

Why did Thomson conclude that electrons are present in all atoms?

Accepted Answer

Thomson studied cathode rays in tubes filled with gases at very low pressure. He found that the nature of the cathode rays was the same whatever material the cathode was made of and whatever gas was filled in the tube. The rays were always streams of the same tiny, negatively charged particles (electrons), with a mass much smaller than that of an atom. Since every material tried gave out exactly the same particles, these particles could not belong to just one kind of substance. They had to be a common part of the atoms of every element . That is why Thomson concluded that electrons are a fundamental component present in all atoms .

Question 4

What do you think would happen if α-particles were replaced with negatively charged particles in Rutherford’s gold foil experiment?

Accepted Answer

α-particles are positive , so the positive nucleus repels them. This repulsion is what pushed a few of them away at large angles and even sent some straight back. If negatively charged particles were used instead: • Most of them would still pass straight through, because most of the atom is empty space . • Particles passing close to a nucleus would be attracted towards it (unlike charges attract), not pushed away. Their paths would bend inwards , curving around the nucleus. • Hardly any would bounce straight back , because there is no repulsion from the nucleus to turn them around. • They would also be repelled by the electrons of the gold atoms, which spread out through the atom; this would scatter them in a more confusing way. So the experiment would not give the sharp, clear evidence of a small, positively charged, dense nucleus that the α-particle experiment gave. Using positive particles was essential to Rutherford’s discovery.

Question 5

Rutherford found that a few α-particles bounced back sharply. How does this single surprising result completely rule out Thomson’s ‘plum pudding model’ of the atom?

Accepted Answer

An α-particle is positively charged , fairly heavy and moves very fast . To turn such a particle completely around, it must meet a very strong repelling force concentrated at one point, from something heavier than itself. In Thomson’s plum pudding model, the positive charge and the mass are spread evenly through the whole atom. A spread-out charge can only give a weak push from every side, so a fast α-particle would pass through with at most a slight deflection. The light electrons are far too small to stop or turn back an α-particle. So in Thomson’s atom nothing could make an α-particle bounce back. Even a few bouncing back proves that the atom contains a tiny region where all the positive charge and most of the mass are packed together — the nucleus. This one result cannot be explained by the plum pudding model at all, so the model had to be rejected.

Question 6

If you could ask Rutherford one question about his work, what would it be?

Accepted Answer

This is an open, personal question — any thoughtful question about his work is a good answer. A strong question usually points at something his model could not explain. Some examples:

•“If an electron moving around the nucleus is accelerating and should lose energy, why doesn’t it spiral into the nucleus and make the atom collapse?” (This is exactly the weakness Bohr later solved.)

•“The nucleus is full of positive protons that repel each other — what holds the nucleus together?”

•“Why did you use a gold foil? Why not a foil of some other metal?” (Gold can be beaten into an extremely thin sheet, only a few atoms thick.)

•“How did you estimate that the nucleus is about one lakh times smaller than the atom?”

Any one such question, with a reason for asking it, is a complete answer.

Question 7

Assertion (A): Rutherford concluded that most of the mass of an atom is concentrated in a small region at the centre called the nucleus. Reason (R): According to Thomson’s model, electrons are embedded in a uniformly distributed positive charge sphere. 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

Correct option: (ii) — both A and R are true, but R is not the correct explanation of A. A: True. From the gold foil experiment Rutherford concluded that the positive charge and most of the mass of an atom are packed into a tiny central region, the nucleus. R: True. It correctly describes Thomson’s plum pudding model: electrons embedded in a uniform sphere of positive charge. Link: R does not explain A. Rutherford’s conclusion came from the bouncing back of a few α-particles in his experiment, not from Thomson’s model. In fact Rutherford’s result disproved the idea in R.

Question 8

Imagine you are a scientist who has discovered a new element. Name this element after yourself and justify that the symbol you have chosen follows the IUPAC rules.

Accepted Answer

Every student’s answer will use their own name. Here is a worked example for a student named Ananya.

Name: Ananyium Symbol: An

Why the symbol follows the rules:

1.It is made from the first two letters of the element’s name (A-n), like Al for aluminium.

2.The first letter is a capital (A) and the second letter is small (n) — written An, never AN or an, just as cobalt is Co and not CO.

3.It is short — one or two letters.

4.It is not already used by any other element (for example, it does not clash with Ar, Au or Al). If the first two letters were already taken, the rule allows the first letter plus another letter from the name (like Cl for chlorine, Zn for zinc).

Finally the name and symbol would have to be approved by IUPAC, the international body that approves the names and symbols of elements.

Question 9

What problems could arise if every scientist used different symbols for the same element?

Accepted Answer

• Confusion and mistakes: a symbol written by one scientist might mean a different element to another, so experiments, formulas and results could be misread. • No common language: scientists in different countries speak different languages. A single agreed set of symbols lets everyone understand each other; without it, they would need to translate every paper. • Chemical formulas would not make sense: a formula like H 2 O is understood worldwide only because H and O mean the same thing to everyone. • Danger in practical work: in medicine, industry and laboratories, mixing up two elements because of a wrong symbol could cause accidents or wrong medicines. • Slower progress: time would be wasted checking what each symbol means, and results could not be compared or repeated easily. This is why IUPAC approves one standard name and symbol for every element.

Question 10

An atom with an atomic number of 26 has 56 nucleons. Find out its number of electrons, protons and neutrons.

Accepted Answer

Given: atomic number $Z = 26$, and the number of nucleons (protons + neutrons) is the mass number $A = 56$. Protons: the atomic number equals the number of protons. $\text{protons} = Z = 26$ Electrons: an atom is neutral, so the number of electrons equals the number of protons. $\text{electrons} = 26$ Neutrons: $\text{neutrons} = A - Z = 56 - 26 = 30$ So the atom has 26 electrons, 26 protons and 30 neutrons . (Atomic number 26 is iron , Fe.)

Question 11

The nucleus of an atom contains 20 protons. If its mass number is 41, find the number of neutrons in it.

Accepted Answer

Mass number = number of protons + number of neutrons, so $\text{neutrons} = A - \text{protons}$ $= 41 - 20 = 21$ The nucleus has 21 neutrons . (20 protons means atomic number 20, i.e. calcium.)

Question 12

An atom has 18 neutrons and an atomic number of 17. What is its mass number?

Accepted Answer

Atomic number 17 means the atom has 17 protons . Mass number is the total number of protons and neutrons: $A = \text{protons} + \text{neutrons}$ $= 17 + 18 = 35$ The mass number is 35 . (This is chlorine-35.)

Question 13

An atom ${}^{23}\text{A}$ has 11 electrons. Find the number of neutrons in it.

Accepted Answer

The superscript 23 is the mass number : $A = 23$. A neutral atom has as many protons as electrons, so it has 11 protons (atomic number 11). $\text{neutrons} = A - \text{protons}$ $= 23 - 11 = 12$ The atom has 12 neutrons . (Atomic number 11 is sodium.)

Question 14

Identify the number of electrons in the outermost shell of the following elements: (i) ${}^{12}_{6}\text{C}$ (ii) ${}^{19}_{9}\text{F}$ (iii) ${}^{28}_{14}\text{Si}$

Accepted Answer

The lower number is the atomic number, which equals the number of electrons in a neutral atom. Fill the shells in order: K holds at most 2, L at most 8, then M. (i) Carbon: 6 electrons → K = 2, L = 4, i.e. configuration 2, 4 . Outermost shell (L) has 4 electrons . (ii) Fluorine: 9 electrons → K = 2, L = 7, i.e. 2, 7 . Outermost shell (L) has 7 electrons . (iii) Silicon: 14 electrons → K = 2, L = 8, M = 4, i.e. 2, 8, 4 . Outermost shell (M) has 4 electrons . The upper numbers (mass numbers 12, 19, 28) do not affect the electron arrangement.

Question 15

Write the electronic configuration of the elements having atomic numbers 12, 16 and 18.

Accepted Answer

Number of electrons = atomic number. Fill K (max 2), then L (max 8), then M. Atomic number (element) Electronic configuration (K, L, M) 12 (magnesium, Mg) 2, 8, 2 16 (sulfur, S) 2, 8, 6 18 (argon, Ar) 2, 8, 8 12: 2 in K, 8 in L, the remaining 12 − 10 = 2 in M. 16: 2 in K, 8 in L, the remaining 16 − 10 = 6 in M. 18: 2 in K, 8 in L, the remaining 18 − 10 = 8 in M — a complete octet, which is why argon is unreactive.

Question 16

Solve this riddle: I am an atom with a mass number of 23 and 11 protons. I am a soft metal and react vigorously with water. Who am I and how many neutrons do I have? You can also create one such riddle.

Accepted Answer

11 protons means atomic number 11, which is sodium (Na) — a soft metal that can be cut with a knife and reacts vigorously with water.

$\text{neutrons} = A - Z = 23 - 11 = 12$

I am sodium, and I have 12 neutrons.

A riddle of my own: “I have 8 protons and 8 neutrons. You breathe me in all day, and I make up about one-fifth of the air. My electrons are arranged as 2, 6. Who am I?” Answer: oxygen, ${}^{16}_{8}\text{O}$.

Question 17

Two different atoms have 11 protons each, but one has 12 neutrons, and the other has 13 neutrons. How do their atomic numbers and mass numbers compare? Are they the same element or different elements?

Accepted Answer

Atomic numbers: both have 11 protons, so both have atomic number 11 — the same . Mass numbers: $A_1 = 11 + 12 = 23$ $A_2 = 11 + 13 = 24$ The mass numbers are different (23 and 24). Because the atomic number decides the identity of an element, both atoms are the same element — sodium . Atoms of the same element with different mass numbers are called isotopes : these are ${}^{23}_{11}\text{Na}$ and ${}^{24}_{11}\text{Na}$. They have the same chemical properties (same electrons, 2, 8, 1) but slightly different masses.

Question 18

If a bromine atom is available in the form of, say two isotopes, ${}^{79}_{35}\text{Br}$ (49.7%) and ${}^{81}_{35}\text{Br}$ (50.3%), calculate the average atomic mass of the bromine atom.

Accepted Answer

The average atomic mass is the weighted average : multiply the mass of each isotope by its fraction in nature and add. $\text{Average atomic mass} = \left(79 \times \frac{49.7}{100} + 81 \times \frac{50.3}{100}\right)\text{ u}$ $= \left(\frac{3926.3}{100} + \frac{4074.3}{100}\right)\text{ u}$ $= \frac{8000.6}{100}\text{ u}$ $= 80.006\text{ u} \approx 80.0\text{ u}$ Quick check: the two percentages add to 100, so we can also write the average as 79 u plus the extra 2 u carried by the heavier isotope’s share: $79 + 2 \times \frac{50.3}{100} = 79 + 1.006 = 80.006\text{ u}$ The average atomic mass of bromine is 80.006 u (about 80 u) . It is almost exactly halfway between 79 and 81 because the two isotopes are present in almost equal amounts.

Question 19

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.

Accepted Answer

Correct options: (ii) and (iii). (i) Incorrect. The gold foil experiment (1911) only revealed a small, positive, dense nucleus. Neutrons have no charge and were discovered much later, in 1932, by James Chadwick . (ii) Correct. Thomson’s spread-out positive charge could not turn back an α-particle. The large deflections and rebounds disproved the plum pudding model and led Rutherford to propose a nucleus at the centre . (iii) Correct. Only a tiny, heavy region carrying all the positive charge can repel a fast, positive α-particle strongly enough to deflect it through a large angle. Since only a few were deflected, this region must be very small . (iv) Incorrect. The α-particles were deflected by the nucleus ; electrons are far too light to deflect them noticeably. The experiment said nothing about how electrons move. Rutherford only proposed that electrons revolve around the nucleus to explain where they could be.

Question 20

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.

Accepted Answer

(i) Incorrect. Bohr’s key postulate is that an electron moving in a fixed orbit (stationary state) does not lose energy, so it does not spiral into the nucleus. This is how his model explains the stability of atoms. (ii) Incorrect. In Bohr’s model electrons can revolve only in certain allowed shells (K, L, M, N …), and each shell has a definite energy . (iii) Correct. This is exactly Bohr’s idea of stationary states: each orbit has a fixed energy, and an electron in it does not lose energy. (iv) Incorrect. Electrons can revolve only in the allowed shells, never in between them. An electron can jump from one shell to another only by absorbing or releasing a fixed amount of energy equal to the difference between the two levels.

Question 21

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

Accepted Answer

(Here X, Y and Z are just the names of the three atoms, not atomic numbers.) First work out the atomic number (= protons) and mass number (= protons + neutrons) of each: $\text{X: atomic number} = 18$ $\text{X: } A = 18 + 19 = 37$ $\text{Y: atomic number} = 17$ $\text{Y: } A = 17 + 18 = 35$ $\text{Z: atomic number} = 17$ $\text{Z: } A = 17 + 20 = 37$ (i) Y and Z are isotopes. Both have atomic number 17 (the same element, chlorine) but different mass numbers, 35 and 37, because they have different numbers of neutrons (18 and 20). They are ${}^{35}_{17}\text{Cl}$ and ${}^{37}_{17}\text{Cl}$. (ii) Z and X are isobars. Both have the same mass number, 37, but different atomic numbers (17 and 18), so they are atoms of different elements — chlorine and argon: ${}^{37}_{17}\text{Cl}$ and ${}^{37}_{18}\text{Ar}$.

Question 22

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?

Accepted Answer

Rutherford concluded that the positive charge of the atom is not spread throughout the atom but is concentrated in a tiny central region, which he called the nucleus.

Position:It sits at the centre of the atom, with electrons revolving around it.

Size:It is extremely small — only a few α-particles came close enough to be strongly deflected. The nucleus is about one lakh times smaller than the atom (diameter about $10^{-15}$ m against about $10^{-10}$ m for the atom).

Charge:It carries all the positive charge of the atom; that is why it strongly repelled the positive α-particles.

Mass:It is very dense and holds almost all the mass of the atom; only something heavy could turn a fast α-particle right back.

Since most α-particles passed straight through, he also concluded that the rest of the atom is mostly empty space.

Question 23

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 ‘plum pudding’ 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.

Accepted Answer

Correct order: (iv) → (ii) → (iii) → (i) 1. (iv) Dalton (1808): matter is made of tiny, indivisible atoms. This was the first scientific atomic theory, but it said nothing about what is inside an atom. 2. (ii) Thomson: after discovering the electron (1897), he showed atoms are divisible and pictured electrons embedded in a sphere of positive charge (plum pudding). 3. (iii) Rutherford (1911): the gold foil experiment disproved the plum pudding model; he proposed a tiny, dense, positive nucleus with electrons revolving around it. 4. (i) Bohr (1913): to explain why atoms are stable, he proposed that electrons move only in fixed orbits (shells) of definite energy without losing energy. Each model was replaced when new experiments showed something it could not explain.

Question 24

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

Accepted Answer

Electrons are negatively charged and the nucleus is positively charged (it contains the protons). Unlike charges attract, so there is a strong electrostatic force of attraction between the nucleus and every electron. This pull always acts towards the nucleus . An electron moving in a circular orbit keeps changing its direction. A force pulling it towards the centre is exactly what is needed to keep bending its path into a circle — just as the Sun’s gravitational pull keeps the planets in their orbits, or the string keeps a whirling stone moving in a circle. Without this attraction the electron would move off in a straight line and leave the atom. So the attraction between the positive nucleus and the negative electrons holds the electrons in the atom. At the same time, according to Bohr, an electron in its fixed orbit does not lose energy, so it does not fall into the nucleus either — it keeps revolving in its shell.

Question 25

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.

Accepted Answer

Correct option: (ii) — both A and R are true, but R is not the correct explanation of A. A: True. The discovery of the electron, proton and neutron showed that atoms are divisible and made it possible to build models of how these particles are arranged. R: True. An atom is electrically neutral, so its number of electrons equals its number of protons. Link: R is a correct fact about atoms, but it does not explain why discovering subatomic particles helped us understand atomic structure. A is explained by the experiments that revealed these particles and their arrangement, not by the equality of electrons and protons.

Question 26

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.

Accepted Answer

Given: $A = 24$ and $Z = 12$. (i) Protons = atomic number = 12 . (ii) Neutrons : $\text{neutrons} = A - Z = 24 - 12 = 12$ (iii) Electrons = protons (the atom is neutral) = 12 . Arrangement of electrons: 12 electrons fill K (2), then L (8), and the remaining 12 − 10 = 2 go into M. So the electronic configuration is 2, 8, 2 : Magnesium has 2 valence electrons , which it loses easily, so its valency is 2 (as in MgCl 2 ).

Question 27

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

Accepted Answer

Method: count the electrons in each shell. The total number of electrons equals the number of protons and the atomic number (the atom is neutral); the atomic number tells us the element. The electrons in the outermost shell are the valence electrons. Valency = valence electrons if there are 1 to 4 of them (they are lost or shared), and 8 − valence electrons if there are 5 to 7 (electrons are gained). (a) Shells: 2, 1 → 3 electrons. Lithium, Li . Valence electrons 1; valency 1 (loses 1 electron). Protons 3, atomic number 3. (b) Shells: 2, 5 → 7 electrons. Nitrogen, N . Valence electrons 5; valency 8 − 5 = 3 (gains 3 electrons, as in NH 3 ). Protons 7, atomic number 7. (c) Shells: 2, 8, 3 → 13 electrons. Aluminium, Al . Valence electrons 3; valency 3 (loses 3 electrons). Protons 13, atomic number 13. (d) Shells: 2, 7 → 9 electrons. Fluorine, F . Valence electrons 7; valency 8 − 7 = 1 (gains 1 electron). Protons 9, atomic number 9. Atom Element Symbol K, L, M (a) Lithium Li 2, 1 (b) Nitrogen N 2, 5 (c) Aluminium Al 2, 8, 3 (d) Fluorine F 2, 7 Atom Z = p = e Valence e Valency (a) Li 3 1 1 (b) N 7 5 3 (c) Al 13 3 3 (d) F 9 7 1

Question 28

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?

Accepted Answer

Why Rutherford’s model failed: an electron moving in a circle keeps changing direction, so it is accelerating . A charged particle that accelerates should keep losing energy . Losing energy, the electron would spiral inwards and finally fall into the nucleus (Fig. 8.6), so the atom would collapse. But real atoms are stable. Rutherford’s model had no way to stop this energy loss, so it could not explain stability. Why Bohr’s model succeeded: Bohr proposed (as a postulate) that electrons can move only in certain fixed orbits, or stationary states , each with a definite energy . While an electron stays in such an orbit it does not lose energy , so it does not spiral into the nucleus. Its energy changes only when it jumps from one shell to another by absorbing or releasing a fixed amount of energy. So the difference is the idea of stationary states : Bohr’s electrons keep constant energy in allowed orbits, while Rutherford’s electrons would continuously lose energy and crash into the nucleus.

Question 29

An atom ${}^{70}\text{X}$ has 31 electrons. How many neutrons are there in its nucleus?

Accepted Answer

The mass number is $A = 70$. A neutral atom with 31 electrons has 31 protons , so its atomic number is 31. $\text{neutrons} = A - Z = 70 - 31 = 39$ There are 39 neutrons in its nucleus. (Atomic number 31 is gallium.)

Question 30

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

Accepted Answer

(i) Neutrons : $\text{neutrons} = A - \text{protons}$ $= 197 - 79 = 118$ (ii) Electrons : a neutral atom has as many electrons as protons, so the number of electrons = 79 . (79 protons means atomic number 79: this is gold , ${}^{197}_{79}\text{Au}$ — the metal used in Rutherford’s foil.)

Question 31

Complete the Table 8.5. (Shown here row by row; “–” means the value is to be found.) Row 1: Atomic number 5, Number of neutrons 6; mass number, protons, electrons and name – Row 2: Mass number 14, Number of electrons 7, Name: Nitrogen; atomic number, neutrons, protons – Row 3: Mass number 24, Number of protons 12; atomic number, neutrons, electrons, name – Row 4: Atomic number 15, Number of neutrons 16; mass number, protons, electrons, name – Row 5: Mass number 1, Number of neutrons 0; atomic number, protons, electrons, name –

Accepted Answer

Use three facts: atomic number = protons = electrons (neutral atom); mass number = protons + neutrons; and the atomic number tells us the element.

Row 1:Z = 5, so protons = electrons = 5.

$A = 5 + 6 = 11$

Atomic number 5 is boron.

Row 2:Electrons = 7, so protons = 7 and Z = 7 (nitrogen, as given).

$\text{neutrons} = 14 - 7 = 7$

Row 3:Protons = 12, so Z = 12 and electrons = 12.

$\text{neutrons} = 24 - 12 = 12$

Atomic number 12 is magnesium.

Row 4:Z = 15, so protons = electrons = 15.

$A = 15 + 16 = 31$

Atomic number 15 is phosphorus.

Row 5:Neutrons = 0, so all the mass number comes from protons.

$\text{protons} = 1 - 0 = 1$

Z = 1 and electrons = 1: this is hydrogen (protium).

Completed Table 8.5 — Z = atomic number (= number of protons = number of electrons), A = mass number, n = number of neutrons:

Element	Z	A	n
Boron	5	11	6
Nitrogen	7	14	7
Magnesium	12	24	12
Phosphorus	15	31	16
Hydrogen	1	1	0

Question 32

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?

Accepted Answer

(i) Protons: $\text{protons} = A - \text{neutrons}$ $= 35 - 18 = 17$ Electrons = protons (neutral atom) = 17 . So X has 17 protons and 17 electrons . (ii) Atomic number = number of protons = 17 . (iii) The element with atomic number 17 is chlorine (Cl) ; X is ${}^{35}_{17}\text{Cl}$. (iv) 17 electrons: K = 2, L = 8, M = 17 − 10 = 7. Configuration 2, 8, 7 . (v) The outermost (M) shell has 7 valence electrons (so its valency is 8 − 7 = 1). (vi) Adding 2 neutrons changes only the mass number: $A_{\text{new}} = 35 + 2 = 37$ (vii) The new atom still has 17 protons, so it is still chlorine, ${}^{37}_{17}\text{Cl}$. Same atomic number, different mass number: X and the new atom are isotopes .

Question 33

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

Accepted Answer

The atom has 12 protons, so it is magnesium with 12 electrons and mass number 12 + 12 = 24. The 12 electrons are replaced by 12 heavier particles with the same charge. (i) Atomic number: no change — still 12. It depends only on the number of protons in the nucleus, which has not changed. (ii) Atomic mass: increases. The mass of ordinary electrons is so small that we ignore it, but particles 500 times heavier are no longer negligible. Taking an electron’s mass as about 1/1836 of a proton’s mass (about 0.00055 u), the 12 new particles add $12 \times 500 \times 0.00055\text{ u} \approx 3.3\text{ u}$ so the atom would weigh about 24 + 3.3 ≈ 27.3 u instead of about 24 u. (iii) Mass number: no change — still 24. Mass number counts only the protons and neutrons (nucleons) in the nucleus; the particles outside the nucleus are not counted. (iv) Overall charge: no change — still neutral (zero). The 12 new particles have the same charge as electrons, so 12 negative charges still balance the 12 positive protons. The mass of an electron (about 1/1836 of a proton) is not given in this chapter; it is used only to show the size of the increase. The key idea is that atomic mass rises while atomic number, mass number and charge do not change.

Journey Inside the Atom

Pause and Ponder — In-text Questions

Revise, Reflect, Refine — End-of-Chapter Questions

Atomic number (element)	Electronic configuration (K, L, M)
12 (magnesium, Mg)	2, 8, 2
16 (sulfur, S)	2, 8, 6
18 (argon, Ar)	2, 8, 8

Atom	Element	Symbol	K, L, M
(a)	Lithium	Li	2, 1
(b)	Nitrogen	N	2, 5
(c)	Aluminium	Al	2, 8, 3
(d)	Fluorine	F	2, 7

Journey Inside the Atom

Pause and Ponder — In-text Questions

Revise, Reflect, Refine — End-of-Chapter Questions

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