List I (Molecule) | List II (Number and types of bond/s between two carbon atoms) | ||
A. | ethane | I. | one σ-bond and two π-bonds |
B. | ethene | II. | two π-bonds |
C. | carbon molecule, C2 | III. | one σ-bonds |
D. | ethyne | IV. | one σ-bond and one π-bond |
Step 1: Analyze the Bonding in Ethane (\( C_2H_6 \))
Ethane contains a single \( \sigma \)-bond between two carbon atoms.
Step 2: Analyze the Bonding in Ethene (\( C_2H_4 \))
Ethene has one \( \sigma \)-bond and one \( \pi \)-bond between the carbon atoms.
Step 3: Analyze the Bonding in Carbon Molecule (\( C_2 \))
The \( C_2 \) molecule contains two \( \pi \)-bonds between carbon atoms, with no \( \sigma \)-bond.
Step 4: Analyze the Bonding in Ethyne (\( C_2H_2 \))
Ethyne has one \( \sigma \)-bond and two \( \pi \)-bonds between carbon atoms.
Step 5: Match with List II
A-III: Ethane has one \( \sigma \)-bond.
B-IV: Ethene has one \( \sigma \)-bond and one \( \pi \)-bond.
C-II: Carbon molecule (\( C_2 \)) has two \( \pi \)-bonds.
D-I: Ethyne has one \( \sigma \)-bond and two \( \pi \)-bonds.
Step 6: Conclusion
The correct match is: A-III, B-IV, C-II, D-I.
The correct increasing order of stability of the complexes based on \( \Delta \) value is:
Match List-I with List-II: List-I
A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is :