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 bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is :