22. Define valency by taking examples of silicon and oxygen.
Solution:
The definite combining capacity of the atoms of each element, wherein electrons are lost, gained or shared to make the octet of electrons present in the outermost shell is defined as valency. To measure valency, we can figure out the number of electrons that are required to complete the shell in which it is contained or losing excess electrons if present, once the filling is complete.
Example : To find the valency of silicon:
The atomic number of silicon is 14
Number of electrons is equal to the number of protons in silicon i.e., 14
The distribution of electrons in silicon atom is K – 2, L – 8, M – 4
Hence, from the distribution of silicon it is clearly evident that to fill the M shell 4 electrons are required. Therefore its valency is 8 - 4 = 4.
To find the valency of oxygen:
The atomic number of oxygen is 8
Number of electrons is equal to the number of protons in oxygen i.e., 8
The distribution of electrons in oxygen atom is K – 2, L – 6
Hence, from the distribution of oxygen it is clearly evident that to fill the M shell 6 more electrons are required. Therefore its valency is 8 - 6 = 2.
23. Explain with examples
(i) Atomic number,
(ii) Mass number,
(iii) Isotopes and
(iv) Isobars.
Give any two uses of isotopes.
Solution:
(i) The number of positively charged protons present in the nucleus of an atom is defined as the atomic number and is denoted by Z. Example: Hydrogen has one proton in its nucleus, hence its atomic number is one.
(ii) The total number of protons and neutrons present in the nucleus of an atom is known as the mass number. It is denoted by A. 20Ca40 . Mass number is 40. Atomic number is 20.
(iii) The atoms which have the same number of protons but different number of neutrons are referred to as isotopes. Hence the mass number varies.
Example: The most simple example is the Carbon molecule which exists as 6C12 and 6C14
(iv) Isobars: Isobars are atoms which have the same mass number but differ in the atomic number.
Examples are, 20Ca40and 18Ar40
Uses of isotopes:
- The isotope of Iodine atom is used to treat goitre and iodine deficient disease.
- In the treatment of cancer, an isotope of cobalt is used.
- Fuel for nuclear reactors is derived from the isotopes of the Uranium atom.
24. Na+ has completely filled K and L shells. Explain.
Solution:
The atomic number of sodium is 11. It has 11 electrons in its orbitals wherein the number of protons is equal to the number of electrons. Hence, its electronic configuration is K-2 ; L-8 ; M-1 ; The one electron in the M shell is lost and it obtains a positive charge since it has one more proton than electrons, and obtains a positive charge, Na+ . The new electronic configuration is K-1 ; L-8 which is the filled state. Hence it is very difficult to eliminate the electron from a filled state as it is very stable.
25. If bromine atom is available in the form of, say, two isotopes 35Br79 (49.7%) and 35Br81 (50.3%), calculate the average atomic mass of Bromine atom.
Solution:
The atomic mass of an element is the mass of one atom of that element. Average atomic mass takes into account the isotopic abundance.
Isotope of bromine with atomic mass 79 u = 49.7%
Therefore, Contribution of 35Br79 to atomic mass = (79 × 49.7)/100
⇒ 39.26 u
Isotope of bromine with atomic mass 81 u = 50.3%
Contribution of 35Br81 to the atomic mass of bromine = (81 × 50.3)/100
⇒ 40.64u
Hence, average atomic mass of the bromine atom = 39.26 + 40.64 u = 79.9u
26. The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes 8X16 and 8X18 in the sample?
Solution:
Let the percentage of 8X16 be ‘a’ and that of 8X18 be ‘100-a’.
As per given data,
16.2u = 16 a / 100 + 18 (100-a) /100
1620 = 16a + 1800 – 18a
1620 = 1800 – 2a
a = 90%
Hence, the percentage of isotope in the sample 8X16 is 90% and that of
8X18 = 100 - a = 100 - 90 = 10%
27. If Z = 3, what would be the valency of the element? Also, name the element.
Solution:
Given: Atomic number, Z = 3
The electronic configuration of the element = K-2; L-1, hence its valency = 1
The element with atomic number 3 is Lithium.
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?
Solution:
Mass number of X: Protons + neutrons = 6 + 6 = 12
Mass number of Y: Protons + neutrons = 6 + 8 = 14
They are the same element as their atomic numbers are the same.
They are isotopes as they differ in the number of neutrons and hence their mass numbers.
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 1/2000 times that of proton.
(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.
Solution:
(a) Statement is False
(b) Statement is False
(c) Statement is True
(d) Statement is False
30. Put a tick(✓) against correct choice and cross(x) against wrong choice in questions 15, 16 and 17.
Rutherford’s alpha – particle scattering experiment was responsible for the discovery of
(a) Atomic nucleus
(b) Electron
(c) Proton
(d) Neutron
Solution:
(a) Atomic nucleus
31. Isotopes of an element have
(a) The same physical properties
(b) Different chemical properties
(c) Different number of neutrons
(d) Different atomic numbers.
Solution:
(c) Different number of neutrons
32. Number of valence electrons in Cl– ion are:
(a) 16
(b) 8
(c) 17
(d) 18
Solution:
(b) 8
Electronic distribution of Cl is K-2, L-8, M-7. Valence electrons are 7, hence chlorine gains one electron for the formation of Cl–. Therefore, its valency is 8.