Question 1: Compare the properties of electrons, protons and neutrons.
|It has -1 charge.||It has +1 charge||It has no charge.|
|Mass is almost zero.||Mass is 1 u||Mass is 1 u|
|Remains outside the nucleus.||Remains in nucleus.||Remains in nucleus.|
Question 2: What are the limitations of J.J. Thomson’s model of the atom?
Answer: Thomson’s model could explain the electrical neutrality of atoms. But results of other experiments could not be explained by Thomson’s model. For example; it could not explain the results of Rutherford’s alpha-scattering experiment. Moreover, Thomson’s model could not be verified experimentally.
Question 3: What are the limitations of Rutherford’s model of the atom?
Answer: Limitations of Rutherford’s model of atom:
When an object is in circular motion, it tends to lose energy. in the event of losing energy, the diameter of circular path (on which the object is moving) keeps in decreasing. Finally, a time comes when the motion stops, and the object falls in centre. This means that electrons should eventually fall in the nucleus; leading to the collapse of the atom. But atoms do not collapse in real life and are stable in nature. Rutherford’s model could not explain this.
Question 4: Describe Bohr’s model of the atom.
Answer: Bohr’s model of atom:
- Electrons revolve round the nucleus in a fixed orbit.
- He called these orbits as ‘stationary orbit’.
- Each stationary orbit is associated with fixed amount of energy, thus electrons do not radiate energy as long as they keep on revolving around the nucleus in fixed orbit.
Question 5: Compare all the proposed models of an atom given in this chapter.
|Thomson’s Model||Rutherford’s Model||Bohr’s Model|
|Atom is a positively charged sphere in which negatively charged electrons are embedded.||Most of the atom is hollow with positively charged nucleus at centre. Electrons revolve around nucleus.||Same as Rutherford’ model on most aspects. Gave the concept of stationary orbits for electrons. Electrons do not lose energy while moving in their fixed orbits.|
Question 6: Summarise the rules for writing of distribution of electrons in various shells for the first eighteen elements.
Answer: Following are the rules for writing the distribution of electrons in various shells:
Maximum number of electrons in a shell is given by 2n2.
So, maximum number of electrons in shells K, L and M are respectively 2, 8 and 18.
Question 7: Define valency by taking examples of silicon and oxygen.
Answer: Combining capacity of an atom of an element is called valency. Electronic configuration of silicon and oxygen are as follows:
Si (14): 2, 8, 4
O (8): 2, 6
There are four electrons in the outermost shell of silicon. It means that silicon can either gain or lose 4 electrons to make a compound. Hence, valency of silicon is 4.
There are 6 electrons in the outermost shell of oxygen. It means that oxygen can gain 2 electrons to complete an octet in its outermost shell. Giving up 6 electrons would be more energy consuming. Hence, valency of oxygen is 2.
Question 8: Explain with examples (i) Atomic number, (ii) Mass number, (iii) Isotopes and iv) Isobars. Give any two uses of isotopes.
Answer: Atomic number is equal to the number of protons present in an atom. Since an atom is electrically neutral, hence number of electrons is equal to number of protons.
So, Atomic Number = Number of protons = Number of electrons
Atomic number of carbon is 6; which means that there are 6 protons and 6 electrons in an atom of carbon.
Mass Number: Mass number of an atom is defined as the sum of the number of protons and number of neutrons. Mass number is nearly equal to the atomic mass of an atom.
Mass number of an atom = Number of protons + Number of neutrons
Example: Atomic mass of carbon is 12; which means there are 6 protons and 6 neutrons in an atom of carbon.
Question 9: Na+ has completely filled K and L shells. Explain.
Answer: Electronic configuration of Na (11): 2, 8, 1
This means that Na+ has 10 electrons because this ion is formed as a result of loss of 1 electron by sodium. Hence, electronic configuration of Na+ can be written as follows: 2, 8. This shows that Na+ has completely filled K and L shells.
Question 10: If bromine atom is available in the form of, say, two isotopes 7935Br (49.7%) and 8135 Br (50.3%), calculate the average atomic mass of bromine atom.
Answer: Average atomic mass can be calculated as follows:
`(79 xx 49.7 + 81 xx 50.3)/100`
`= (3926.3 + 4074.3)/100 = 8000.6/100 = 80.006 u`
Question 11: The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes 168X and 188X in the sample?
Answer: Let us assume that percentage of 168X is A and percentage of 188X is (100 – A).
Then, average atomic mass can be given as follows:
`(16 xx A + 18 xx (100 – A))/100 = 16.2`
Or, `16A + 1800 – 18A = 1620`
Or, `1800 – 2A = 1620`
Or, `2A = 1800 – 1620 = 180`
Or, `A = 90`
So, `100 – A = 100 – 90 = 10`
Hence, percentages are respectively 90% and 10%
Question 12: If Z = 3, what would be the valency of the element? Also, name the element.
Answer: Electronic configuration for element with Z = 3 is 2, 1
Hence, valency = 1
This element is Lithium.
Question 13: Composition of the nuclei of two atomic species X and Y are given as under
X: Protons (6), Neutrons (6)
Y: Protons (6), Neutrons (8)
Give the mass numbers of X and Y. What is the relation between the two species?
Answer: Mass number of X = number of protons + number of neutrons `= 6 + 6 = 12`
Mass number of `Y = 6 + 8 = 14`
Since number of protons, i.e. atomic number is same for both species, they are isotopes.
Question 14: 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.
(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.
(c) The mass of an electron is about 1/2000 times that of proton.
(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.
Answer: (a) F, (b) F, (c) T, (d) T
Question 15: Rutherford’s alpha-particle scattering experiment was responsible for the discovery of
(a) Atomic Nucleus
Answer: (a) Atomic nucleus