Atoms, Molecules and Radicals Test

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Atoms, Molecules and Radicals Test - 1

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Weekly Quiz Competition

Question 1
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Which of the following elements has atomicity 2?

A
Helium

B
Nitrogen

C
Aluminium

D
Phosphorus

Solution

The number of atoms constituting a molecule is known as its atomicity. A molecule of nitrogen molecule consists of two atoms of nitrogen. Hence, it is a diatomic molecule and its atomicity is 2.
Question 2
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What kind of configuration does every atom want to achieve?

A
Halogen configuration

B
Noble gas configuration

C
Alkali metal configuration

D
Alkaline metal configuration

Solution

In aspect of atoms, stability is mainly attributed to attain the noble gas configuration.
Question 3
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Which of the following rules is observed while writing a chemical formula?

A
Polyatomic ions are enclosed in a bracket before writing the number to indicate the ratio.

B
The charge on cation should be more than the charge on anion.

C
Atomic numbers of combining atoms should be crossed.

D
Non-metal is written first followed by the metal.

Solution

In compounds that are formed from polyatomic ions such as ammonium ions, the ions are first enclosed in a bracket and then the number indicating the ratio is written. For example, calcium hydroxide is written as Ca(OH)₂.

The correct answer is A.

Why alternative B is wrong:

The formation of a compound does not depend on the charges of cations and anions that are present in it. However, a molecule as an entity is neutral. Thus, in a molecule, the negative and positive charges balance each other completely.

Why alternative C is wrong:

While writing the chemical formula of a compound, the valency and not the atomic numbers of combining atoms should be crossed.

Why alternative D is wrong:

While writing the chemical formula of a compound, the metal is written first followed by the non-metal.
Question 4
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The positive and negative ions present in the compound Pb₃(PO₄)₂ are

A
Pb³⁺ and PO₄^2-

B
Pb²⁺ and PO₄^3-

C
Pb⁺ and PO₄^3-

D
Pb⁺ and PO₄^2-

Solution

Ionic compounds are represented in such a manner that there is no charge on the molecule as a whole. If the charges on Pb and PO₄are +2 and −3 respectively, then the positive and negative charges on molecule will be +6 and −6 respectively, thereby making the charge on the molecule neutral.

The correct answer is B.

Why alternative A is wrong:

In Pb₃(PO₄)_2, if the charges on Pb and PO₄are +3 and −2 respectively, then the positive and negative charges on the molecule will be +9 and −4 respectively. On the whole, the molecule will be positive, which is not possible.

Why alternative C is wrong:

In Pb₃(PO₄)_2, if the charges on Pb and PO₄are +1 and −3 respectively, then the positive and negative charges on the molecule will be +3 and −6 respectively. On the whole, the molecule will be negative, which is not possible.

Why alternative D is wrong:

In Pb₃(PO₄)_2, if the charges on Pb and PO₄are +1 and −2 respectively, then the positive and negative charges on the molecule will be +3 and −4 respectively. On the whole, the molecule will be negative, which is not possible.
Question 5
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Which of the following ions is mono-atomic?

A
Nitrate

B
Sulphite

C
Sulphide

D
Hydroxide

Solution

Sulphide (S²⁻) ion is mono-atomic since it consists of only one single charged atom.
Question 6
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What are the respective numbers of carbon, hydrogen, and oxygen atoms in a molecule of glucose?

A
6, 6, and 6

B
6, 12, and 6

C
12, 6, and 12

D
12, 12, and 12

Solution

The chemical formula of glucose is C₆H₁₂O₆. The formula of glucose shows that a molecule of glucose contains six carbon atoms, twelve hydrogen atoms, and six oxygen atoms.
Question 7
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A chemical compound is composed of

A
a single atom

B
a single molecule

C
different atoms

D
different molecules

Solution

Two or more atoms of different elements joined together to form a chemical

compound.
Question 8
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Which of the following configurations is most stable?

A
1s² 2s²2p⁶3s²

B
1s² 2s²2p⁶3s¹

C
1s² 2s²2p⁶

D
1s² 2s²2p⁵

Solution

The configuration 1s² 2s²2p⁶corresponds to neon, which is a noble gas and is hence, stable.
Question 9
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What is the electronic configuration of sulphur after it gains two electrons to form sulphide (S^–2) ion?

A
2, 8

B
2, 10

C
2, 8, 8

D
2, 10, 6

Solution

The atomic number of sulphur is 16 and its electronic configuration is 2, 8, 6.

If an atom gains electrons, the electrons add to the outermost shell. Thus, the electronic configuration of sulphur after gaining two electrons will become 2, 8, 8. Thus, the electronic configuration of sulphide (S^–2) ion is 2, 8, 8.
Question 10
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Use the following information to answer the next question.

Four boxes A, B, C, and D contain the number atoms which are present in one molecule each of sugar (C₁₂H₂₂O₁₁), carbon dioxide (CO₂), sodium carbonate (Na₂CO₃), and glucose (C₆H₁₂O₆) respectively.

If the boxes are arranged in the increasing order of the number of atoms they contain, then the correct order will be

A
A < B < D < C

B
B < D < C < A

C
B < C < D < A

D
C < D < B < A

Solution

The number of atom in the boxes A, B, C, and D are 45, 3, 6, and 24 respectively.

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Atoms, Molecules and Radicals Test - 1

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Question 1

Helium

Nitrogen

Aluminium

Phosphorus

Solution

Question 2

Halogen configuration

Noble gas configuration

Alkali metal configuration

Alkaline metal configuration

Solution

Question 3

Polyatomic ions are enclosed in a bracket before writing the number to indicate the ratio.

The charge on cation should be more than the charge on anion.

Atomic numbers of combining atoms should be crossed.

Non-metal is written first followed by the metal.

Solution

Question 4

Pb3+ and PO42-

Pb2+ and PO43-

Pb+ and PO43-

Pb+ and PO42-

Solution

Question 5

Nitrate

Sulphite

Sulphide

Hydroxide

Solution

Question 6

6, 6, and 6

6, 12, and 6

12, 6, and 12

12, 12, and 12

Solution

Question 7

a single atom

a single molecule

different atoms

different molecules

Solution

Question 8

1s2 2s2 2p6 3s2

1s2 2s2 2p6 3s1

1s2 2s2 2p6

1s2 2s2 2p5

Solution

Question 9

2, 8

2, 10

2, 8, 8

2, 10, 6

Solution

Question 10

A < B < D < C

B < D < C < A

B < C < D < A

C < D < B < A

Solution

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Pb³⁺ and PO₄^2-

Pb²⁺ and PO₄^3-

Pb⁺ and PO₄^3-

Pb⁺ and PO₄^2-

1s² 2s²2p⁶3s²

1s² 2s²2p⁶3s¹

1s² 2s²2p⁶

1s² 2s²2p⁵