Class 9 Chemistry Worksheet on Chapter 4 Structure of the Atom- Set 2 - Free Printable

Problem: Solve the given chemistry worksheet questions related to the structure of the atom.

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#### Q-1: Boron occurs naturally as two isotopes with atomic masses of 10.01 and 11.01. Boron has an atomic mass of 10.81. Determine the percentage of each isotope in natural Boron.

Solution:
Let the percentage of the isotope with atomic mass 10.01 be \( x \) and the percentage of the isotope with atomic mass 11.01 be \( y \). Since these are the only two isotopes, we have:
\[ x + y = 100 \]

The weighted average atomic mass of Boron is given by:
\[ 10.01x + 11.01y = 10.81 \times 100 \]

Substitute \( y = 100 - x \) into the equation:
\[ 10.01x + 11.01(100 - x) = 1081 \]
\[ 10.01x + 1101 - 11.01x = 1081 \]
\[ -1.00x + 1101 = 1081 \]
\[ -1.00x = 1081 - 1101 \]
\[ -1.00x = -20 \]
\[ x = 20 \]

So, \( y = 100 - x = 100 - 20 = 80 \).

Thus, the percentage of the isotope with atomic mass 10.01 is 20%, and the percentage of the isotope with atomic mass 11.01 is 80%.

Answer:
\[ \boxed{20\%, 80\%} \]

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#### Q-2: Which of the following statements about a proton is incorrect?

a) It is a positively charged particle
b) The mass of a proton is equal to the mass of a neutron
c) It resides in the nucleus of an atom
d) It is a constituent of cathode rays

Solution:
- Option (a): Protons are positively charged particles. This statement is correct.
- Option (b): The mass of a proton is approximately equal to the mass of a neutron. This statement is correct.
- Option (c): Protons reside in the nucleus of an atom. This statement is correct.
- Option (d): Cathode rays consist of electrons, not protons. This statement is incorrect.

Answer:
\[ \boxed{\text{d}} \]

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#### Q-3: Isobars are two atoms that have

a) the same mass number but different atomic numbers
b) the same number of protons but different electrons
c) the same number of protons and electrons
d) the same atomic number but different mass numbers

Solution:
Isobars are atoms with the same mass number but different atomic numbers. For example, Carbon-14 (\(^{14}_6C\)) and Nitrogen-14 (\(^{14}_7N\)) are isobars.

Answer:
\[ \boxed{\text{a}} \]

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#### Q-4: An atom with the atomic mass number 39 contains 20 neutrons. What is the atom's atomic number?

Solution:
The atomic mass number (\(A\)) is the sum of the number of protons (\(Z\)) and neutrons (\(N\)):
\[ A = Z + N \]

Given:
\[ A = 39 \]
\[ N = 20 \]

So,
\[ 39 = Z + 20 \]
\[ Z = 39 - 20 \]
\[ Z = 19 \]

Answer:
\[ \boxed{19} \]

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#### Q-5: Give the electronic configurations of the following ions:

i) H
ii) S²⁻
iii) N³⁻
iv) N²⁺

Solution:
- i) H: Hydrogen has 1 electron. Its electronic configuration is \(1s^1\).
- ii) S²⁻: Sulfur (atomic number 16) has 16 electrons. When it gains 2 electrons to form S²⁻, it has 18 electrons. The electronic configuration is:
\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \]
- iii) N³⁻: Nitrogen (atomic number 7) has 7 electrons. When it gains 3 electrons to form N³⁻, it has 10 electrons. The electronic configuration is:
\[ 1s^2 \, 2s^2 \, 2p^6 \]
- iv) N²⁺: Nitrogen (atomic number 7) has 7 electrons. When it loses 2 electrons to form N²⁺, it has 5 electrons. The electronic configuration is:
\[ 1s^2 \, 2s^2 \, 2p^1 \]

Answer:
\[ \boxed{\text{i) } 1s^1, \text{ ii) } 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6, \text{ iii) } 1s^2 \, 2s^2 \, 2p^6, \text{ iv) } 1s^2 \, 2s^2 \, 2p^1} \]

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#### Q-6: An element's atomic mass is equal to twice its atomic number. If the L-shell contains six electrons, then predict its valency.

Solution:
- The L-shell corresponds to the second shell, which can hold up to 8 electrons. If it contains 6 electrons, the element must be Oxygen (O), which has an atomic number of 8.
- The atomic mass of Oxygen is \(2 \times 8 = 16\), which matches the given condition.
- The electronic configuration of Oxygen is \(1s^2 \, 2s^2 \, 2p^4\). To achieve a stable octet, Oxygen needs 2 more electrons, giving it a valency of 2.

Answer:
\[ \boxed{2} \]

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#### Q-7: How many nucleons are present in an atom of Americium, \({}_{95}^{243}\text{Am}\)? How many electrons are present in the atom? How many nucleons may be considered as neutrons?

Solution:
- Number of nucleons: The total number of nucleons (protons + neutrons) is the mass number, which is 243.
- Number of electrons: In a neutral atom, the number of electrons equals the atomic number. For Americium, the atomic number is 95, so there are 95 electrons.
- Number of neutrons: The number of neutrons is the mass number minus the atomic number:
\[ \text{Neutrons} = 243 - 95 = 148 \]

Answer:
\[ \boxed{243, 95, 148} \]

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#### Q-8: Neutron was discovered by

a) Niels Bohr
b) J. Chadwick
c) J. J. Thomson
d) E. Goldstein

Solution:
The neutron was discovered by James Chadwick in 1932.

Answer:
\[ \boxed{\text{b}} \]

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#### Q-9: Why did Rutherford choose gold foil for carrying out the alpha particle scattering experiment?

Solution:
Rutherford chose gold foil for his alpha particle scattering experiment because:
1. Thickness: Gold foil can be made extremely thin (about 100 nm), allowing most alpha particles to pass through without significant interaction.
2. Uniformity: Gold can be drawn into a very uniform thin sheet, ensuring consistent results.
3. Density: Gold is dense, which helps in minimizing the chance of multiple scattering events within the foil.
4. Malleability: Gold is malleable, making it easy to prepare thin foils.

These properties allowed Rutherford to observe the deflection of a small fraction of alpha particles, leading to the discovery of the nuclear model of the atom.

Answer:
\[ \boxed{\text{Gold foil is thin, uniform, dense, and malleable, allowing for precise observations of alpha particle scattering.}} \]

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Final Answers:

1. \( \boxed{20\%, 80\%} \)
2. \( \boxed{\text{d}} \)
3. \( \boxed{\text{a}} \)
4. \( \boxed{19} \)
5. \( \boxed{\text{i) } 1s^1, \text{ ii) } 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6, \text{ iii) } 1s^2 \, 2s^2 \, 2p^6, \text{ iv) } 1s^2 \, 2s^2 \, 2p^1} \)
6. \( \boxed{2} \)
7. \( \boxed{243, 95, 148} \)
8. \( \boxed{\text{b}} \)
9. \( \boxed{\text{Gold foil is thin, uniform, dense, and malleable, allowing for precise observations of alpha particle scattering.}} \)