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

- (i) The model would carry a net negative charge — it would no longer stand for a neutral atom. - Thomson's atom is neutral only because the positive charge of the sphere exactly cancels the total negative charge of the electrons inside it. - With less positive charge on the clay than the beads carry between them, some negative charge is left unbalanced. - What you would be showing is a negatively charged particle, not an atom. - (ii) No — it would still not represent a neutral atom. - If the clay itself is a bit negative, then everything in the model is negative: negative clay plus negative beads, with nothing positive to cancel either. - It also destroys Thomson's central idea, which needs a sphere of positive charge for the electrons to be embedded in. - To repair the model, the clay must carry a positive charge exactly equal in size to the total negative charge on all the beads.

Question 2

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

(ii), (iii) are correct. - Correct: (ii) and (iii). Incorrect: (i) and (iv). - (ii) Correct — Thomson's evenly spread positive charge predicted that every α-particle would pass through or be deflected only slightly. Large deflections and bounce-backs contradicted that prediction outright, and forced the idea of a nucleus at the centre. - (iii) Correct — only a very small region carrying a large positive charge and most of the atom's mass can turn a fast α-particle through a large angle. That such deflections were rare shows how tiny that centre is. - (i) Incorrect — the experiment revealed nothing about neutrons. The neutron was discovered by James Chadwick in 1932, twenty-one years later; being uncharged, it could not have repelled α-particles at all. - (iv) Incorrect — the scattering only recorded what happens to positive particles near the positive centre. Electrons are far too light to deflect an α-particle noticeably, so the experiment gave no evidence about how electrons move. The orbiting-electron picture was a proposal Rutherford added to his model, not something the deflections showed — and it was in fact the part his model could not defend, since it failed to explain atomic stability.

Question 3

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

(iii) is correct. - Only (iii) is correct. (i), (ii) and (iv) are all incorrect. - (i) Incorrect — Bohr's model says the opposite. In a fixed orbit, a stationary state, the electron's energy stays constant and it does not lose energy, so it never spirals in. The spiralling problem belonged to Rutherford's model, and Bohr's model exists precisely to remove it. - (ii) Incorrect — electrons cannot exist just anywhere. Each shell (K, L, M, N... or n = 1, 2, 3, 4... ) has a definite energy, and an electron must occupy one of them. - (iii) Correct — this is Bohr's stationary-state postulate exactly: electrons revolve in orbits of fixed energy without losing energy. It is what explains why atoms are stable and do not collapse. - (iv) Incorrect — the space between two shells is forbidden. An electron changes shell only by absorbing or releasing a fixed amount of energy equal to the difference between the two levels; it is never found in between.

Question 4

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

- Yes, it works as a rough comparison, but it is a weaker one than the watermelon the chapter uses. - Where it matches: the soft pulp plays the part of the positively charged matter that fills the whole atom. - Where it matches: the seeds play the part of the electrons, held inside that matter rather than stuck on the outside. - Where it matches: the fruit is a single round body, like Thomson's single sphere. - Where it falls short: an orange is built of segments, so its pulp is not the smooth, evenly spread positive charge Thomson imagined. - Where it falls short: the seeds sit in a few fixed places near the centre, whereas Thomson's electrons are spread throughout the sphere. - Where it falls short: the seeds are heavy compared with the pulp, while an electron's mass is almost negligible compared with the atom. - Where it falls short: no part of a fruit is charged at all, so it cannot show the balance of positive against negative charge — which is the entire point of the model.

Question 5

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

Accepted Answer

(i) Isotopes (ii) Isobars. - (i) Y and Z are isotopes. (ii) Z and X are isobars. - First work out the atomic number and mass number of each species, using Z = protons and A = protons + neutrons: - X: atomic number = 18, mass number = 18 + 19 = 37 - Y: atomic number = 17, mass number = 17 + 18 = 35 - Z: atomic number = 17, mass number = 17 + 20 = 37 - (i) Y and Z both have 17 protons, so the same atomic number, but mass numbers 35 and 37. Same Z, different A → isotopes. They are the same element (chlorine), with the same electronic configuration and so the same chemical properties. - (ii) Z and X both have mass number 37, but atomic numbers 17 and 18. Same A, different Z → isobars. Because the atomic number differs, these are different elements.

Question 6

Why did Thomson conclude that electrons are present in all atoms? 8.2.2 Testing Thomson’s model: The gold foil experiment In 1911, Geiger and Marsden, working under Ernest Rutherford, tested Thomson’s model of the atom through what became famous as the gold foil experiment. They aimed a narrow beam of alpha particles at an extremely thin sheet of gold foil. Alpha (symbol α) particles are tiny, positively charged particles emitted from certain radioactive elements. Later in this chapter, you will learn that an alpha particle is actually a nucleus of a helium atom containing two protons and two neutrons. According to Thomson’s model, the positive charge in the atom was spread out evenly. So they expected the alpha particles to pass straight through the gold foil or be deflected only slightly. But to their surprise, while most particles passed through undeflected, some were sharply deflected (Fig. 8.4), and a few even bounced back. This deflection from the straight path is called scattering. Hence, the gold foil experiment is also called an α-ray scattering experiment. T homson’s model failed to explain the results of the gold foil experiment, particularly the deflection of some α-particles through large angles and that most of the α-particles passed undeflected.

Accepted Answer

- Because cathode rays turned out to be exactly the same whatever atoms they came from. - Thomson changed the material of the cathode and changed the gas filled in the tube, and the nature of the cathode rays did not change. - In every case the rays were streams of negatively charged particles with a mass much smaller than that of an atom. - These particles were being emitted out of atoms, so they had to be inside them to begin with. - Since identical particles came out of every element tried, the electron must be a fundamental component of all atoms, not a feature of one particular substance.

Question 7

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 is not spread through the atom at all, but packed into a tiny, dense centre he called the nucleus. - Position — at the centre of the atom, occupying an extremely small part of its volume. - Size — diameter about 10^-15 m, against an atom's 10^-10 m; that is roughly 10^5 times smaller than the atom itself. - Charge — it carries all the positive charge of the atom, concentrated enough to repel an approaching α-particle violently. - Mass — it is dense, holding almost the entire mass of the atom. - Why the bounce-backs prove this: an α-particle is itself positive, fast and heavy, so only a near head-on approach to a small region of large positive charge and large mass can push it straight back the way it came. - Why so few bounced back: such a target is a very small mark to hit. Most α-particles missed it completely and went straight through, which is what shows that the rest of the atom is empty space.

Question 8

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

Accepted Answer

- The deflections would reverse in direction: negative particles would be pulled towards the nucleus instead of being pushed away from it. - The nucleus is positively charged. It repels a positive α-particle, but unlike charges attract, so it would attract a negative particle. - Most particles would still pass straight through, because the atom is mostly empty space — that part of the result would not change at all. - The few that came close to a nucleus would swerve inward, bending round the far side of it, instead of being pushed outward from the near side. - Large-angle scattering, right up to nearly 180^°, would still be seen. A particle aimed almost straight at the nucleus is whipped tightly around it and comes back out roughly the way it came — a slingshot, the way a comet swinging past the Sun is turned right around by attraction. - For the same off-centre aim, an attracting nucleus turns a particle through the same angle as a repelling one; only the side it passes on differs. So the experiment would still show a tiny, dense, charged nucleus in a mostly empty atom. - The atom's electrons would repel them slightly, but electrons have far too little mass to deflect a fast particle by much.

Question 9

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.

Accepted Answer

- Correct chronological order: (iv) → (ii) → (iii) → (i). - (iv) Dalton, 1808 — all matter is made of indivisible particles called atoms, the fundamental building blocks that cannot be broken down. The first scientific description of matter, and the starting point for everything after it. - (ii) Thomson, after his 1897 discovery of the electron — cathode-ray work showed atoms contain smaller negative particles, so atoms are not indivisible. To keep the atom neutral he placed electrons inside a sphere of positive charge: the plum pudding model. - (iii) Rutherford, 1911 — the gold foil experiment showed most α-particles pass straight through while a few are thrown sharply back. The positive charge and most of the mass must therefore sit in a dense central nucleus, with the rest of the atom empty. - (i) Bohr, 1913 — Rutherford's orbiting electrons should have lost energy and spiralled in. Bohr fixed this by allowing only certain shells of definite energy, in which an electron moves without losing energy — which explains why atoms are stable. - Each step was forced by the failure of the one before it: this is how the chapter shows science moving forward, keeping what works and repairing what does not.

Question 10

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

Accepted Answer

- Because a thinly spread positive charge can never turn a fast α-particle around — and thin spreading is the whole of Thomson's model. - In the plum pudding model the positive charge is smeared evenly over the entire atom, so an α-particle crossing it meets only a weak push at any one point. - The embedded electrons cannot do it either: they are negative (so they attract, not repel) and far too light to reverse a heavy α-particle. - To send a fast, massive, positively charged particle straight back, it must run almost head-on into something that is very small, very massive and strongly positive all at once. - So even one bounce-back is fatal evidence: it proves such a concentrated centre exists, and Thomson's model contains no such thing anywhere. - This single observation is what forced the idea of the nucleus.

Question 11

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

- They are held in by the electrostatic force of attraction between the negatively charged electron and the positively charged nucleus. - The nucleus contains protons, which carry positive charge; the electron carries negative charge. Unlike charges attract, so the electron is continuously pulled towards the centre. - That inward pull is what bends the electron's path into a closed orbit instead of letting it fly off in a straight line. - To actually escape the atom, an electron would have to be supplied with enough energy from outside to overcome this attraction. - The opposite worry — that the pull should make the electron fall into the nucleus — is answered by Bohr: in a stationary state the electron keeps a fixed amount of energy and loses none while moving. - With no energy lost, the electron neither escapes outward nor spirals inward. It stays in its shell, and can change only by absorbing or releasing a fixed amount of energy to jump to another shell.

Question 12

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

Accepted Answer

- This asks for your own question, so there is no single answer — but the chapter tells you where the good questions are. - "Why did you take the few α-particles that bounced back seriously, instead of writing them off as experimental error?" — they were rare, yet they carried the entire discovery. - "How did you decide the nucleus is positively charged, rather than merely small and heavy?" - "You found the nucleus to be about 10^5 times smaller than the atom — what convinced you that the atom really is that empty?" - "Your model could not explain why atoms are stable. Did you already know this when you proposed it in 1911?" - "Why did you have Geiger and Marsden fire the particles at gold foil in particular?" - A strong question points either at something the chapter says his model could not explain, or at how he read evidence that others might have discarded.

Question 13

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

(ii). - (ii) Both A and R are true, but R is not the correct explanation of A. - A is true — the electron (Thomson, 1897), the proton (Rutherford) and the neutron (Chadwick, 1932) are exactly the particles out of which atomic structure is built. Each discovery forced a better model. - R is true as well — in a neutral atom the number of electrons equals the number of protons: helium has 2 and 2, sodium has 11 and 11. - But R is only one particular fact about two of those particles. It does not say why discovering subatomic particles advanced our understanding of atomic structure. - Indeed R cannot explain A, because the neutron — one of the three discoveries — has nothing to do with the electron-proton balance at all. - Both statements true, R not the explanation → option (ii).

Question 14

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

(ii). - (ii) Both A and R are true, but R is not the correct explanation of A. - A is true: from the gold foil experiment Rutherford concluded that the positive charge and most of the mass are concentrated in a tiny central region, the nucleus. - R is also true: Thomson's model really does describe electrons embedded in a uniformly distributed sphere of positive charge. - But R describes the older model that this experiment overturned. It gives no reason at all for Rutherford's conclusion. - The real reason behind A is the observation that a few α-particles were deflected through large angles and a few bounced back. - So both statements stand on their own, but R does not explain A — which is option (ii).

Question 15

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

- Worked example: a student named Anaya discovers an element, names it anayium, and chooses the symbol An. - Rule 1 — the symbol is taken from the first letters of the name: An ayium gives An. ✓ - Rule 2 — the first letter is a capital and the second is small: An, never AN or an (just as cobalt is Co, not CO). ✓ - Rule 3 — if those first two letters are already in use, take the first letter with a later letter of the name instead, the way chlorine gives Cl and zinc gives Zn; so "Ay" or "Ai" would be the fallback here. Not "Am" — that is already americium. ✓ - Rule 4 — the symbol must not clash with an existing element's symbol, or the same two letters would stand for two elements at once. ✓ - Rule 5 — the choice is not yours alone to make final: IUPAC approves the names and symbols of all elements internationally. - Now repeat this with your own name and check your symbol against the same five points.

Question 16

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

(i) Protons 12 (ii) Neutrons 12 (iii) Electrons 12; Electronic configuration 2, 8, 2. - (i) protons = 12 (ii) neutrons = 12 (iii) electrons = 12 - (i) Atomic number Z = 12, and Z is the number of protons → 12 protons. - (ii) neutrons = A - Z = 24 - 12 = 12 → 12 neutrons. - (iii) A magnesium atom is electrically neutral, so electrons = protons = 12 electrons. - Electronic configuration: fill K first (maximum 2×1^2 = 2 ), then L (maximum 2×2^2 = 8 ), and the 2 left over go into M → 2, 8, 2. - What your drawing must contain: - a small circle at the centre for the nucleus, labelled inside with "12 p, 12 n"; - three concentric circles drawn around it, labelled K (n = 1), L (n = 2) and M (n = 3), with K nearest the nucleus; - 2 dots on the K circle, 8 dots on the L circle, 2 dots on the M circle — check that 2 + 8 + 2 = 12; - the 2 electrons on the outermost M shell marked as the valence electrons; - a title, "Magnesium, ^24 12 Mg ".

Question 17

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 (a)

Accepted Answer

(a) (i) Lithium; (ii) Li; (iii) 3; (iv) 1; (v) 1; (vi) 3; (vii) 3 (b) (i) Nitrogen; (ii) N; (iii) 7; (iv) 5; (v) 3; (vi) 7; (vii) 7 (c) (i) Aluminium; (ii) Al; (iii) 13; (iv) 3; (v) 3; (vi) 13; (vii) 13 (d) (i) Fluorine; (ii) F; (iii) 9; (iv) 7; (v) 1; (vi) 9; (vii) 9. - The four shell diagrams in Fig. 8.17 carry 3, 7, 13 and 9 electrons respectively, and everything asked follows from those counts. - (a) Lithium — (i) lithium; (ii) Li; (iii) 3 electrons, arranged 2, 1; (iv) 1 valence electron; (v) valency 1; (vi) 3 protons; (vii) atomic number 3. - (b) Nitrogen — (i) nitrogen; (ii) N; (iii) 7 electrons, arranged 2, 5; (iv) 5 valence electrons; (v) valency 3; (vi) 7 protons; (vii) atomic number 7. - (c) Aluminium — (i) aluminium; (ii) Al; (iii) 13 electrons, arranged 2, 8, 3; (iv) 3 valence electrons; (v) valency 3; (vi) 13 protons; (vii) atomic number 13. - (d) Fluorine — (i) fluorine; (ii) F; (iii) 9 electrons, arranged 2, 7; (iv) 7 valence electrons; (v) valency 1; (vi) 9 protons; (vii) atomic number 9. - How the valencies were fixed: fewer than four valence electrons → the atom loses them (Li loses 1, so valency 1; Al loses 3, so valency 3). - More than four valence electrons → the atom gains enough to complete its octet (N gains 8 - 5 = 3, so valency 3; F gains 8 - 7 = 1, so valency 1). - Protons and atomic number are equal to the electron count in each case, because a neutral atom has as many protons as electrons.

Question 18

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

Accepted Answer

- The core problem: scientists would no longer understand one another. - The same element would be written differently in different countries and languages, so a formula or equation written in one place could not be read in another — the very language barrier that symbols were invented to cross. - Worse, one symbol could stand for two different elements, so a written formula would be genuinely ambiguous rather than merely unfamiliar. - Results could not be checked, compared or repeated by other laboratories, which is how science is supposed to work. - Textbooks, data tables and research papers would all need translating before use, wasting effort and introducing errors. - In medicine, industry and the laboratory a mistaken element is not a small error — it can be dangerous. - This is exactly why names and symbols are internationally agreed and approved by IUPAC, so that one symbol means one element everywhere.

Question 19

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

Accepted Answer

Electrons 26; Protons 26; Neutrons 30. - Electrons 26, protons 26, neutrons 30. - Atomic number Z = 26, and Z is the number of protons, so protons = 26. - The atom is electrically neutral, so electrons = protons = 26. - Nucleons means protons and neutrons together, so the mass number is A = 56. - neutrons = A - Z = 56 - 26 = 30 - (This is iron, which the chapter lists as having 26 protons and 30 neutrons.)

Question 20

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

- Rutherford had no answer to the accelerating electron; Bohr blocked the problem by postulating stationary states. - Why Rutherford's model failed: an electron moving in a circular path is constantly changing direction, which means it is accelerating. - An accelerating charged particle loses energy. As it lost energy the electron would spiral inward and eventually fall into the nucleus. - Every atom would then collapse — but real atoms are stable, which is why the matter around us stays intact. So the model had to be incomplete. - Why Bohr's model succeeded: he allowed the electron only certain fixed circular paths, the stationary states or shells K, L, M, N... ( n = 1, 2, 3, 4... ). - He postulated that while an electron moves in one of these shells its energy stays constant, even though it is in motion — it radiates no energy at all. - With no energy lost there is no spiral and no collapse. The electron changes only by absorbing or releasing a fixed amount of energy to jump between allowed shells. - Worth noting honestly: Bohr did not prove this, he assumed it as a postulate; it was accepted because it explained many experimental observations. Later work showed even Bohr's model is not entirely correct.

Question 21

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

Accepted Answer

Neutrons 39. - 39 neutrons. - The atom is neutral, so protons = electrons = 31, giving Z = 31. - The superscript is the mass number, A = 70. - neutrons = A - Z = 70 - 31 = 39

Question 22

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

Accepted Answer

Neutrons 21. - 21 neutrons. - Protons = 20, so Z = 20; mass number A = 41. - A = protons + neutrons - neutrons = A - Z = 41 - 20 = 21

Question 23

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

Accepted Answer

Mass number 35. - Mass number = 35. - Atomic number Z = 17, so the nucleus holds 17 protons. - A = protons + neutrons = 17 + 18 = 35 - (An atom with Z = 17 is chlorine.)

Question 24

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 118 (ii) Electrons 79. - (i) 118 neutrons (ii) 79 electrons - Protons = 79, so Z = 79; mass number A = 197. - (i) neutrons = A - Z = 197 - 79 = 118 - (ii) The atom is electrically neutral, so electrons = protons = 79. - ( Z = 79 is gold, symbol Au from the Latin aurum.)

Question 25

An atom electrons. number of neutrons in it. 150 Exploration Grade 9

Accepted Answer

Neutrons 12. - 12 neutrons. - The atom is neutral, so protons = electrons = 11, which makes Z = 11. - The superscript gives the mass number, A = 23. - neutrons = A - Z = 23 - 11 = 12 - ( Z = 11 with A = 23 is sodium.)

Question 26

Complete the Table 8.5: Atomic number 5 - - 15 -

Accepted Answer

Atomic Mass Number of Number of Number of Name of the number number neutrons protons electrons 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. - Completed table: Atomic number Mass number Neutrons Protons Electrons 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 - Two relations do all the work: Z = protons = electrons in a neutral atom, and A = protons + neutrons. - Row 1 — Z = 5 gives 5 protons and 5 electrons; A = 5 + 6 = 11; Z = 5 is boron. - Row 2 — 7 electrons gives 7 protons and Z = 7; neutrons = 14 - 7 = 7; this confirms nitrogen. - Row 3 — 12 protons gives Z = 12 and 12 electrons; neutrons = 24 - 12 = 12; Z = 12 is magnesium. - Row 4 — Z = 15 gives 15 protons and 15 electrons; A = 15 + 16 = 31; Z = 15 is phosphorus. - Row 5 — A = 1 with 0 neutrons leaves 1 proton, so Z = 1 and 1 electron; this is hydrogen, the only atom whose nucleus has no neutron.

Question 27

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) Electrons 17; Protons 17 (ii) Atomic number 17 (iii) Chlorine (iv) Electronic configuration 2, 8, 7 (v) Valence electrons 7 (vi) Mass number 37 (vii) Isotopes. - (i) 17 electrons and 17 protons - protons = A - neutrons = 35 - 18 = 17; the atom is neutral, so electrons = protons = 17. - (ii) Atomic number = 17 — the atomic number is simply the number of protons. - (iii) X is chlorine (Cl) — the atomic number identifies the element, and Z = 17 is chlorine. - (iv) Electronic configuration: 2, 8, 7 — K takes 2, L takes 8, and the remaining 17 - 10 = 7 go into M. - (v) 7 valence electrons — the outermost shell is M, and it holds 7. - (vi) New mass number = 37 — adding neutrons does not change the proton count: A = 17 + (18 + 2) = 37. - (vii) The two are isotopes — same atomic number 17, different mass numbers 35 and 37. Being isotopes, they have the same electronic configuration and therefore the same chemical properties.

Question 28

Identify the number of electrons in the outermost shell of the following elements: (i) 12 6 C (ii) 19 9 F (iii) 28 14 Si

Accepted Answer

(i) 4 (ii) 7 (iii) 4. - (i) 4 (ii) 7 (iii) 4 - Shells fill from the nucleus outward — K first (maximum 2×1^2=2 ), then L (maximum 2×2^2=8 ), then M. - (i) Carbon, Z = 6: 2 electrons in K, the remaining 4 in L → configuration 2, 4 → outermost shell has 4 electrons. - (ii) Fluorine, Z = 9: 2 in K, the remaining 7 in L → configuration 2, 7 → outermost shell has 7 electrons. - (iii) Silicon, Z = 14: 2 in K, 8 in L, the remaining 4 in M → configuration 2, 8, 4 → outermost shell has 4 electrons. - The subscript is the atomic number, which equals the number of electrons in a neutral atom; the superscript (mass number) plays no part here.

Question 29

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

Accepted Answer

2, 8, 2; 2, 8, 6; 2, 8, 8. - Z = 12: 2, 8, 2 - Z = 16: 2, 8, 6 - Z = 18: 2, 8, 8 - Rule used: a shell holds at most 2n^2 electrons — K = 2×1^2 = 2, L = 2×2^2 = 8, M = 2×3^2 = 18 — and shells are filled in order K, L, M. - 12 = 2 + 8 + 2 (magnesium), 16 = 2 + 8 + 6 (sulfur), 18 = 2 + 8 + 8 (argon). - For Z = 18 the M shell stops at 8, not 18, because the outermost shell can never hold more than 8 electrons — argon therefore has a complete octet.

Question 30

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

- (i) Atomic number: unchanged, still 12. - The atomic number counts protons in the nucleus, and the nucleus was not touched — the substitution happened outside it. - (ii) Atomic mass: it increases — the atom becomes measurably heavier. - Normally the 12 electrons can be ignored in mass calculations because an electron's mass is almost negligible. - Here the 12 replacement particles together carry 12 × 500 = 6000 times an electron's mass, which is no longer something you may throw away. - (iii) Mass number: unchanged, still 24. - Mass number counts nucleons only: A = protons + neutrons = 12 + 12 = 24. Particles outside the nucleus never enter this count, however heavy they are. - (iv) Overall charge: unchanged — the atom is still neutral, total charge zero. - The hypothetical particles carry the same charge as electrons, so 12 of them still cancel the +12 of the protons exactly. - Charge and mass are independent properties: making a particle heavier does not make it more negative.

Question 31

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

Sodium; 12. - Sodium (Na), with 12 neutrons. - 11 protons means Z = 11, and the element of atomic number 11 is sodium. - neutrons = A - Z = 23 - 11 = 12, which is the neutron count Table 8.4 lists for sodium. - The chapter's sodium fits: its configuration is 2, 8, 1, and it is worked as the atom that completes its octet by losing that one electron, so its valency is 1. - "Soft metal" and "reacts vigorously with water" are the riddle's own clues, not the chapter's — but a single, loosely held valence electron is exactly the chapter-level reason such a metal is so reactive. - Your own riddle, built the same way: "My single shell is full with just 2 electrons, my mass number is 4, and I refuse to react with anything at all. Who am I, and how many neutrons do I have?" - Answer to it: helium, Z = 2, neutrons = 4 - 2 = 2 — unreactive because 2 electrons already complete its only shell.

Question 32

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

- Their atomic numbers are the same (11 each); their mass numbers differ (23 and 24). They are the same element — two isotopes of sodium. - Atomic number counts protons only, and both atoms have 11 protons, so Z = 11 for both. - Mass number counts protons and neutrons: 11 + 12 = 23 and 11 + 13 = 24. - It is the atomic number, not the neutron count, that decides which element an atom is — so both are sodium. - Atoms of the same element with the same Z but different A are called isotopes. - Both have 11 electrons and the same configuration 2, 8, 1, so their chemical properties are the same; only physical properties such as mass differ.

Question 33

If a bromine atom is available in the form of, say two isotopes, average atomic mass of the bromine atom. You have learnt that the two atoms that have the same As we end our journey into the structure of the atom, likely to be, not exactly where they are. You will learn all these details in higher grades. The journey of exploring the mysteries of the atom is far from over,

Accepted Answer

80.006 u. - Average atomic mass of bromine = 80.006 u. - Use the weighted average: multiply each isotope's mass by its percentage abundance, then add. - average = 79 × 49.7/100 + 81 × 50.3/100 - = 39.263 + 40.743 - = 80.006 u - The value sits almost exactly midway between 79 and 81 because the two isotopes occur in nearly equal numbers. - No single bromine atom weighs 80.006 u; the figure is what a large collection of bromine atoms averages out to.

Atomic number	Mass number	Neutrons	Protons	Electrons	Element
$\displaystyle 5$	$\displaystyle 11$	$\displaystyle 6$	$\displaystyle 5$	$\displaystyle 5$	Boron
$\displaystyle 7$	$\displaystyle 14$	$\displaystyle 7$	$\displaystyle 7$	$\displaystyle 7$	Nitrogen
$\displaystyle 12$	$\displaystyle 24$	$\displaystyle 12$	$\displaystyle 12$	$\displaystyle 12$	Magnesium
$\displaystyle 15$	$\displaystyle 31$	$\displaystyle 16$	$\displaystyle 15$	$\displaystyle 15$	Phosphorus
$\displaystyle 1$	$\displaystyle 1$	$\displaystyle 0$	$\displaystyle 1$	$\displaystyle 1$	Hydrogen

Pause and Ponder 8.1 (part 1 of 23)

Exercise 8.1