Wednesday, February 4, 2015

Solved Question Bank chapter: Structure of an Atom class 9

1. Q. The description of atomic particles of two elements X and Y is given below


(i) What is the atomic number of Y?
(ii) What is the mass number of X?
(iii) What is the relation between X and Y?
(iv) Which element/elements do they represent?
(v) Write the electronic configuration of X?
(vi) Write the cation/anion formed by the element

Ans: (i) Atomic number of y =  8            
(ii) Mass number of x – 16         
 (iii) x and y are isotopes  
(iv) x and y represent – oxygen              
(v) 2, 6                                      
(vi) It will form an anion – O-2

2. Q. Which of the following are isotopes and which are isobars?

Argon,  Protium, Calcium, Deuterium. Explain why the isotopes have similar chemical properties but they differ in physical properties?

Ans: Isotopes – Protium, Diuterium                       

Isobars – Argon and calcium

Since isotopes have identical electronic configuration containing same number of valence electrons they have similar chemical properties.

Since the masses are slightly different the physical properties (density, melting pt, boiling pt, etc) are different

3. Q. (a) Explain Bohr and Bury rules for distribution of electrons into different shells. (b) Draw the electronic structure of element X with atomic number 17 and element Y with atomic number 16 ?

Ans: (a) Bohr Bury Rules 
(i) The maximum no. of electrons present in a shell is given by the formula 2n2 (where n is shell no.)

(ii) The maximum no. of electrons that can be accommodated in the outer most orbit is 8. (iii) 

Electron are not accommodated in a given shell, unless the inner shells are filled :

(b) (i) X at No. 17 E.C.= 2 , 8, 7 (ii) Y At No. 16 E.C = 2, 8, 6

4. Q. The atomic number and mass number of an element are 16 and 32 respectively. Find the number of protons, electrons and neutrons in it. State its valency. Is this element a metal or a non – metal. Justify your answer.

Ans: (a) No. of protons = 16      
No. of electrons = 16     
No. of neutrons = 16
(b) Electronic configuration 2, 8, 6  
Valency = 8-6 = 2 
(c) It is a non metal because it has 6 valence electrons

5. Q. (a) The composition of nuclei of two atomic species X and Y are given below


Find the mass number of X and Y. State the relationship between X and Y (b) The K and L shells of an atom are completely filled. Find the number of electrons present in it. State the name of this element.

Ans: (a) Mass number of X = 35 Mass number of Y = 37  

Relationship between the two species since number of protons is same (same atomic number) they are isotopes of same element   

(b) Numbers of electrons = 10 Name of the element = Neon  k L 2 8

6. Q. State the observations in a - particle scattering experiment which led Rutherford to make the following conclusions
(i) Most of the space in an atom is empty. 
(ii) Whole mass of an atom is concentrated in its centre.
(iii) Centre is positively charged.

(i) Most of the alpha particles passed through gold foil with getting deflected.
(ii) Very few particles were deflected from their path by 1800 indicating that whole mass of the atom is present in its centre.
(iii) Few particles deflected at small and large angle from their path indicating that centre is positively charged.

7. Q.(i) State the limitations of J.J. Thomson‟s model of an atom. 
(ii) Define valency by taking the examples of magnesium (At. no = 12)  and oxygen (At. no=8) 
(iii) S -2 has completely filled K,L and M shells. Find its atomic number.

Ans: (i) The results of experiments carried out by other scientists could not be explained by J.J. Thomson’s model of atom.

(ii) The combining capacity of an element is called its valency. Magnesium has atomic number 12 and electronic configuration is 2,8,2. It can lose 2 electrons to get octet configuration thus its valency is 2.
Oxygen has atomic number 8 and its electronic configuration is 2, 6. It can gain 2 electrons to get octet configuration thus its valency is 8-6=2
(iii) The atomic number is equal to number of protons thus atomic number of S-2 ion is 16.

8. Q. State one use each of an isotope of (i) uranium , (ii) iodine.

Ans: Isotope of uranium is used in nuclear reactions and Isotope of iodine is used in treatment of goiter

9. Q. Is it possible for the atom of an element to have one electron, one proton and no neutron? If so, name the element.
Ans: Yes, it is true for hydrogen atom which is represented as 1H1

10. Q. Why did Rutherford select a gold foil in his α–ray scattering experiment?

Ans it is because gold has high malleability can be hammered into thin sheet

11. Q. Will Cl-35  and Cl-37 have different valences?

Ans: No, It is because these are isotopes of chlorine that have same atomic number but different mass number

12. Q. Calculate the number of neutrons present in the nucleus of an element X which is represented as 31 X15 . 

Ans: 31 X15 .indicate that No. of proton=15 and mass number =31
         Mass number = No. of protons + No. of neutrons = 31
         Number of neutrons = 31– number of protons = 31–15 = 16

13. Q. The atomic number of calcium and argon are 20 and 18 respectively, but the mass number of both these elements is 40. What is the name given to such a pair of elements?               

Ans: Isobars

14. Q. Why do Helium, Neon and Argon have a zero valency?

Ans: Helium, Neon and Argon have 2, 8 and 8 electron in outermost cell so they are having no need to gain or loss electrons. Hence they have zero valency.

15.Q. In what way the Rutherford proposed atomic model?

Ans: Rutherford proposed a model in which electrons revolve around the nucleus in well-defined orbits. There is a positively charged centre in an atom called the nucleus. He also proposed that the size of the nucleus is very small as compared to the size of the atom and nearly all the mass of an atom is centered in the nucleus.

Structure of atom for CBSE class 9(IX)
Solved Summative Assessment Paper

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