Welcome dear students! Today we are going to learn about Structure_of_the_Atom from Class 9 Science. In the previous chapter, we have learnt that atoms and molecules are the fundamental building blocks of matter. The existence of different kinds of matter is due to different atoms constituting them. Now the questions arise: What makes the atom of one element different from the atom of another element? And are atoms really indivisible, as proposed by Dalton, or are there smaller constituents inside the atom? We shall find out the answers to these questions in this chapter. We will learn about sub-atomic particles and the various models that have been proposed to explain how these particles are arranged within the atom.
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A major challenge before the scientists at the end of the nineteenth century was to reveal the structure of the atom as well as to explain its important properties. The elucidation of the structure of atoms is based on a series of experiments. One of the first indications that atoms are not indivisible, comes from studying static electricity and the condition under which electricity is conducted by different substances. From these activities, can we conclude that on rubbing two objects together, they become electrically charged? Where does this charge come from? This question can be answered by knowing that an atom is divisible and consists of charged particles. Many scientists contributed in revealing the presence of charged particles in an atom. It was known by nineteen hundred that the atom was not an indivisible particle but contained at least one sub-atomic particle, the electron identified by J. J. Thomson. Even before the electron was identified, E. Goldstein in eighteen eighty six discovered the presence of new radiations in a gas discharge and called them canal rays. These rays were positively charged radiations which ultimately led to the discovery of another sub-atomic particle. This sub-atomic particle had a charge, equal in magnitude but opposite in sign to that of the electron. Its mass was approximately two thousand times as that of the electron. It was given the name of proton. In general, an electron is represented as e⁻ and a proton as p⁺. The mass of a proton is taken as one unit and its charge as plus one. The mass of an electron is considered to be negligible and its charge is minus one. It seemed that an atom was composed of protons and electrons, mutually balancing their charges. It also appeared that the protons were in the interior of the atom, for whereas electrons could easily be removed off but not protons. Now the big question was: what sort of structure did these particles of the atom form?
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For understanding the nature of charged particles in matter, let us carry out Activity four point one. Part A: Comb dry hair. Does the comb then attract small pieces of paper? Part B: Rub a glass rod with a silk cloth and bring the rod near an inflated balloon. Observe what happens. The answers to the in-text questions are as follows. Canal rays are positively charged radiations discovered by E. Goldstein in eighteen eighty six during gas discharge experiments. If an atom contains one electron and one proton, it will not carry any charge because the negative charge of the electron and the positive charge of the proton are equal in magnitude and opposite in sign, thus balancing each other out to make the atom electrically neutral.
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We have learnt Dalton's atomic theory, which suggested that the atom was indivisible and indestructible. But the discovery of two fundamental particles, electrons and protons, inside the atom, led to the failure of this aspect of Dalton's atomic theory. It was then considered necessary to know how electrons and protons are arranged within an atom. For explaining this, many scientists proposed various atomic models. J. J. Thomson was the first one to propose a model for the structure of an atom. Thomson proposed the model of an atom to be similar to that of a Christmas pudding. The electrons, in a sphere of positive charge, were like currants in a spherical Christmas pudding. We can also think of a watermelon. The positive charge in the atom is spread all over like the red edible part of the watermelon, while the electrons are studded in the positively charged sphere, like the seeds in the watermelon. In this diagram, we see a large positively charged sphere with negatively charged electrons embedded inside it like seeds. J. J. Thomson proposed that an atom consists of a positively charged sphere and the electrons are embedded in it. The negative and positive charges are equal in magnitude. So, the atom as a whole is electrically neutral. Although Thomson's model explained that atoms are electrically neutral, the results of experiments carried out by other scientists could not be explained by this model.
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Ernest Rutherford was interested in knowing how the electrons are arranged within an atom. Rutherford designed an experiment for this. In this experiment, fast moving alpha particles were made to fall on a thin gold foil. He selected a gold foil because he wanted as thin a layer as possible. This gold foil was about one thousand atoms thick. Alpha particles are doubly-charged helium ions. Since they have a mass of four u, the fast-moving alpha particles have a considerable amount of energy. It was expected that alpha particles would be deflected by the sub-atomic particles in the gold atoms. Since the alpha particles were much heavier than the protons, he did not expect to see large deflections. But, the alpha particle scattering experiment gave totally unexpected results. The following observations were made: Most of the fast moving alpha particles passed straight through the gold foil. Some of the alpha particles were deflected by the foil by small angles. Surprisingly one out of every twelve thousand particles appeared to rebound. In the words of Rutherford, this result was almost as incredible as if you fire a fifteen inch shell at a piece of tissue paper and it comes back and hits you.
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Let us think of an activity in an open field to understand the implications of this experiment. Let a child stand in front of a wall with his eyes closed. Let him throw stones at the wall from a distance. He will hear a sound when each stone strikes the wall. If he repeats this ten times, he will hear the sound ten times. But if a blind-folded child were to throw stones at a barbed-wire fence, most of the stones would not hit the fencing and no sound would be heard. This is because there are lots of gaps in the fence which allow the stone to pass through them. Following a similar reasoning, Rutherford concluded from the alpha particle scattering experiment that most of the space inside the atom is empty because most of the alpha particles passed through the gold foil without getting deflected. Very few particles were deflected from their path, indicating that the positive charge of the atom occupies very little space. A very small fraction of alpha particles were deflected by one hundred eighty degrees, indicating that all the positive charge and mass of the gold atom were concentrated in a very small volume within the atom. From the data he also calculated that the radius of the nucleus is about 10⁵ times less than the radius of the atom.
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On the basis of his experiment, Rutherford put forward the nuclear model of an atom, which had the following features: There is a positively charged centre in an atom called the nucleus. Nearly all the mass of an atom resides in the nucleus. The electrons revolve around the nucleus in circular paths. The size of the nucleus is very small as compared to the size of the atom. However, there are drawbacks to Rutherford's model. The revolution of the electron in a circular orbit is not expected to be stable. Any particle in a circular orbit would undergo acceleration. During acceleration, charged particles would radiate energy. Thus, the revolving electron would lose energy and finally fall into the nucleus. If this were so, the atom should be highly unstable and hence matter would not exist in the form that we know. We know that atoms are quite stable.
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In order to overcome the objections raised against Rutherford's model of the atom, Neils Bohr put forward the following postulates about the model of an atom: Only certain special orbits known as discrete orbits of electrons, are allowed inside the atom. While revolving in discrete orbits the electrons do not radiate energy. These orbits or shells are called energy levels. In the accompanying figure, we see a central nucleus with concentric circular paths around it, representing energy levels. These orbits or shells are represented by the letters K, L, M, N, or the numbers, n equals one, two, three, four, and so on. Let us answer the in-text questions. Based on Thomson's model, the atom is neutral as a whole because the negative and positive charges are equal in magnitude. Based on Rutherford's model, the proton is present in the nucleus of an atom. A sketch of Bohr's model with three shells would show a central nucleus, an innermost K shell, a middle L shell, and an outer M shell, with electrons placed on these circular paths. If the alpha particle scattering experiment is carried out using a foil of a metal other than gold, the observations would be similar, provided the foil is thin enough, as the fundamental atomic structure remains the same, though heavier nuclei might cause slightly more deflections.
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In nineteen thirty two, J. Chadwick discovered another sub-atomic particle which had no charge and a mass nearly equal to that of a proton. It was eventually named as neutron. Neutrons are present in the nucleus of all atoms, except hydrogen. In general, a neutron is represented as n. The mass of an atom is therefore given by the sum of the masses of protons and neutrons present in the nucleus. The three sub-atomic particles of an atom are electrons, protons, and neutrons. For the helium atom question: Helium has an atomic mass of four u and two protons. Since mass number equals protons plus neutrons, four equals two plus neutrons, so it has two neutrons.
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The distribution of electrons into different orbits of an atom was suggested by Bohr and Bury. The following rules are followed for writing the number of electrons in different energy levels or shells: The maximum number of electrons present in a shell is given by the formula 2n², where n is the orbit number or energy level index, one, two, three, and so on. Hence the maximum number of electrons in different shells are as follows: first orbit or K shell will be 2 × 1² = 2, second orbit or L shell will be 2 × 2² = 8, third orbit or M shell will be 2 × 3² = 18, fourth orbit or N shell will be 2 × 4² = 32. The maximum number of electrons that can be accommodated in the outermost orbit is eight. Electrons are not accommodated in a given shell, unless the inner shells are filled. That is, the shells are filled in a step-wise manner. In the schematic figure, we see the atomic structures of the first eighteen elements from Hydrogen to Argon, showing their nuclei and electron shells with specific electron counts. Activity four point two instructs you to make a static atomic model displaying electronic configuration of the first eighteen elements using the composition data.
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Let us go through the composition of atoms of the first eighteen elements. Hydrogen has atomic number one, one proton, zero neutrons, one electron, configuration K one, valency one. Helium has atomic number two, two protons, two neutrons, two electrons, configuration K two, valency zero. Lithium has atomic number three, three protons, four neutrons, three electrons, configuration K two L one, valency one. Beryllium has atomic number four, four protons, five neutrons, four electrons, configuration K two L two, valency two. Boron has atomic number five, five protons, six neutrons, five electrons, configuration K two L three, valency three. Carbon has atomic number six, six protons, six neutrons, six electrons, configuration K two L four, valency four. Nitrogen has atomic number seven, seven protons, seven neutrons, seven electrons, configuration K two L five, valency three. Oxygen has atomic number eight, eight protons, eight neutrons, eight electrons, configuration K two L six, valency two. Fluorine has atomic number nine, nine protons, ten neutrons, nine electrons, configuration K two L seven, valency one. Neon has atomic number ten, ten protons, ten neutrons, ten electrons, configuration K two L eight, valency zero.
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Continuing the table: Sodium has atomic number eleven, eleven protons, twelve neutrons, eleven electrons, configuration K two L eight M one, valency one. Magnesium has atomic number twelve, twelve protons, twelve neutrons, twelve electrons, configuration K two L eight M two, valency two. Aluminium has atomic number thirteen, thirteen protons, fourteen neutrons, thirteen electrons, configuration K two L eight M three, valency three. Silicon has atomic number fourteen, fourteen protons, fourteen neutrons, fourteen electrons, configuration K two L eight M four, valency four. Phosphorus has atomic number fifteen, fifteen protons, sixteen neutrons, fifteen electrons, configuration K two L eight M five, valency three, five. Sulphur has atomic number sixteen, sixteen protons, sixteen neutrons, sixteen electrons, configuration K two L eight M six, valency two. Chlorine has atomic number seventeen, seventeen protons, eighteen neutrons, seventeen electrons, configuration K two L eight M seven, valency one. Argon has atomic number eighteen, eighteen protons, twenty two neutrons, eighteen electrons, configuration K two L eight M eight, valency zero. The distribution of electrons in carbon is two, four. In sodium it is two, eight, one. If K and L shells of an atom are full, the total number of electrons would be two plus eight, which equals ten electrons.
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We have learnt how the electrons in an atom are arranged in different shells. The electrons present in the outermost shell of an atom are known as the valence electrons. From the Bohr Bury scheme, we also know that the outermost shell of an atom can accommodate a maximum of eight electrons. It was observed that the atoms of elements, completely filled with eight electrons in the outermost shell show little chemical activity. In other words, their combining capacity or valency is zero. Of these inert elements, the helium atom has two electrons in its outermost shell and all other elements have atoms with eight electrons in the outermost shell. The combining capacity of the atoms of elements, that is, their tendency to react and form molecules with atoms of the same or different elements, was thus explained as an attempt to attain a fully-filled outermost shell. An outermost-shell, which had eight electrons was said to possess an octet. Atoms would thus react, so as to achieve an octet in the outermost shell. This was done by sharing, gaining or losing electrons. The number of electrons gained, lost or shared so as to make the octet of electrons in the outermost shell, gives us directly the combining capacity of the element, that is, the valency discussed in the previous chapter.
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For example, hydrogen, lithium, and sodium atoms contain one electron each in their outermost shell, therefore each one of them can lose one electron. So, they are said to have valency of one. The valency of magnesium and aluminium is two and three, respectively, because magnesium has two electrons in its outermost shell and aluminium has three electrons in its outermost shell. If the number of electrons in the outermost shell of an atom is close to its full capacity, then valency is determined in a different way. For example, the fluorine atom has seven electrons in the outermost shell, and its valency could be seven. But it is easier for fluorine to gain one electron instead of losing seven electrons. Hence, its valency is determined by subtracting seven electrons from the octet and this gives you a valency of one for fluorine. Valency can be calculated in a similar manner for oxygen. Therefore, an atom of each element has a definite combining capacity, called its valency. To find the valency of chlorine, sulphur and magnesium: Chlorine has seven valence electrons, so it gains one, valency is one. Sulphur has six, gains two, valency is two. Magnesium has two, loses two, valency is two.
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We know that protons are present in the nucleus of an atom. It is the number of protons of an atom, which determines its atomic number. It is denoted by Z. All atoms of an element have the same atomic number, Z. In fact, elements are defined by the number of protons they possess. For hydrogen, Z equals one, because in hydrogen atom, only one proton is present in the nucleus. Similarly, for carbon, Z equals six. Therefore, the atomic number is defined as the total number of protons present in the nucleus of an atom. After studying the properties of the sub atomic particles of an atom, we can conclude that mass of an atom is practically due to protons and neutrons alone. These are present in the nucleus of an atom. Hence protons and neutrons are also called nucleons. Therefore, the mass of an atom resides in its nucleus. For example, mass of carbon is twelve u because it has six protons and six neutrons, six u plus six u equals twelve u. Similarly, the mass of aluminium is twenty seven u, which is thirteen protons plus fourteen neutrons. The mass number is defined as the sum of the total number of protons and neutrons present in the nucleus of an atom. It is denoted by A. In the notation for an atom, the atomic number, mass number and symbol of the element are to be written as: Mass number A at the top left, Atomic number Z at the bottom left, and the symbol of the element X in the middle. For example, nitrogen is written as ¹⁴₇N.
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Let us answer the questions for this section. If number of electrons in an atom is eight and number of protons is also eight, then the atomic number of the atom is eight, and the charge on the atom is zero because the positive and negative charges balance. With the help of the table, the mass number of oxygen atom is sixteen, which is eight protons plus eight neutrons. The mass number of sulphur atom is thirty two, which is sixteen protons plus sixteen neutrons. In nature, a number of atoms of some elements have been identified, which have the same atomic number but different mass numbers. For example, take the case of hydrogen atom, it has three atomic species, namely protium, deuterium or D, and tritium or T. The atomic number of each one is one, but the mass number is one, two and three, respectively. Other such examples are carbon, carbon twelve and carbon fourteen, and chlorine, chlorine thirty five and chlorine thirty seven. On the basis of these examples, isotopes are defined as the atoms of the same element, having the same atomic number but different mass numbers. Therefore, we can say that there are three isotopes of hydrogen atom, namely protium, deuterium and tritium.
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Many elements consist of a mixture of isotopes. Each isotope of an element is a pure substance. The chemical properties of isotopes are similar but their physical properties are different. Chlorine occurs in nature in two isotopic forms, with masses thirty five u and thirty seven u in the ratio of three to one. Obviously, the question arises: what should we take as the mass of chlorine atom? Let us find out. The average atomic mass of chlorine atom, on the basis of above data, will be (35 × 75/100 + 37 × 25/100) = (105/4 + 37/4) = 142/4 = 35.5 u. The mass of an atom of any natural element is taken as the average mass of all the naturally occurring atoms of that element. If an element has no isotopes, then the mass of its atom would be the same as the sum of protons and neutrons in it. But if an element occurs in isotopic forms, then we have to know the percentage of each isotopic form and then the average mass is calculated. This does not mean that any one atom of chlorine has a fractional mass of thirty five point five u. It means that if you take a certain amount of chlorine, it will contain both isotopes of chlorine and the average mass is thirty five point five u.
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Applications of isotopes are important. An isotope of uranium is used as a fuel in nuclear reactors. An isotope of cobalt is used in the treatment of cancer. An isotope of iodine is used in the treatment of goitre. Let us consider two elements, calcium, atomic number twenty, and argon, atomic number eighteen. The number of protons in these atoms is different, but the mass number of both these elements is forty. That is, the total number of nucleons is the same in the atoms of this pair of elements. Atoms of different elements with different atomic numbers, which have the same mass number, are known as isobars. For the symbol H, D and T, the three sub-atomic particles in each are: Protium H has one proton, zero neutrons, one electron. Deuterium D has one proton, one neutron, one electron. Tritium T has one proton, two neutrons, one electron. For the electronic configuration of a pair of isotopes, take carbon twelve and carbon fourteen. Both have atomic number six, so configuration is two, four. For isobars, take calcium forty and argon forty. Calcium has atomic number twenty, configuration two, eight, eight, two. Argon has atomic number eighteen, configuration two, eight, eight.
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Let us summarize what you have learnt. Credit for the discovery of electron and proton goes to J. J. Thomson and E. Goldstein, respectively. J. J. Thomson proposed that electrons are embedded in a positive sphere. Rutherford's alpha particle scattering experiment led to the discovery of the atomic nucleus. Rutherford's model of the atom proposed that a very tiny nucleus is present inside the atom and electrons revolve around this nucleus. The stability of the atom could not be explained by this model. Neils Bohr's model of the atom was more successful. He proposed that electrons are distributed in different shells with discrete energy around the nucleus. If the atomic shells are complete, then the atom will be stable and less reactive. J. Chadwick discovered presence of neutrons in the nucleus of an atom. So, the three sub-atomic particles of an atom are electrons, protons and neutrons. Electrons are negatively charged, protons are positively charged and neutrons have no charges. The mass of an electron is about 1/2000 times the mass of an hydrogen atom. The mass of a proton and a neutron is taken as one unit each. Shells of an atom are designated as K, L, M, N. Valency is the combining capacity of an atom. The atomic number of an element is the same as the number of protons in the nucleus of its atom. The mass number of an atom is equal to the number of nucleons in its nucleus. Isotopes are atoms of the same element, which have different mass numbers. Isobars are atoms having the same mass number but different atomic numbers. Elements are defined by the number of protons they possess.
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Now, we will solve the exercises completely. Question one: Compare the properties of electrons, protons and neutrons. Electrons are negatively charged, have negligible mass compared to protons and neutrons, and are found outside the nucleus. Protons are positively charged, have a mass of one unit, and are located in the nucleus. Neutrons have no charge, have a mass of one unit, and are located in the nucleus. Question two: What are the limitations of J. J. Thomson's model of the atom? It could not explain the results of the alpha particle scattering experiment. It failed to account for the stability of the atom and did not propose a concentrated nucleus. Question three: What are the limitations of Rutherford's model of the atom? It could not explain the stability of the atom. According to classical physics, an accelerating charged electron should radiate energy, lose speed, and spiral into the nucleus, making the atom unstable, which contradicts reality.
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Question four: Describe Bohr's model of the atom. Bohr proposed that electrons revolve around the nucleus in certain discrete orbits without radiating energy. These orbits are called energy levels or shells, represented by K, L, M, N or n equals one, two, three, four. Electrons only jump between these fixed orbits by absorbing or emitting specific amounts of energy. Question five: Compare all the proposed models of an atom given in this chapter. Thomson's model described a uniform positive sphere with embedded electrons, explaining neutrality but failing scattering results. Rutherford's model introduced a dense, positively charged nucleus with orbiting electrons, explaining scattering but failing stability. Bohr's model added fixed energy levels where electrons do not radiate energy, successfully explaining stability and spectral lines. Question six: Summarise the rules for writing of distribution of electrons in various shells for the first eighteen elements. The maximum electrons in a shell is 2n². The outermost shell holds a maximum of eight electrons. Shells are filled step-wise, meaning inner shells must be full before filling outer ones.
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Question seven: Define valency by taking examples of silicon and oxygen. Valency is the combining capacity of an atom, determined by the number of electrons it needs to lose, gain, or share to achieve an octet. Silicon has four electrons in its outermost shell, so it shares four electrons, giving it a valency of four. Oxygen has six electrons in its outermost shell, so it gains two electrons to complete its octet, giving it a valency of two. Question eight: Explain with examples Atomic number, Mass number, Isotopes and Isobars. Give any two uses of isotopes. Atomic number is the total number of protons, for example carbon has atomic number six. Mass number is the sum of protons and neutrons, for example carbon twelve has mass number twelve. Isotopes are atoms of the same element with same atomic number but different mass numbers, for example carbon twelve and carbon fourteen. Isobars are atoms of different elements with same mass number but different atomic numbers, for example calcium forty and argon forty. Two uses of isotopes: uranium isotope for nuclear reactors, and iodine isotope for treating goitre.
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Question nine: Na plus has completely filled K and L shells. Explain. Sodium atom has atomic number eleven, so its configuration is two, eight, one. When it loses one electron to form Na plus ion, it has ten electrons. These ten electrons are arranged as two in the K shell and eight in the L shell. Both shells are now completely filled. Question ten: If bromine atom is available in the form of two isotopes, bromine seventy nine over thirty five at forty nine point seven percent and bromine eighty one over thirty five at fifty point three percent, calculate the average atomic mass of bromine atom. The calculation is: (79 multiplied by 49.7 divided by 100) plus (81 multiplied by 50.3 divided by 100) equals 39.263 plus 40.743 equals 80.006 u, which is approximately 80 u. Question eleven: The average atomic mass of a sample of an element X is sixteen point two u. What are the percentages of isotopes sixteen over eight X and eighteen over eight X in the sample? Let the percentage of X sixteen be y. Then percentage of X eighteen is one hundred minus y. The equation is: sixteen times y over one hundred plus eighteen times one hundred minus y over one hundred equals sixteen point two. Solving gives sixteen y plus one thousand eight hundred minus eighteen y equals one thousand six hundred twenty. Minus two y equals minus one hundred eighty. So y equals ninety. Therefore, X sixteen is ninety percent and X eighteen is ten percent.
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Question twelve: If Z equals three, what would be the valency of the element? Also, name the element. Atomic number three means three electrons. Configuration is two, one. The outermost shell has one electron, so it loses one to be stable. Valency is one. The element is Lithium. Question thirteen: Composition of the nuclei of two atomic species X and Y are given as: X has six protons and six neutrons. Y has six protons and eight neutrons. Give the mass numbers of X and Y. What is the relation between the two species? Mass number of X is six plus six equals twelve. Mass number of Y is six plus eight equals fourteen. Since they have the same atomic number but different mass numbers, they are isotopes. Question fourteen: For the following statements, write T for True and F for False. (a) J. J. Thomson proposed that the nucleus of an atom contains only nucleons. False. (b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral. False. (c) The mass of an electron is about 1/2000 times that of proton. True. (d) An isotope of iodine is used for making tincture iodine, which is used as a medicine. False.
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Question fifteen: Rutherford's alpha particle scattering experiment was responsible for the discovery of Atomic Nucleus. Tick against option a. Question sixteen: Isotopes of an element have different number of neutrons. Tick against option c. Question seventeen: Number of valence electrons in Cl minus ion are eight. Chlorine has seventeen electrons, configuration two, eight, seven. Gaining one electron makes it Cl minus with eighteen electrons, configuration two, eight, eight. The outermost shell has eight electrons. Tick against option b. Question eighteen: Which one of the following is a correct electronic configuration of sodium? Sodium has eleven electrons: two, eight, one. Tick against option d. Question nineteen: Complete the following table. First row: Atomic number nine, neutrons ten, so mass number is nineteen. Protons nine, electrons nine. Species is Fluorine. Second row: Atomic number sixteen, mass number thirty two, neutrons sixteen, protons sixteen, electrons sixteen. Species is Sulphur. Third row: Mass number twenty four, protons twelve, so atomic number twelve. Neutrons twelve, electrons twelve. Species is Magnesium. Fourth row: Mass number two, protons one, so atomic number one. Neutrons one, electrons one. Species is Deuterium. Fifth row: Mass number one, protons one, neutrons zero, electrons zero. Species is Hydrogen ion.
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This completes our comprehensive coverage of the chapter. We have explored the historical development of atomic models, understood the properties and arrangements of sub-atomic particles, mastered electron distribution rules, and clarified concepts like valency, atomic number, mass number, isotopes, and isobars. Remember to practice writing electronic configurations and calculating atomic masses, as these are frequently tested in examinations. Review the definitions carefully, as they must be written exactly as stated in your textbook. Keep practicing the numerical problems on isotopes and mass calculations to build confidence.
Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]