Atomic Structure Class-8 Dalal Simplified ICSE Chemistry

Atomic Structure Class-8 Dalal ICSE New Simplified Chemistry Dr Viraf J Dalal Middle School Allied Publishers Solutions. Chapter-4 . We Provide Step by Step Solutions of Exercise/Lesson -4 Atomic Structure with Objective Type Questions, Fill in the blanks and Give reason , Match the following of Dr Viraf J Dalal Middle School Chemistry Allied Publishers. Visit official Website CISCE for detail information about ICSE Board Class-8.

Atomic Structure Class-8 Dalal ICSE New Simplified Chemistry Dr Viraf J Dalal Middle School Allied Publishers Solutions. Chapter-4


EXERCISE-4

Question 1.
State the main postulates of – Dalton’s atomic theory. Explain how the modern atomic theory contradicted Dalton’s atomic theory.
Answer 1:

The main postulates of Dalton’s atomic theory are:

  • Atoms are neither created nor destroyed. This implies that during chemical reactions, no atoms are created nor destroyed.
  • The formation of new products (compounds) results from the rearrangement of existing atoms (reactants).
  • Atoms of an element are identical in mass, size and many other chemical or physical properties, but atoms of two-different elements differ in mass, size, and many other chemical or physical properties.
  • The matter is made up of indivisible particles known as atoms.
  • The properties of all the atoms of a given element are the same including mass. This can also be stated as all the atoms of an element have identical mass while the atoms of different elements have different masses.
  • Atoms of different elements combine in fixed ratios to form compounds.

The modern atomic theory contradicts dalton’s atomic theory :

  • Atoms can combine in any ratio and not just whole numbers. Example – Sugar  C12H22O11 (Sugar).
  • Atoms can be destroyed and converted into energy.
  • Atoms are divisible into protons, neutrons, electrons.
  • Atoms of the same element have different properties, such atoms are isotopes.
  • Atoms of different elements have the same properties, such atoms are isobars.

Question 2.

With reference to the discovery of the structure of an atom, explain in brief – William Crookes experiment for the discovery of cathode rays, followed by – J.J. Thomsons experiment pertaining to the constituents of the cathode rays. State which sub-atomic particle was discovered from his experiment.
Answer 2:

The three subatomic particles that constitute an atom are:

  • Electrons
  • Protons
  • Neutrons

The discovery of cathode rays lead to the discovery of electrons :

William Crookes experiment for the discovery of cathode rays:

Discovery of cathode rays leading to the discover of ‘electrons

Experiment – The figure shows a vacuum tube which has a gas at a very low pressure. An electric discharge is passed through the tube.

Emission of blue rays can be seen from the cathode.

Conclusion – The blue rays emitted from cathode are cathode rays.

J.J. Thomson’s experiment on constituents of cathode rays:

new-simplified-chemistry-class-8-icse-solutions-atomic-structure - 5

Experiment – To test the properties of the particles, Thomson placed two oppositely-charged electric plates around the cathode ray. The cathode ray was deflected away from the negatively-charged electric plate and towards the positively-charged plate. This indicated that the cathode ray was composed of negatively-charged particles.

Thomson also placed two magnets on either side of the tube, and observed that this magnetic field also deflected the cathode ray. These subatomic particles can be found within atoms of all elements.

Conclusion – The cathode rays are made up of negatively charged particles known as electrons.

Question 3.
Explain in brief – Goldstein’s experiment which led to the discovery of the proton and – Lod Rutherford’s experiment which led to the discovery of the atomic nucleus.
Answer 3:

Goldstein’s experiment which led to discovery of protons:

new-simplified-chemistry-class-8-icse-solutions-atomic-structure - 5.1

Experiment – The figure shows a modified cathode ray tube with a perforated cathode.

When an electric discharge is passed through this tube, a new type of rays is produced from the anode passing through the holes of positive cathode, they are anode rays..

Conclusion – Anode rays are made of positively charged particles because they were affected by magnetic fields in a direction opposite to that of cathode rays. These positively charged particles are protons.

Lord Rutherford’s experiment on discovery of atomic nucleus:

new-simplified-chemistry-class-8-icse-solutions-atomic-structure - 6

Experiment – To direct energetic alpha particles at a thin metal foil and measure how an alpha particle beam is scattered when it strikes a thin metal foil. A narrow collimated beam of alpha particles was aimed at a gold foil of approximately 1 μm thickness (about 10,000 atoms thick). Alpha particles are energetic nuclei of helium (usually about 6 MeV). Alpha particles, which are about 7300 times more massive than electrons, have a positive charge of +2e. Because of their relatively much greater mass, alpha particles are not significantly deflected from their paths by the electrons in the metal’s atoms.

If an alpha particle collides with an atom, it would fly straight through, its path being deflected by at most a fraction of a degree. To deflect the alpha particle backward, there must be a very large force. This force could be provided only result from a collision with a massive target or from an interaction with an electric or magnetic field of great strength.

Conclusion – It was discovered that every atom contains a nucleus (whose diameter is of the order 10-14m) where all of its positive charge and most of its mass are concentrated in a small region called an atomic nucleus.

Question-4

‘Electrons revolve around the nucleus in fixed orbits or shells called energy levels’. State how these energy levels are represented.

Answer 4:

Electrons revolve around the nucleus in fixed orbits or shells called energy levels. Electrons rotate around the nucleus in one or more energy levels.

These are represented as 1, 2, 3 i.e. an integer ‘n’ or as K, L, M, N.

Question 5.
Draw a neat labelled diagram representing an atom. Name the three sub-atomic particles in the atom & represent them symbolically showing the mass & charge of each. State where the sub-atomic particles are present in the atom.
Answer 5:

new-simplified-chemistry-class-8-icse-solutions-atomic-structure - 7

Question 6.
Define the term – ‘atomic number’ of an atom. If an atom ‘A’ has an atomic number of – eleven, state the number of protons & electrons it contains.
Answer 6:

Atomic number is the number of protons in the atom. An atom is electrically neutral, i.e. it doesn’t have any charge.

Atomic number = Number of electrons = Number of protons

In the atom ‘A’, atomic number = 11, that is the number of protons and electrons is 11.

Atomic number = z = p = e

Question 7.
Define the term – ‘mass number ’ of an atom. If an atom t ‘B’ has mass number 35 & atomic number 17, state the number of protons, electrons & neutrons it contains.
Answer 7:

Mass number of an atom is the sum of the number of protons and neutrons in the nucleus of the atom.

Mass number = Number of protons + Number of electrons

In the atom ‘B’, mass number = 35 and atomic number = 17

Number of protons = Atomic Number = 17 = p

Mass number = 17 + Number of neutrons

35 = 17 + n

n = 18

Number of neutrons = 18

Number of electrons = number of protons = 17 =

Question 8.
State why the atomic weight of an element is also termed – relative atomic mass.
Answer 8:

  • Atomic weight : Atomic weight is the mass of an atom compared to an atom of hydrogen. For example – carbon is 12 times heavier than one hydrogen atom.
  • Relative mass : Relative atomic mass is the atomic weight compared to 1/12 the mass of one atom of carbon. Since it is a comparison of atomic weight with 1/12 the mass of an atom of carbon it is termed relative atomic mass.

Question 9.
State how electrons are distributed in an atom. Explain in brief the rules which govern their distribution.
Answer 9:
Electrons rotate around the nucleus in one or more orbits/energy levels.

new-simplified-chemistry-class-8-icse-solutions-atomic-structure - 9
Rules : Maximum number of electrons in a shell is given by 2n2. Where n is the number of shell i.e. 1st shell can have maximum of 2 electrons.
2n2 = 2(1)2 = 2 × 1 = 2
2nd shell can have maximum of 8 electrons
2n2 = 2(2)2 = 2 × 4 = 8
3rd shell can have maximum of 18 electrons
2n2 = 2 (3)2 = 2 × 9 = 18 and so on….

The outermost shell cannot have more than 18 electrons.

A new shell cannot be started until the previous one has reached its maximum limit.

Question 10.
If an atom ‘A’ has atomic number 19 & mass number 39, state –

  1. Its electronic configuration.
  2. The number of valence electrons it possesses.

Answer 10:

Atomic number = 19

Number of protons = Number of electrons = 19

Mass number = Number of protons + Number of neutrons

39 = 19 + N

N = 20

Electronic configuration

19 = (2, 8, 8, 1)

Electrons in K shell (1st) = 2

Electrons in L shell (2nd) = 8

Electrons in M shell (3rd) = 8

Electrons in N shell (4th) = 1

Valence electrons is the number of electrons in the outer shell, i.e. 1 in this atom.

Question 11.
Draw the atomic diagrams of the following elements showing the distribution of – protons, neutrons & the electrons in the various shells of the atoms.

new-simplified-chemistry-class-8-icse-solutions-atomic-structure - 10
[The upper number represents the – mass number & the lower number the – atomic number e.g. calcium – mass number = 40, atomic number = 20] Answer 11:

a. Carbon

Atomic number = 6

Mass Number = 12

No. of protons = 6

No. of electrons = 6

No. of neutrons = 12 – 6 = 6

Electronic configuration – (K, L) = (2, 4)

b. Oxygen

Atomic number = 8

Mass Number = 16

No. of protons = 8

No. of electrons = 8

No. of neutrons = 16 – 8 = 8

Electronic configuration – (K, L) = (2, 6)

c. Phosphorus

Atomic number = 15

Mass Number = 31

No. of protons = 15

No. of electrons = 15

No. of neutrons = 31 – 15 = 16

Electronic configuration – (K, L, M) = (2, 8, 5)

d. Argon

Atomic number = 18

Mass Number = 40

No. of protons = 18

No. of electrons = 18

No. of neutrons = 40 – 18 = 12

Electronic configuration – (K, L, M) = (2, 8, 8)

e. Calcium

Atomic number = 20

Mass Number = 40

No. of protons = 20

No. of electrons = 20

No. of neutrons = 40 – 20 = 20

Electronic configuration – (K, L, M, N) = (2, 8, 8, 2)

Question 12.
‘ Valency is the number of hydrogen atoms which can combine with [or displace] one atom of the element [or radical] forming a compound’. With reference to the above definition of valency, state the valency of chlorine in hydrogen chloride, giving reasons.
Answer 12:
In Hydrogen chloride [HCl], one atom of chlorine has combined with one atom of hydrogen and also 1 atom of hydrogen can be replaced by metals like potassium, sodium. Hence valency of chlorine in one.

Question 13.
‘ Valency is also the number of electrons – donated or accepted by an atom so as to achieve stable electronic configuration of the nearest noble gas’. With reference to this definition –

(a) State what is meant by ‘stable electronic configuration’.
(b) State why the valency of –

  1. sodium, magnesium & aluminium is : +1, +2 & +3 respectively.
  2. chlorine, oxygen & nitrogen is : -1, -2 & -3 respectively.

Answer 13:

(a) Stable electronic configuration means an atom in which the outermost (valence) shell is complete.

(b)

  1. Valency is the number of electrons lost or gained from the outer shell of the atom. Sodium loses 1 electron, Magnesium loses 2 and aluminium loses 3, hence their valency is +1, +2 & +3 respectively.
  2. Valency is the number of electrons lost or gained from the outer shell of the atom. Chlorine gains 1 electron, oxygen gains 2 and nitrogen gains 3, hence their valency is -1, -2 & -3 respectively.

Question 14.
With reference to formation of compounds from atoms by electron transfer – electro valency, state the basic steps in the conversion of sodium & chlorine atoms to sodium & chloride ions leading to the formation of the compound – sodium chloride.
[electronic configuration of : Na = 2, 8, 1 & Cl = 2, 8, 7]
Answer 14:
Electronic configuration

conversion of sodium & chlorine atoms to sodium & chloride ions


OBJECTIVE TYPE QUESTIONS

Atomic Structure Class-8 ICSE New Simplified Chemistry Dr Viraf J Dalal Middle School Allied Publishers Solutions. Chapter-4

Question.1. 

Match the statements in List I with the correct answer from List II.

List I List II
1. Mass number of an atom is the number of protons and A: Electron
2. The subatomic particles with negligible mass B: Argon
3. An atom have stable electronic configuration C: Nitrogen
4. A molecule formed by sharing of electrons (covalency) D: Sodium
5. A metallic atom have unstable electronic configuration E: Neutrons

Answer -1:

List I List II
1. Mass number of an atom is the number of protons and E: Neutrons
2. The subatomic particles with negligible mass A: Electron
3. An atom have stable electronic configuration B: Argon
4. A molecule formed by sharing of electrons (covalency) C: Nitrogen
5. A metallic atom have unstable electronic configuration D: Sodium

Question-.2. 

Select the correct answer from the choice in bracket to complete each sentence :

  1. An element ‘X’ has six electrons in its outer or valence shell. Its valency is______ (+2 / -2 / -1)
  2. An element ‘Y’ has electronic configuration 2, 8, 6. The element ‘Y’ is a _______ (metal/ non-metal/ noble gas)
  3. A _________ is a sub-atomic particle with no charge and unit mass.
  4. An element Z with zero valency is a _________ (metal / noble gas/ non-metal )
  5. magnesium atom with electronic configuration 2, 8, 2 achives stable electronic configuration by losing two electrons, thereby achieving stable electronic configuration of the nearest noble gas ________ ( neon / argon )

Answer-2

  1. -2
  2. non-metal
  3. Neutron
  4. Noble gas
  5. Neon

Question-.3. 

The diagram represents an isotope of hydrogen [H]. Answer the following :

new-simplified-chemistry-class-8-icse-solutions-atomic-structure - 15

  1. Are isotopes atoms of the same element or different elements?
  2. Do isotopes have the same atomic number or the same mass number?
  3. If an isotope of ‘H’ has mass no. = 2, how many electrons does it have?
  4. If an isotope of ‘H’ has mass no. = 3, how many neutrons does it have?
  5. Which subatomic particles in the 3 isotopes of ‘H’ are the same?

Answer-3

  1. Isotopes are atoms of the same element.
  2. Same atomic number
  3. One electron
  4. Two neutrons (A= P + n)
  5. Protons and electrons in each isotope are the same.

Question-.4.

 State the electronic configuration for each of the following :

  1. Hydrogen [p =1]
  2. Boron [p=5]
  3. Nitrogen [p=7]
  4. Neon [p=10]
  5. Magnesium [p=12]
  6. Aluminium [p=13]
  7. Sulphur [p=16]
  8. Argon [p=18]
  9. Potassium [p=19]
  10. Calcium [p=20]

Answer-4:
Electronic configuration of :

  1. Hydrogen – (K) – (1)
  2. Boron – (K, L) – (2, 3)
  3. Nitrogen – (K, L) – (2, 5)
  4. Neon – (K, L) – (2, 8)
  5. Magnesium – (K, L, M) – (2, 8, 2)
  6. Aluminium – (K, L, M) – (2, 8, 3)
  7. Sulphur – (K, L, M) – (2, 8, 8)
  8. Argon – (K, L, M, N) – (2, 8, 8, 2)
  9. Potassium – (K, L, M, N) – (2, 8, 8, 3)
  10. Calcium – (K, L, M, N) – (2, 8, 8, 4)

Question-.5. 

Draw the structure of the following atoms showing the nucleus containing – protons, neutrons and the orbits with the respective electrons :

  1. Lithium [At. no. = 3, Mass no. = 7]
  2. Carbon [At. no. = 6, Mass no. = 12]
  3. Silicon [At. no. = 14, Mass no. = 28]
  4. Sodium [At. no. = 11, Mass no. = 23]
  5. Isotopes of hydrogen [11H, 21H , 31H]

Answer 5:
Structure of atoms :

1. Lithium

Z = 3

P = 3

E = 3 = (2, 1)

A = 7

P + n = A

3 + n = 7

N = 4

structure of Lithium

2. Carbon

Z = 6

P = 6

E = 6 = (2, 4)

A = 12

P + n = A

6 + n = 12

N = 6

structure of carbon

3.  Silicon

Z = 14

P = 14

E = 14 = (2, 8, 4)

A = 28

P + n = A

14 + n = 28

N = 14

structure of silicon

4.  Sodium

Z = 11

P = 11

E = 11 = (2, 8, 1)

A = 23

P + n = A

11 + n = 23

N = 12

structure of sodium

5.- Isotope of Hydrogen

Z = 1

P = 1

E = 1 = (1)

A = 1

P + n = A

1 + n = 1

N = 0

structure of hydrogen

.– : End of Atomic Structure Class-8 Dalal Solutions :–


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