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The Vault
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The Vault
CBSE Class 9 Chemistry · 12 questions · 29 marks
For a long time the atom was believed to be the smallest indivisible piece of matter, until experiments with electric discharge and radioactivity revealed particles hiding inside it. In this chapter you follow that detective story — from Thomson's soft sphere of positive charge to Rutherford's tiny dense nucleus and Bohr's fixed energy shells — and learn to write the electronic configuration and valency of any element from its atomic number.
In the α-particle scattering experiment, the observation that the vast majority of α-particles passed straight through the gold foil without any deflection led to the conclusion that:
Answer
Most of the space inside an atom is empty. If matter inside the atom were spread out evenly, almost every α-particle would have been deflected; instead nearly all of them sailed through undisturbed, which can only happen if they met nothing on the way. The separate observation that a very small fraction rebounded through large angles is what pointed to a dense positively charged nucleus, and the idea of fixed orbits came later from Bohr, not from this experiment.
The electronic configuration of a neutral atom whose atomic number is 19 is:
Answer
2, 8, 8, 1. The neutral atom has 19 electrons. The K shell takes 2 and the L shell takes 8, leaving 9. Although the M shell can eventually hold 2 × 3² = 18 electrons, it cannot be the outermost shell with 9 electrons, because the outermost shell can never hold more than 8. So the M shell takes 8 and the remaining single electron starts the N shell, giving 2, 8, 8, 1.
Two atoms P and Q have the composition: P has 18 protons and 22 neutrons, Q has 20 protons and 20 neutrons. P and Q are:
Answer
Isobars. Their mass numbers are equal — P has 18 + 22 = 40 and Q has 20 + 20 = 40 — but their atomic numbers differ (18 and 20), so they are atoms of two different elements with the same mass number, which is exactly the definition of isobars. They cannot be isotopes, because isotopes must share the same atomic number and differ in mass number.
According to the Bohr–Bury scheme, the maximum number of electrons that the N shell of an atom can accommodate is:
Answer
32. The maximum capacity of a shell is given by 2n², where n is the shell number. For the N shell, n = 4, so the capacity is 2 × 4² = 2 × 16 = 32 electrons. Note that this is only the theoretical capacity: if the N shell happens to be the outermost shell of an atom, it can still hold no more than 8 electrons.
Assertion (A): All the isotopes of a given element show identical chemical behaviour. Reason (R): Isotopes of an element have the same number of protons and therefore the same number and arrangement of electrons.
Answer
Both A and R are true and R is the correct explanation of A. Chemical behaviour is decided entirely by the valence electrons of an atom. Isotopes differ only in the number of neutrons in the nucleus, so they have the same atomic number, the same number of electrons and the same electronic configuration — hence identical chemical properties. Their physical properties, such as density and rate of diffusion, do differ because their masses differ.
Why is almost the entire mass of an atom said to be concentrated in its nucleus?
Answer
The nucleus contains the protons and the neutrons, each of which has a mass of about 1 u, while an electron has a mass of only about 1/1836 u — so light that it is treated as negligible in calculations. Since all the heavy particles sit inside the nucleus and only the near-massless electrons lie outside it, the sum of the masses of the protons and neutrons accounts for practically the whole mass of the atom. This is also why the mass number of an atom is simply the number of protons plus the number of neutrons.
An atom of element X has a mass number of 35 and an atomic number of 17. Find the number of protons, electrons and neutrons in a neutral atom of X, write its electronic configuration, and state its valency.
Answer
Number of protons = atomic number = 17. In a neutral atom the number of electrons equals the number of protons, so number of electrons = 17. Number of neutrons = mass number − atomic number = 35 − 17 = 18. Electronic configuration: the K shell takes 2, the L shell takes 8, and the remaining 17 − 10 = 7 electrons go into the M shell, giving 2, 8, 7. Valency: the outermost shell has 7 valence electrons, which is more than 4, so valency = 8 − 7 = 1. Element X therefore needs to gain one electron to complete its octet and shows a valency of 1.
State two practical applications of isotopes, one in medicine and one outside medicine.
Answer
In medicine, an isotope of cobalt is used as a source of penetrating radiation to destroy the rapidly dividing cells of a tumour in radiotherapy, and an isotope of iodine is used in the diagnosis and treatment of disorders of the thyroid gland, which naturally absorbs iodine. Outside medicine, an isotope of uranium serves as the fuel in nuclear reactors, where its controlled fission releases energy that is used to generate electricity. Radioactive isotopes are also used to estimate the age of very old rocks and fossils.
State two shortcomings of Rutherford's nuclear model of the atom and explain how Bohr's model overcame the more serious of them.
Answer
Shortcoming 1: The model could not explain the stability of the atom. According to classical theory, a charged particle moving in a circular path is constantly accelerating and must continuously radiate energy; a revolving electron would therefore lose energy, spiral inwards and crash into the nucleus, so the atom would collapse almost instantly. Shortcoming 2: The model said nothing about how the electrons are arranged around the nucleus, so it could not account for the observed line spectra of elements. Bohr resolved the stability problem by proposing that electrons revolve only in certain permitted discrete orbits, called energy levels or shells, and that while an electron remains in one of these shells it does not radiate energy at all. Energy is absorbed or emitted only when an electron jumps from one shell to another, which keeps the atom stable and also explains why the light emitted by an element appears as sharp lines rather than a continuous band.
Describe Rutherford's α-particle scattering experiment. State the three main observations, the conclusion drawn from each, and explain how the resulting model differed from Thomson's model of the atom.
Answer
The experiment: A stream of fast-moving, positively charged α-particles from a radioactive source was directed at an extremely thin sheet of gold foil, only a few hundred atoms thick. A fluorescent screen surrounding the foil produced a tiny flash of light wherever an α-particle struck it, so the direction taken by each particle could be traced. Observation 1: The overwhelming majority of α-particles passed straight through the foil with no deflection at all. Conclusion: most of the space inside an atom is empty, since the particles met nothing to obstruct them. Observation 2: A small fraction of the particles were deflected through small angles. Conclusion: the positive charge of the atom is not spread evenly but is concentrated in a very small region, whose repulsion pushed the positively charged α-particles off course when they passed close to it. Observation 3: A very small number of particles — roughly one in twenty thousand — were deflected through very large angles, some bouncing almost straight back. Conclusion: this tiny central region must be extremely dense and massive, because only a head-on collision with something both heavy and highly charged could reverse a fast α-particle. Rutherford named it the nucleus, and from the rarity of such rebounds he concluded that the nucleus is minute compared with the atom as a whole. Comparison with Thomson's model: Thomson had pictured the atom as a uniform sphere of positive charge with electrons embedded throughout it, so charge and mass were spread evenly over the whole atom. Such an atom could never have deflected an α-particle sharply backwards. Rutherford's model instead placed all the positive charge and nearly all the mass in a tiny central nucleus, with the electrons revolving around it at a relatively enormous distance, leaving the bulk of the atom empty. It was, however, an incomplete picture, since it could not explain why the revolving electrons do not lose energy and collapse into the nucleus.
State the Bohr–Bury rules for the distribution of electrons in shells. Apply them to write the electronic configurations of elements with atomic numbers 12, 16 and 20, and determine the valency of each, explaining your reasoning.
Answer
The Bohr–Bury rules: (1) The maximum number of electrons that the nth shell can hold is 2n², so the K shell (n = 1) holds 2, the L shell (n = 2) holds 8, the M shell (n = 3) holds 18 and the N shell (n = 4) holds 32. (2) Whatever the shell number, the outermost shell of an atom can never contain more than 8 electrons. (3) Shells are filled in a stepwise manner from the innermost outwards, so electrons are not placed in a new shell until the shell before it has received its permitted share. Atomic number 12: There are 12 electrons. K takes 2, L takes 8, and the remaining 2 go into M, giving 2, 8, 2. The outermost shell holds 2 electrons, which is fewer than 4, so the valency equals the number of valence electrons, that is 2. The atom achieves a stable configuration most easily by losing these 2 electrons. Atomic number 16: There are 16 electrons. K takes 2, L takes 8, and the remaining 6 go into M, giving 2, 8, 6. The outermost shell holds 6 valence electrons, which is more than 4, so the valency = 8 − 6 = 2. The atom needs to gain 2 electrons to complete its octet. Atomic number 20: There are 20 electrons. K takes 2 and L takes 8, leaving 10. The M shell could in principle hold 18, but if it were the outermost shell it could not hold 10, since that exceeds the limit of 8. So M takes 8 and the last 2 electrons begin the N shell, giving 2, 8, 8, 2. The outermost shell has 2 valence electrons, so the valency is 2. The pattern to notice is that valency depends only on the outermost shell: elements with 1, 2 or 3 valence electrons tend to lose them, elements with 5, 6 or 7 tend to gain the shortfall, and an atom with a complete outermost shell of 8 (or 2 in the case of helium) has a valency of zero and is chemically inert.
A laboratory sample of a metal Q is found to be a mixture of just two isotopes. One isotope has a mass number of 63 and the other has a mass number of 65, and the number of atoms of the two isotopes in the sample is in the ratio 7 : 3. The atomic number of Q is 29. Answer the following: (a) What are isotopes, and what makes these two atoms isotopes of the same element? (b) Calculate the number of neutrons in each isotope. (c) Calculate the average atomic mass of Q in the sample. (d) Explain why the average atomic mass obtained is not a whole number, even though no single atom of Q has this mass.
Answer
(a) Isotopes are atoms of the same element that have the same atomic number but different mass numbers. Both these atoms have 29 protons and therefore 29 electrons and the same electronic configuration and chemistry; they differ only in the number of neutrons in the nucleus, which is why their mass numbers are 63 and 65. (b) Neutrons = mass number − atomic number. For the lighter isotope, 63 − 29 = 34 neutrons. For the heavier isotope, 65 − 29 = 36 neutrons. (c) The average atomic mass is the weighted mean of the isotopic masses. Out of every 10 atoms, 7 have mass 63 u and 3 have mass 65 u. Average atomic mass = [(63 × 7) + (65 × 3)] ÷ 10 = (441 + 195) ÷ 10 = 636 ÷ 10 = 63.6 u. (d) The value 63.6 u is not the mass of any real atom of Q — every individual atom weighs either 63 u or 65 u. It is a statistical average that takes into account the relative abundance of each isotope in the sample, and since the two isotopes are not present in equal numbers, the weighted mean lands between them at a fractional value. This is exactly why the atomic masses of many elements in the periodic table are not whole numbers.
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