"AKC SCIENCE CLASSES"
CLASS 09 TH (CBSE AND MP BOARD)
CHAPTER 04
STRUCTURE OF THE ATOM
EXERCISES QUESTIONS AND ANSWERS
(PART 01)
Q.01:- Compare the properties of electrons, protons and neutrons.
Ans:-
Q.02:- What are the limitations of J.J Thomson's model of an atom ?
Ans:- Limitations of Thomson's model of an atom
(ⅰ) It could not explain the result of the scattering experiment performed by Rutherford.
(ⅱ) It did not have any experimental evidence in its support.
Q.03:- What are the limitations of Rutherford's model of an atom?
Ans:- The limitations of Rutherford's model of an atoms are :-
(ⅰ) Any charged particle when acceleration is expected to radiate energy. To remain in a circular orbit, the electron would need to undergo acceleration. Therefore, it 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. Therefore, matter would not exist, but we know matter exists. It means that atoms are quite stable.
(ⅱ) Rutherford's model could not explain the distribution of electrons in the extra nuclear portion of the atom.
(ⅲ) It could not explain the stability of an atom when charged electrons are moving under the attractive force of positively charged nucleus.
Q.04:- Describe Bohr's model of an atom.
Ans:- According to Bohr atomic model, a small positively charged nucleus is surrounded by revolving negatively charged electrons is fixed orbits. He concluded that electron will have more energy if it is located away from the nucleus whereas the electrons will have less energy if it located near the nucleus.
The main postulates of the theory are listed :-
- In the extra nuclear portion of an atom, the electrons revolve in well-defined circular orbits.
- These circular orbits are also known as energy levels or energy shells.
- These have been designated as K, L, M, N, O, ..... ( or as 1, 2, 3, 4, 5, .....) based on the energy present.
- The order of the energy of these energy shells is :-
or 1 < 2 < 3 < 4 < 5 <.......
- While revolving in an orbit, the electron is not in a position to either lose or gain energy. In other words, its energy remains stationary. Therefore, these energy states of the electrons are also known as stationary states.
Q.05:- Compare all the proposed models of an atom given in this chapter.
Ans:-
| SL No. | Thomson's model | Rutherford's model | Bohr's model |
|---|---|---|---|
| 1 | An atom consists of a positively charged sphere and the electrons are embedded in it. | An atom consists of a positively charged center in the atom called the nucleus. The mass of the atom is contributed mainly by the nucleus. | Bohr agreed with almost all points as shaid by Rutherford except regarding the revolution of electrons for which he added that there are only certain orbits known as discrete orbits inside the atom in which electrons revolve around the nucleus. |
| 2 | The negative and positive charges are equal in magnitude. As a result the atom is electrically neutral. | The size of the nucleus is very small as compared to the size of the atom. | While revolving in its discrete orbits the electrons do not radiate energy. |
| 3 | The electrons revolve around the nucleus in well-defined orbits. |
Q.06:- Summarise the the rules for writing of distribution of electrons in various shells for the first eighteen elements.
Ans:- (1) The maximum number of electrons present in an orbit is given by formula 2n² , where n is the orbit number counted from the nucleus to outside.
First energy level can have 2n² = 2 × 1² = 2 electrons.
Second energy level can have 2n² = 2 × 2² = 8 electrons.
Third energy level can have 2n = 2 × 3² = 18 electrons.
(2) The maximum number of electron that can be accommodated in the outermost orbit is 8.
Q.07:- Define valency by taking examples of silicon and oxygen.
Ans:- Valency of an element may be defined as the combining capacity of its atom with atoms of other elements in order to acquire 8 electrons (2 in some exceptional cases).
Valency of silicon (Si) :- Atomic number of the elements is 14. Its electronic distribution is ; K(2), L(8), M(4). As silicon atom has four valence electrons, it can lose all of them. At the same time, it can also gain four electrons to have a complete octet. Therefore, the valency of silicon is 4.
Valency of Oxygen (O) :- Atomic number of the element is 8. Its electronic distribution is ; K(2), L(6). As oxygen atom has six valence electrons, it needs two or more electrons to complete its octet (8 - 6 = 2). Therefore, valency of oxygen is 2.
Q.08:- Explain with example
(ⅰ) Atomic number
(ⅱ) Mass number
(ⅲ) Isotopes
(ⅳ) Isobars. Give any two uses of isotopes.
Ans:- (ⅰ) ATOMIC NUMBER :- "The total number of positive charge i.e., the number of protons present in the nucleus of an atom of an element is called atomic number of that element. It is denoted by Z. i.e.,
Atomic No. (Z) = No. of protons (p)
(ⅱ) MASS NUMBER OR ATOMIC MASS :- "The total sum of number of protons and neutrons (i.e., nucleus) present in the nucleus of an atom of an element is called mass or atomic mass of that element". It is denoted by A i.e.,
Mass No. (A) = No. of protons (p) + No. of Neutrons (n)
(ⅲ) ISOTOPES :- "The atoms of various elements which have same atomic number but different mass number (atomic mass) are called isotopes."
Example :- ³⁵₁₇Cl and ³⁷₁₇Cl are isotopes.
(ⅳ) ISOBARS :- "The atoms of various elements which have different atomic number but similar atomic mass (mass number) are called isobars."
Examples :- ⁴⁰₁₈Ar and ⁴⁰₂₀Ca are isobars.
Applications of Isotopes :-
(ⅰ) An isotopes of cobalt is used in the treatment of cancer.
(ⅱ) An isotopes of iodine is used in the treatment of goitre.
(ⅲ) An isotopes of sodium is used in the study of blood circulation.
Q.09:- Na⁺ has completely filled K and L shells. Explain.
Ans:- Na⁺ ion is formed when Na atoms loses one electron from the valence shell i.e., M shell. The Na⁺ ION formed has only two shells which are complete
Na : K(2) L(8) M(1) ; Na⁺ : K(2) L(8)
Q.10:- If bromine atom is available in the form of, say, two isotopes Br (49.7%) and Br (50.3%), calculate the average atomic mass of bromine atom.
Ans:- Percentage of Br-79 isotope = 49.7
Percentage of Br-81 isotope = 50.3
∴ Average atomic mass of bromine
(79 × 49.7) + (81 × 50.3)
= ----------------------------
(49.7 + 50.3)
3926.3 + 4074.3
= --------------------
100
8000.6
= ---------- = 80.OO6 u.
100
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