Learn NCERT Exercise 4 with Free Lessons & Tips

(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

Isotopes are atoms of the same element having the same atomic number, but different mass numbers. For example, hydrogen has three isotopes. They are protium, deuterium,and tritium.

(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, andare isobars.

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.

According to J.J. Thomson’s model of an atom, an atom consists of a positively charged sphere with electrons embedded in it. However, it was later found that the positively charged particles reside at the centre of the atom called the nucleus, and the electrons revolve around the nucleus.

According to Rutherford’s model of an atom, electrons revolve around the nucleus in fixed orbits. But, an electron revolving in circular orbits will not be stable because during revolution, it will experience acceleration. Due to acceleration, the electrons will lose energy in the form of radiation and fall into the nucleus. In such a case, the atom would be highly unstable and collapse.

Bohr’s model of the atom

Niels Bohr proposed the following postulates regarding the model of the atom.

(i) Only certain orbits known as discrete orbits of electrons are allowed inside the atom.

(ii) While revolving in these discrete orbits, the electrons do not radiate energy.

These discrete orbits or shells are shown in the following diagram.

The first orbit (i.e., for n = 1) is represented by letter K. Similarly, for n = 2, it is L − shell, for n = 3, it is M − shell and for n = 4, it is N − shell. These orbits or shells are also called energy levels.

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 in an 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.

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.

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.

It is given that the average atomic mass of the sample of element X is 16.2 u.

Let the percentage of isotope be y%. Thus, the percentage of isotope will be (100 − y) %.

Therefore,

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.

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.

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 times that of proton. (T)

(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine. (F)

Rutherford’s alpha-particle scattering experiment was responsible for the discovery of

Isotopes of an element have

Number of valence electrons in Cl⁻ ion are:

(d) The correct electronic configuration of sodium is 2, 8, 1.