Chapter 4 Structure Of The Atom Solutions
Question - 21 : - Summarize the rules for writing of distribution of electrons in various shells for the first eighteen elements.
Answer - 21 : -
The rules for writing of the distribution of electrons in various shells for the first eighteen elements are given below.
(i) The maximum number of electrons that a shell can accommodate is given by the formula ‘2n2’, where ‘n’ is the orbit number or energy level index (n = 1, 2, 3…).
The maximum number of electrons present inan orbit of n =1 is given by 2n2 = 2×12 =2
Similarly, for second orbit, it is 2n2 =2×22 = 8
For third orbit, it is 2n2 =2×32 = 18
And so on……
(ii) The outermost orbit can be accommodated by a maximum number of 8 electrons.
(iii) Shells are filled with electrons in a stepwise manner i.e., the outer shell is not occupied with electrons unless the inner shells are completely filled with electrons.
Define valency by taking examples of silicon and oxygen.
Answer - 22 : -
The valency of an element is the combining capacity of that element. The valency of an element is determined by the number of valence electrons present in the atom of that element.
If the number of valence electrons of the atom of an element is less than or equal to four, then the valency of that element is equal to the number of valence electrons. For example, the atom of silicon has four valence electrons. Thus, the valency of silicon is four.
On the other hand, if the number of valence electrons of the atom of an element is greater than four, then the valency of that element is obtained by subtracting the number of valence electrons from eight. For example, the atom of oxygen has six valence electrons. Thus, the valency of oxygen is (8 − 6) i.e., two.
Question - 23 : - Explain with examples (i) Atomic number, (ii) Mass number, (iii) Isotopes and (iv) Isobars. Give any two uses of isotopes.
Answer - 23 : -
(i) Atomic number
The atomic number of an element is the total number of protons present in the atom of that element. For example, nitrogen has 7 protons in its atom. Thus, the atomic number of nitrogen is 7.
(ii) Mass number
The mass number of an element is the sum of the number of protons and neutrons present in the atom of that element. For example, the atom of boron has 5 protons and 6 neutrons. So, the mass number of boron is 5 + 6 = 11.
(iii) Isotopes
(iv) Isobars
Isobars are atoms having the same mass number, but different atomic numbers i.e., isobars are atoms of different elements having the same mass number.
For example, 
and
areisobars.Two uses of isotopes are:
(i) One isotope of uranium is used as a fuel in nuclear reactors.
(ii) One isotope of cobalt is used in the treatment of cancer.
Na+ has completely filled K and L shells. Explain.
Answer - 24 : -
An atom of Na has a total of 11 electrons. Its electronic configuration is 2, 8, 1. But, Na+ ion has one electron less than Na atom i.e., it has 10 electrons. Therefore, 2 electrons go to K-shell and 8 electrons go to L-shell, thereby completely filling K and L shells.
Question - 25 : - If bromine atom isavailable in the form of, say, two isotopes 
(49.7%) and 
(50.3%), calculate the average atomic mass of bromineatom.
Answer - 25 : - It is given that twoisotopes of bromine are (49.7%) and 
(50.3%). Then, the average atomic mass of bromine atom isgiven by:
Question - 26 : - The average atomic mass of a sample of an element X is16.2 u. What are the percentages of isotopes 
and 
in the sample?
Answer - 26 : -
It is given that the average atomic mass of the sample ofelement X is 16.2 u.
Let the percentage of isotope be y%. Thus, the percentage ofisotope will be (100 − y) %.Therefore,
Therefore, the percentageof isotopeis 10%.
And, the percentage of isotopeis (100 − 10) % = 90%.
Question - 27 : - If Z = 3, what would be the valency of the element? Also, name the element.
Answer - 27 : -
By Z = 3, we mean that the atomic number of the element is 3. Its electronic configuration is 2, 1. Hence, the valency of the element is 1 (since the outermost shell has only one electron).
Therefore, the element with Z = 3 is lithium.
Question - 28 : - Composition of the nuclei of two atomic species X and Y are given as under
X Y
Protons = 6 6
Neutrons = 6 8
Give the mass numbers of X and Y. What is the relation between the two species?
Answer - 28 : -
Mass number of X = Number of protons + Number of neutrons
= 6 + 6
= 12
Mass number of Y = Number of protons + Number of neutrons
= 6 + 8
= 14
These two atomic species X and Y have the same atomic number, but different mass numbers. Hence, they are isotopes.
Question - 29 : - 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 1/2000times that of proton.
(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.
Answer - 29 : -
(a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons. (F)
(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral. (F)
(c) The mass of an electron is about 1/2000 times that of proton. (T)
(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine. (F)
Question - 30 : - Put tick (
) against correct choice and cross (
) against wrong choice in the following question:Rutherford’s alpha-particle scattering experiment was responsible for the discovery of
(a) Atomic nucleus
(b) Electron
(c) Proton
(d) Neutron
Answer - 30 : -
Rutherford’s alpha-particle scattering experiment was responsible for the discovery of
(a) Atomic nucleus
(b) Electron (c) Proton
(d) Neutron