Atom Discovery & Structure: Dalton, Thomson, Rutherford, Bohr Models & Isotopes, Cheat Sheet of Chemistry

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APNI KAKSHA

APNI KAKSHA

Dalton considered atom to be an indivisible particle but this concept is rejected at the end of

19th Century. When Scientists contributed in revealing the presence of charge particle in an

atom.

These particles were called the Sub-Atomic particle.

• Discovery of Electrons J.J Thomson
• Discovery of Protons Rutherford
• Discovery of Neutron Chadwick
• Discovery of Canal Rays E. Goldstein

Particle Charge

Mass

−

− 1. 6 × 10

− 19

9. 1 × 10

− 31

+ 1. 6 × 10

− 19

1. 67 × 10

− 27

No charge

1. 67 × 10

− 27

Canal Ray (Or Anode Rays)

Stream of Positively charged particles, which move towards negatively charge electrode.

Structure of an Atom

Thomson’s Model of an Atom

• Compared to Christmas Pudding: Electrons are dry fruits of Christmas pudding, inside

the positive sphere.

• Compared with Watermelon: Red part of watermelon is positive part of Atom and

electrons are studded in positive part in the water melon.

Structrue of Atom

APNI KAKSHA

Q. Element ‘X’ has a proton number of 𝟕. It also have seven neutrons.

a) Deduce the number of electrons and nucleons of ‘X’

b) Represent ‘X’ by writing the chemical symbol [NCERT Exercise]

Sol. 𝑝 = 7 , 𝑛 = 7

a) Number of electrons = Number of protons = 7

Number of nucleons = number of 𝑝 + number of 𝑛

b) 𝑋

7

14

Bohr’s Model of Atom

To overcome drawbacks of Rutherford’s Model, Neil Bohr in 1912 proposed modified structure

of Atom.

Postulates of Bohr Model

1. Only certain special orbit known as discrete orbits of electrons, are allowed inside the

atom.

2. While revolving in discrete orbit, electrons do not radiate energy.
3. These orbits or shells are called Energy

These orbits OR shells are Represented by the letters

𝐾, 𝐿, 𝑀, 𝑁 or numbers 𝑛 = 1 , 2 , 3 , 4

Mass Number

• It is denoted by ‘A’
• Mass number of total sum of total number of protons and no. of Neutrons lying in the

nucleus of an atom.

• Mass number = No. of Protons + No. of Neutron

Atomic Number

“Total number of protons present in the Nucleus of any Atom is called the atomic Number.”

→ Atomic Number is denoted by ‘z’.

Distribution of electrons in various shells

Distribution of electrons into different orbits of an atom was suggested by

“Bohr – Bury Scheme”

APNI KAKSHA

1. Filling of electrons in an atom is done by 2n

2

Rule where n = no. of shell.

If 𝑛 = 1 , K (

st

shell) = 2 𝑛

2

= 2 × ( 1 )

2

= 2 × 1 = 2 𝑒

−

If 𝑛 = 2 , K (

nd

shell) = 2 𝑛

2

= 2 × ( 2 )

2

= 2 × 4

−

2. The outermost shell; Can’t hold more than 8 electrons

Name of Element Atomic No. Electronic Configuration

Valency

“Combining capacity of any Element is known as Valency”

• From Bohr – Bury Scheme, we also know that If outermost shell of an atom has of 8e

−

in

outermost shell it become very stable.

• It means if in any Element outermost shell contain 8 electron, It is already stable so, it

will not combine with other element so it’s valency will be zero.

• These Elements are know as inert Elements.
• In case of Helium, It has two electrons in it’s outer-most shell, due to small size of

Helium. We cannot comp. it’s octate so 𝐻 and 𝐻𝑒 are exceptions of octate Rule.

APNI KAKSHA

Q. The average Atomic mass of sample of an element ‘𝑿’ is 𝟏𝟔. 𝟐 𝒖. What are the % age

of isotopes 𝑿

𝟖

𝟏𝟖

and 𝑿

𝟖

𝟏𝟖

in the sample? [NCERT Exercise]

Sol. Let the % age of isotope 𝑋

8

16

= 𝑋 % age of isotope 𝑋 = 100 − 𝑋

8

18

Average atomic mass

𝑀𝑎𝑠𝑠 𝑜𝑓 𝑋

8

16

×𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑜𝑓 𝑋

8

16

+𝑀𝑎𝑠𝑠 𝑋

8

18

×𝑝𝑒𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑜𝑓 𝑋

8

18

100

𝑥× 16 +( 100 −𝑥)× 18

100

16 𝑥+ 1800 − 18 𝑥

100

1620 = − 2 𝑥 + 1800 or 2 𝑥 = 1800 − 1620

180

2

Thus % of isotope 𝑋

8

16

% of isotope 𝑋

8

18

Notes End

APNI KAKSHA

Important NCERT Questions

Q 1. Which of the following are true for an element?

i) Atomic number = number of protons + number of electron.

ii) Mass number = number of protons + number of neutrons

iii) Atomic number = number of protons = number of neutrons

iv) Atomic number = number of protons = number of electrons

a) (i) and (ii) b) (i) and (iii)

c) (ii) and (iii) d) (ii) and (iv)

Sol. Points (ii) and (iv) are correct.

Q2. Will 𝑪

𝟑𝟓

𝒍 and 𝑪

𝟑𝟕

𝒍 have different valencies? Justify your answer.

Sol. No. 𝐶

35

𝑙 and 𝐶

37

𝑙 have same valency. 𝐶

35

𝑙 and 𝐶

37

𝑙 are the isotopes. So, they have same

number of protons and electrons and have the same atomic number viz. 17

Electronic configuration = 2 ,

𝐾

𝐿

𝑀

Valency 8 − 7 = 1

Therefore, both of them have valency = 1

Q3. Why did Rutherford select a gold foil in his 𝜶 - ray scattering experiment?

Sol. Gold is a heavy metal with high mass number. A light metal cannot be used because on

being hit by fast moving 𝛼-particle, the atom of light metal will be simply pushed forward

and no scattering can occur. Moreover, gold is the best malleable metal. A very thin foil

(≈ 1000 atoms thick) can be made from gold to get the clear observations.

Q4. Calculate the number of neutrons present in the nucleus of an element 𝑿 which is

represented as 𝑿

𝟏𝟓

𝟑𝟏

Sol. 𝑋

15

31

represents

Atomic number, 𝑍 = 15

Mass number, 𝐴 = 31

∴ Number of neutrons = 𝐴 − 𝑍 = 31 − 15 = 16

Q5. Why do helium, neon and argon have a zero valency?

Sol. Helium

, neon

and argon

have completely filled outermost shell, i.e.,

2

10

18

Thus, they have stable electronic configuration. They neither lose electrons nor gain

electrons. Hence, their valency is zero.

Q8. a) If the number of electrons in anion is 𝟏𝟎 and number of proton is 𝟗 then