class 9 (Chem) Ch.4 Structure of Atom.pptx

grade 9 ch. Structure of an atom

1. CHAPTER - 4 STRUCTURE OF THE ATOM MS.ANITA SHARMA

2. Charged particles in M atter :- Atoms have three types of sub atomic particles. They are electrons, protons and neutrons. Electrons are negatively charged (e - ), protons are positively charged (p + ) and neutrons have no charge (n). The mass of an electron is 1/2000 the mass of a hydrogen atom. The mass of a proton is equal to the mass of a hydrogen atom and is taken as 1 unit. The mass of a neutron is equal to the mass of a hydrogen atom and is and is taken as 1 unit.

3. Relative mass is 1/2000 times of Hydrogen. Outside Nucleus Relative Mass of Proton & Neutron is 1 U (2000 times as the mass of an electron)

4. Discovery of sub atomic particles :- In 1900, J.J.Thomson discovered the presence of the negatively charged particles called electrons in the atom. In 1886, E.Goldstein discovered new radiations in gas discharge and called them canal rays. These rays were positively charged. This later led to the discovery of the positively charged particles called protons in the atom. In 1932 Chadwick discovered the presence of particles having no charge in the atom called neutrons.Neutrons has a mass nearly equal to that of a proton,present in the nucleus of all atoms.

5. The Structure of an atom :- a) Thomson’s model of an atom :- Thomson proposed the model of an atom to be similar to that of a Christmas pudding. The electrons in a sphere of positive charge were like dry fruits in a spherical Christmas pudding. . He proposed that :- An atom consists of a positively charged sphere and the electrons are embedded in it. The negative and positive charges are equal in magnitude So the atom as a whole is electrically neutral. But the results of experiments carried by other scientists could not be explained by this model.

6. b) Rutherford’s Model of an atom :-

7. Rutherford’s alpha scattering experiment :- Rutherford allowed a beam of fast moving alpha particles (Helium ions) having positive charge to fall on a thin gold foil. He observed that :- Most of the α – particles passed straight through the gold foil. Some of the α – particles were slightly deflected by small angles. Very few α – particles appeared to rebound.

8. Observations & Conclusions from Rutherford’s alpha scattering experiment :- Observation- Most of the space inside an atom is empt y. Conclusion - because most of the α lpha particles passed straight through the gold foil . Observation - The atom had a small nucleus having positive charge Conclusion - because some of the α lpha particles having positive charge were slightly deflected by small angles . iii) Observation - The size of the nucleus is very small compared to the size of the atom. Conclusion - because very few α lpha particles appeared to rebound and most of the positive charge and mass of the atom is in the nucleus .

9. What do you think would be the observation if the α-particle scattering experiment is carried out using a foil of a metal other than gold ? Answer : If the α-scattering experiment is carried out using a foil of a metal rather than gold, there would be no change in the observation. In the α-scattering experiment, a gold foil was taken because gold is malleable and a thin foil of gold can be easily made. It is difficult to make such foils from other metals.

10. Rutherford’s model of an atom :- An atom has a positively charged nucleus at its centre and most of the mass of the atom is in the nucleus. The electrons revolve around the nucleus in different orbits. The size of the nucleus is very small compared to the size of the atom. ELECTRONS NUCLEUS RUTHERFORD NUCLEAR MODEL OF AN ATOM

11. Defects of Rutherford’s model of the atom :- Negatively charged electron Positively charged nucleus Any particle in a circular orbit would undergo acceleration and during acceleration the charged particle would radiate energy. So the revolving electrons would lose energy and fall into the nucleus and the atom would be unstable. We know that atoms are stable. Rutherford’s model of an atom Very small positively charged nucleus Negatively charged electrons in orbits around the nucleus - - + -

12. CHECK POINT QS.1 NAME THE SCIENTIST WHO DISCOVERED ELECTRON. QS.2 WHAT ARE CANAL RAYS? QS.3 WHICH MODEL OF AN ATOM IS CALLED CHRISTMAS PUDDING MODEL & WHY? QS.4 STATE THE LOCATION OF ELECTRON ,PROTONS & NEUTRONS IN AN ATOM. QS.5 NAME THE SUB-ATOMIC PARTICLES PRESENT IN AN ATOM.

13. Bohr’s model of an atom :- An atom has a positively charged nucleus at its centre and most of the mass of the atom is in the nucleus. The electrons revolve around the nucleus in special orbits called discrete orbits. These orbits are called shells or energy levels and are represented by the letters K, L, M, N etc. or numbered as 1, 2, 3, 4, etc. While revolving in the discrete orbits the electrons do not radiate energy. Shells or energy levels in an atom

14. 3) Distribution of electrons in different shells :- The distribution of electrons in the different shells was suggested by B oh r and Bury. The following are the rules for filling electrons in the different shells. i) The maximum number of electrons in a shell is given by the formula 2n 2 where n is the number of the shell 1, 2, 3 etc. First shell or K shell can have = 2n 2 = 2 x 1 2 = 2x1x1 = 2 electrons Second shell or L shell can have = 2n 2 = 2 x 2 2 = 2x2x2 = 8 electrons = 2x3x3 = 18 electrons = 2x4x4 = 32 electrons Third shell or M shell can have = 2n 2 = 2 x 3 2 Fourth shell or N shell can have = 2n 2 = 2 x 4 2 and so on. The maximum number of electrons that can be filled in the outermost shell is 8. Electrons cannot be filled in a shell unless the inner shells are filled.

15. Electronic Configuration-It is the process of distribution of electrons into various energy shells. N a me of S y m b ol element Atomic N u m b er Number of Protons Number of Neutrons Number of Electrons D i s t r i bu t i o n Va l ency Of Electrons K L M N Hydrogen H 1 1 - 1 1 - - - 1 Helium He 2 2 2 2 2 - - - Lithium Li 3 3 4 3 2 1 - - 1 Beryllium Be 4 4 5 4 2 2 - - 2 Boron B 5 5 6 5 2 3 - - 3 Carbon C 6 6 6 6 2 4 - - 4 Nitrogen N 7 7 7 7 2 5 - - 3 Oxygen O 8 8 8 8 2 6 - - 2 Fluorine F 9 9 10 9 2 7 - - 1 Neon Ne 10 10 10 10 2 8 - - Sodium Na 11 11 12 11 2 8 1 - 1 Magnesium Mg 12 12 12 12 2 8 2 - 2 Aluminium Al 13 13 14 13 2 8 3 - 3 Silicon Si 14 14 14 14 2 8 4 - 4 Phosphorus P 15 15 16 15 2 8 5 - 3,5 Sulphur S 16 16 16 16 2 8 6 - 2 Chlorine Cl 17 17 18 17 2 8 7 - 1

16. Atomic structure of the first eighteen elements :- H He Li Be B C N O F Ne Na Mg A l Si P S Cl A r

17. 4) Valency :- Valency is the combining capacity of an atom of an element. The electrons present in the outermost shell of an atom are called valence electrons. If an atom’s outermost shell is completely filled, they are inert or least reactive and their combining capacity or valency is zero. Of the inert elements Helium atom has 2 electrons in the outermost shell and the atoms of other elements have 8 electrons in their outermost shell. Atoms having 8 electrons in their outermost shell is having octet configuration and are stable. BOHR-BURY RULE- If an atom’s outermost shell is not completely filled it is not stable. It will try to attain stability by losing, gaining or sharing electrons with other atoms to attain octet configuration like noble gases. The number of electrons lost, gained or shared by an atom to attain octet configuration is the combining capacity or valency of the element Eg :- Hydrogen, Lithium, Sodium atoms can easily lose 1 electron and become stable. So their valency is 1. Magnesium can easily lose 2 electrons. So its valency is 2. Aluminiun can easily lose 3 electrons. So its valency is 3. Carbon shares 4 electrons. So its valency is 4. Fluorine can easily gain 1 electron and become stable. So its valency is 1. Oxygen can easily gain 2 electrons. So its valency is 2. Nitrogen can easily gain 3 electrons. So its valency is 3.

18. HOW TO CALCULATE VALENCY OF ELEMENTS ? IF VALENCE ELECTRONS ARE 1,2,3,4 SO VALENCY WILL BE REMAIN SAME ,FOR EXAMPLE HYDROGEN(AT.NO.1) , VE. =1 VALENCY=1 IF VALENCE ELECTRONS ARE MORE THAN 4 ,THAN SUBTRACT VALENCE ELECTRONS FROM 8(OCTET RULE) EG. OXYGEN(AT.NO. 8) ITS CONFIGURATION IS 2,6 V.E =6 SO ITS VALENCY WILL BE 8-6=2

19. a) Atomic number (Z) :- The atomic number of an element is the number of protons present in the nucleus of the atom of the element. All the atoms of an element have the same atomic number. Eg :- Hydrogen – Atomic number = 1 (1 proton) Helium L i th i um Atomic number = 2 (2 protons) Atomic number = 3 (3 protons) b) Mass number (A) :- The mass number of an element is the sum of the number of protons and neutrons (nucleons) present in the nucleus of an atom of the element. The mass of an atom is mainly the mass of the protons and neutrons in the nucleus of the atom. Eg :- Carbon – Mass number = 12 (6 protons + 6 neutrons) Mass = 12u Aluminium – Mass number = 27 (13 protons + 14 neutrons) Mass = 27u Sulphur – Mass number = 32 (16 protons + 16 neutrons) Mass = 32u In the notation of an atom the atomic number and mass number are written as :- Mass number Eg :- N Atomic number Symbol of element 14 7

20. ATOMIC NO.=NO.OF PROTONS=NO.OF ELECTRONS MASS NO.=NO.OF PROTONS+NO.OF NEUTRONS NO.OF NEUTRONS= MASS NO- ATOMIC NO.

21. Isotopes :- Isotopes are atoms of the same element having the same atomic numbers but different mass numbers. Eg :- Hydrogen has three isotopes. They are Protium, Deuterium (D) and Tritium (T). P r otium D eu t e ri u m T ri t i u m 1 H 1 H H 1 1 2 3 Carbon has two isotopes. They are :- 12 C 14 C 6 6 Chlorine has two isotopes They are :- 35 37 17 Cl 17 Cl

22. APPLICATIONS OF ISOTOPES

23. HOW TO CALCULATE AVERAGE ATOMIC MASS?

24. Question : If bromine atom is available in the form of say, two isotopes  79 35 Br (49.7%) and  81 35 Br (50.3%), calculate the average atomic mass of bromine atom.

25. Question:The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes  16 8 X and  18 8 X in the sample? Answer:  Let the percentage of  16 8 X be x and the percentage of  16 8 X be 100 – x .

26. Isobars Isobars are atoms of different elements having different atomic numbers but same mass numbers. These pairs of elements have the same number of nucleons . Eg - Calcium (Ca) – atomic number - 20 and Argon ( Ar ) atomic number 18 have different atomic numbers but have the same mass numbers – 40. 40 Ca 40 Ar 20 18 Iron (Fe) and Nickel (Ni) have different atomic numbers but have the same atomic mass numbers – 58. 58 Fe 58 Ni 26 27

27. Complete the following table. nswer:   COMPLETE THE FOLLOWING TABLE

28. THANK YOU MS.ANITA SHARMA