Unit 3: Chemical Bonding and Molecular Structure — Online MCQ Test
CHEMISTRY · CLASS 11 INTERMEDIATE 1 YEAR · Telangana State Board
Practice Unit 3: Chemical Bonding and Molecular Structure with a free chapter-wise online MCQ test.
This chapter covers: Focusing on chemical bonding this chapter covers ionic covalent metallic bonds Lewis structures VSEPR theory and hybridization. Students examine valence bond theory molecular orbit....
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Unit 3: Chemical Bonding and Molecular Structure — Important Questions & Answers
Which of the following is a correct statement about an ionic bond?
- A. It is formed by sharing of electron pairs equally
- B. It is formed by complete transfer of electrons between atoms
- C. It is formed only between two non-metals
- D. It is always weaker than hydrogen bond
Answer: B. It is formed by complete transfer of electrons between atoms
An ionic bond is formed due to complete transfer of electron(s) from one atom to another, producing oppositely charged ions. The electrostatic attraction between these ions holds the compound together.
An ionic bond is formed due to complete transfer of electron(s) from one atom to another, producing oppositely charged ions. The electrostatic attraction between these ions holds the compound together.
The Lewis symbol of sodium is:
- A. Na••
- B. Na•
- C. Na:
- D. Na
Answer: B. Na•
Sodium has one valence electron in its outermost shell, so its Lewis symbol is Na•. Lewis symbols show only valence electrons.
Sodium has one valence electron in its outermost shell, so its Lewis symbol is Na•. Lewis symbols show only valence electrons.
According to VSEPR theory, the shape of CH4 is:
- A. Trigonal planar
- B. Tetrahedral
- C. Pyramidal
- D. Bent
Answer: B. Tetrahedral
Carbon in CH4 has four bond pairs and no lone pairs, so the electron pair geometry and molecular shape are tetrahedral. The H–C–H bond angle is about 109.5°.
Carbon in CH4 has four bond pairs and no lone pairs, so the electron pair geometry and molecular shape are tetrahedral. The H–C–H bond angle is about 109.5°.
Which of the following is NOT a reason for the stability of the oxygen molecule (O2) according to molecular orbital theory?
- A. It has more bonding electrons than antibonding electrons
- B. It has a positive bond order
- C. It has two unpaired electrons
- D. Its bond order is 2
Answer: C. It has two unpaired electrons
The presence of two unpaired electrons explains the paramagnetic nature of O2, not its stability. Stability is due to bond order being positive and bonding electrons exceeding antibonding electrons.
The presence of two unpaired electrons explains the paramagnetic nature of O2, not its stability. Stability is due to bond order being positive and bonding electrons exceeding antibonding electrons.
In hydrogen bonding, which of the following conditions is essential?
- A. Hydrogen must be attached to a highly electronegative atom like F, O or N
- B. Hydrogen must be attached to carbon only
- C. Hydrogen bonding occurs only in ionic compounds
- D. Hydrogen bonding is the same as covalent bonding
Answer: A. Hydrogen must be attached to a highly electronegative atom like F, O or N
Hydrogen bonding occurs when hydrogen is bonded to highly electronegative atoms such as fluorine, oxygen or nitrogen. This creates strong intermolecular attraction with lone pairs on nearby molecules.
Hydrogen bonding occurs when hydrogen is bonded to highly electronegative atoms such as fluorine, oxygen or nitrogen. This creates strong intermolecular attraction with lone pairs on nearby molecules.