NCERT SOLUTIONS Class 9 Chapter 4

Question 1:

Compare the properties of electrons, protons and neutrons.

Answer:

Electron

Proton

Neutron

  1. Electrons are present outside the nucleus.

Protons are present in the nucleus.

Neutrons are present in the nucleus of the atom.

  1. They are negatively charged.

They are positively charged.

They are neutral in charge.

  1. Mass of electrons is negligible.

The mass of a proton is approximately 2000 times as the mass of an electron. The mass of a neutron is nearly equal to the mass of a proton.

Question 2:

What are the limitations of J.J. Thomson’s model of the atom?

Answer:

According to J.J. Thomson’s model, an atom consists of a positively charged sphere with electrons embedded in it.

This model did not have any experimental evidence. Therefore, it was opposed by his co-scientists and was rejected.

However, his prediction that an atom is electrically neutral and has no net charge is still accepted.

Question 3:

What are the limitations of Rutherford’s model of the atom?

Answer:

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 the 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. But we know that the atom is highly stable.

Question 4:

Describe Bohr’s model of the atom.

Answer:

In order to overcome the objections raised against Rutherford’s model of the atom, Neil’s Bohr put forward the following postulates about the model of an atom:

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

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

Class_9_Chemistry_Structure_Of_Atom_Orbits

These orbits or shells are called energy levels.

These orbits or shells are represented by the letters K, L, M, N… or the numbers, n=1, 2, 3, 4…

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.

Question 5:

Compare all the proposed models of an atom given in this chapter.

Answer:

Thompson’s Model-

  1. An atom consists of a sphere of positive charge with negatively charged electrons embed in it.
  2. The positive and the negative charges in an atom are equal in magnitude due to which an atom is electrically neutral.

Rutherford’s Model-

  1. There is a positively charged, highly dense centre in an atom called the nucleus. Nearly the whole mass of an atom resides in the nucleus.
  2. The electrons revolve around the nucleus in circular path.

Bohr’s Model

  1. An atom is made up of 3 particles electrons, protons and neutrons. Due to the presence of equal number of electrons and protons, atom on the whole is electrically neutral.
  2. The protons and neutrons are located inside the nucleus due to which nucleus is positively charged.

3.The size of the nucleus is very small compared to the size of an atom.

  1. The electrons revolve around the nucleus in fixed paths called orbits or shells or energy levels. The energy levels are represented as K, L, M, N or 1, 2, 3, 4, 5,
  2. There is a limit to the number of electrons each energy level can hold. It is given by the formula 2n2 where n is the orbit number.
  3. Each energy level is associated with fixed amount of energy, the shell nearest to the nucleus having minimum amount of energy and the one farthest having maximum energy.
  4. There is no change in the energy of electrons as they keeping revolving in the same shell and the atom remains stable.

Question 6:

Summarise the rules for writing of distribution of electrons in various shells for the first eighteen elements.

Answer:

The following rules are followed for writing the number of electrons in different energy levels or shells:

(i) The maximum number of electrons present in a shell is given by the formula 2n2, where ‘n’ is the orbit number or energy level index, 1, 2, 3 …

Hence the maximum number of electrons in different shells are as follows:

First orbit or K-shell will be =2n2 = 2 × 12 = 2, (n =1)

Second orbit or L-shell will be = 2n2 =2 × 22 = 8, (n=2)

Third orbit or M-shell will be = 2n2 =2 × 32 = 18, (n=3)

Fourth orbit or N-shell will be = 2n2 =2 × 42= 32,(n=4)  and so on.

(ii) The maximum number of electrons that can be accommodated in the outermost orbit is 8.

(iii) Electrons are not accommodated in a given shell, unless the inner shells are filled. That is, the shells are filled in a step-wise manner.

Question 7:

Define valency by taking examples of silicon and oxygen.

Answer:

Valency is the combining capacity of the element.

The valency of an element is determined by the number of valence electrons present in the atom of that element.

For example, the atom of silicon has four valence electrons. Thus, the valency of silicon is four.

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.

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.

Class_9_Chemistry_Structure_Of_Atom_Silicon_&_Oxygen

 

Question 8:

Explain with examples

(I) Atomic number, (ii) Mass number, (iii) Isotopes and (iv) Isobars.

Give any two uses of isotopes.

Answer:

  1. i) Atomic number: The atomic number of an element is equal to the number of protons in the nucleus of its atom.

e.g., Oxygen has 6 protons hence atomic no. = 6.

(ii) Mass number: The mass number of an atom is equal to the number of protons and neutrons in its nucleus.

Nucleons = number of protons + number of neutrons Example: Protons + Neutrons = Nucleus = Mass number (6 + 6) = 12

Class_9_Chemistry_Structure_Of_Atom_Nucleons

(iii) Isotopes: They are atoms of the same element and have same atomic number but different mass number/atomic mass.

For example: Carbon: 12C6 and 614C

(iv) Isobars

They are atoms of different elements having same mass number but different atomic number.

For example calcium, atomic number 20 and argon, atomic number 18.

The number of electrons in these atoms is different, but the mass number of both these elements is 40.

That is, the total number of neutrons is the same in the atoms of this pair of elements.

Two uses of isotopes are as follows:

(i) An isotope of uranium is used as a fuel in nuclear reactors.

(ii) An isotope of cobalt is used in the treatment of cancer.

 Question 9:

Na+ has completely filled K and L shells. Explain.

Answer:

Na has atomic number 11,

Number of Protons = 11

Number of Electrons = 11,

Electronic configuration is = 2, 8, 1

When it loses its outermost shell single electron it changes to Na+ =10= 2, 8

The above configuration indicates completely filled K, L shells.

Question 10:

If bromine atom is available in the form of, say two isotopes 79 Br35 (49.7%) and 81 Br35 (50.3%), calculate the average atomic mass of bromine atom.

Answer:

The average atomic mass of bromine atom = (79 x (49.7/100)) + (81 x (50.3/100))

=39.263 + 40.743

=80.006 u

Question 11:

The average atomic mass of a sample of element X is 16.2 u. What are the percentages of isotopes 168X and 188X in the sample?

Answer:

Let the percentage of 16 8X be x and the percentage of 18 8X be (100 – x).

Therefore, (16 x (x/100)) + ((18(100-x))/100) = 16.2

(16x/100) + (1800 – 18x)/ (100) = 16.2

Therefore, (16x – 18x +1800)/ (100) = 16.2

(-2x + 1800) = (16.2 x 100)

-2x = (1620 – 1800)

-2x = -180

x = (180/2)

x=90

16 8X = 90% and 18 8X = 10%

Question 12:

If Z = 3, what would be the valency of the element? Also, name the element.

Answer:

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 13:

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:

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

As the atomic number is same i.e., = 6. [Atomic number = Number of protons].

Both X and Y is isotopes of same element.

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) 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 a proton. (T)

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

Put a tick (√) against correct choice and cross (×) against the wrong choice in questions 15, 16 and 17

Question15:

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

Atomic Nucleus (b) Electron (c) Proton (d) Neutron

Answer:

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

(a) Atomic nucleus (√)

(b) Electron (x)

(c) Proton (x)

(d) Neutron (x)

Question16:

Isotopes of an element have

 (a) The same physical properties

(b) Different chemical properties

(c) Different number of neutrons

(d) Different atomic numbers.

Answer:

Isotopes of an element have

(a) The same physical properties (x)

(b) Different chemical properties (x)

(c) Different number of neutrons (√)

(d) Different atomic numbers (x)

Question 17:

Number of valence electrons in Cl– ion is:

(a) 16 (b) 8 (c) 17 (d) 18

Answer:

Number of valence electrons in Cl− ion is:

(a) 16 (x)

(b) 8 (√)

(c) 17 (x)

(d) 18 (x)

Question 18:

Which one of the following is a correct electronic configuration of sodium?

(a) 2,8 (b) 8,2,1 (c) 2,1,8 (d) 2,8,1.

Answer:

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