Unit 11: p-Block Elements - Group 14 — Online MCQ Test
CHEMISTRY · CLASS 11 INTERMEDIATE 1 YEAR · Telangana State Board
Practice Unit 11: p-Block Elements - Group 14 with a free chapter-wise online MCQ test.
This chapter covers: Focusing on carbon family this chapter covers oxidation states property trends allotropes of carbon anomalous carbon behavior and industrial compounds. Students examine silicates s....
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Unit 11: p-Block Elements - Group 14 — Important Questions & Answers
Which element shows the maximum tendency to form catenated compounds in Group 14?
- A. Carbon
- B. Silicon
- C. Germanium
- D. Tin
Answer: A. Carbon
Carbon has the strongest C-C bond and the smallest atomic size, so it shows the maximum catenation among Group 14 elements.
Carbon has the strongest C-C bond and the smallest atomic size, so it shows the maximum catenation among Group 14 elements.
The common oxidation states shown by Group 14 elements are:
- A. +2 and +4
- B. +1 and +3
- C. -2 and +6
- D. 0 and +1
Answer: A. +2 and +4
Due to the inert pair effect, heavier Group 14 elements commonly show +2 and +4 oxidation states.
Due to the inert pair effect, heavier Group 14 elements commonly show +2 and +4 oxidation states.
Why does carbon differ markedly from the other Group 14 elements?
- A. It has the largest atomic size in the group
- B. It has the ability to form strong pπ-pπ multiple bonds and catenation
- C. It readily expands its octet
- D. It shows only +4 oxidation state
Answer: B. It has the ability to form strong pπ-pπ multiple bonds and catenation
Carbon shows anomalous behavior because of its small size, high electronegativity, and strong ability to form multiple bonds and chains.
Carbon shows anomalous behavior because of its small size, high electronegativity, and strong ability to form multiple bonds and chains.
Which of the following best explains why lead prefers the +2 oxidation state more than the +4 state?
- A. High bond enthalpy of Pb-Pb bonds
- B. Inert pair effect
- C. Low electronegativity of oxygen
- D. Strong pπ-pπ bonding
Answer: B. Inert pair effect
In lead, the 6s electrons are less available for bonding due to the inert pair effect, making +2 more stable than +4.
In lead, the 6s electrons are less available for bonding due to the inert pair effect, making +2 more stable than +4.
Which one is the most appropriate reason for the non-formation of strong pπ-pπ bonds by silicon compared to carbon?
- A. Silicon has smaller atomic size than carbon
- B. Silicon has larger atomic size and diffuse orbitals leading to poor overlap
- C. Silicon has higher electronegativity than carbon
- D. Silicon has no valence p-orbitals
Answer: B. Silicon has larger atomic size and diffuse orbitals leading to poor overlap
Silicon atoms are larger and their orbitals overlap less effectively, so stable pπ-pπ multiple bonding is much less common than in carbon.
Silicon atoms are larger and their orbitals overlap less effectively, so stable pπ-pπ multiple bonding is much less common than in carbon.