Chhattisgarh State Board Class 9 Science Chapter 3 Atomic Structure Exercise Multiple Choice, Fill in the Blanks, Questions and Answers here.
Chhattisgarh State Class 9 Science Chapter 3 Solution
1) Choose the correct option:
(i) Isotopes have different:
(a) electrons (b) protons
(c) neutrons (d) both electrons and neutrons.
Ans: – option (d).
(ii) Who first included electrons in his atomic model?
(a) Dalton (b) Thomson
(c) Rutherford (d) Bohr.
Ans: – option (b) Thomson.
(iii) Which of the following statements is true for 39K19 atom
(a) The atom has 39 electrons (b) The atom has 39 protons
(c) The atom has 19 protons (d) None of the above.
Ans: – option (c) is correct.
2) Choose the appropriate option to fill in the blanks:
(i) All atoms of an element are ………. similar………………….. (similar/different).
(ii) The number of ………….. electron………………………. (electrons/neutrons) is equal to the number ofprotons in a neutral atom.
(iii) 14V6 and 14N7are ………… isobars………………………. (isotopes/isobars) of each other.
3) How was the atom proposed by Thomson different from the atom proposed by Dalton?
Ans: – Thomson model is different from the Dalton atomic theory. Dalton States that an atom consistwith the many tiny molecules whereas the Thomson model talked about the spherical shape of the atomic models.
4)16O8 and 16N7 are isobars. Use this example to explain isobars.
Ans: – The elements when called isobars then the mass no of the elements must be equal. In this example the mass no of oxygen and nitrogen are same so these are called isobars.
5) Bromine-79 and bromine-81 are found in nature in the ratios 50.69 and 49.31. What will be theaverage atomic weight of bromine?
Ans: – The average atomic weight of bromine is,
= (79×50.69 + 81×49.31)/ (50.69 + 49.31)
=79.98 or 80.
6) Find out the number of valence electrons in 16O8 and 14N7.
Ans: – The valence electron is the no free of electron present in their last cell. So,in oxygen no of free electron are 6 so valence electron is6 and in nitrogen free electron are 5 so valence electron is 5.
7) Describe Bohr’s atomic model.
Ans: – The Bohr’s atomic model isknown as like the solar system. In a solar system the sun is fixed and different planets are moving around it likewise in this model the nucleolus is fixed and the electron are moving outside it.
8) Apart from oxygen-16, oxygen-17 and oxygen-18 are also known to exist in nature. Are these atoms isotopes or isobars? Explain.
Ans: – As we all know that the element which has same no of proton are called isotopes. In this example we can see all these are isotope of oxygen atom. As all have same atomic no and different no of neutron this are called isotopes.
9) Explain Dalton’s atomic theory. What are its limitations?
Ans: – Dalton’s atomic theory states that tiny small molecules make an atom or other ward in the atom are consist of different types of small atom together. In this theory there are some limitations likes –
- The subatomic particles are no explained.
- Concept of isotopes are not included.
- All atoms have identical mass and density. etc.
10) What was the alpha particle scattering experiment carried out by Rutherford? What conclusions did he draw from this experiment regarding the structure of an atom?
Ans: – For the measuring the atomic structure Rutherford made an experiment in which he takes a gold sheet and passes the alpha particles through it. Then he follows that the most of the alpha particles are passes through the gold sheet so he came to this conclusion that most of part in an atom are empty. This experiment at last published by the name of Rutherford atomic models.
11) Write the rules proposed for electron distribution under the Bohr-Bury scheme. Use the rules to write the electronic configuration of given atoms. Also give the number of neutrons present in each atom. 19F9, 24Mg12, 28Si14, 31P15, 35Cl17.
Ans: – The Bohr Bury Scheme States that the electron of any atom is distributed according to their different energy levels. As we all know from this model that the lowest energy levels are field first then it goes to the upper region. The removing of electron from this different energy level energy is required.
The no of neutrons in these atoms are – for F neutron is (19-9=10) 10, for Mg is 12, for Si is 14, for P is 16 and for chlorine is 18.