Step 1: Understanding the Topic:
The topic of this question is "Chemical Bonding and Molecular Structure." It specifically focuses on the types of covalent bonds ($\sigma$ and $\pi$) and the presence of lone pairs in various simple molecules. Sigma bonds result from the head-on overlap of orbitals and are found in all single bonds, while pi bonds result from sideways overlap and are found in double and triple bonds. Lone pairs are non-bonding valence electrons that influence molecular geometry.
Step 2: Key Formulas and Approach:
We evaluate the Lewis structure of each molecule:
A single bond is 1 $\sigma$ bond.
A double bond is 1 $\sigma$ + 1 $\pi$ bond.
A triple bond is 1 $\sigma$ + 2 $\pi$ bonds.
Lone pairs = (Valence electrons of central atom - used electrons) / 2.
Step 3: Detailed Explanation:
A. C$_2$H$_4$ (Ethene): Structure: $H_2C=CH_2$. There is one $C=C$ double bond and four $C-H$ single bonds. Total: $(1+4) = 5$ $\sigma$ bonds and $1$ $\pi$ bond. A matches with IV.
B. C$_2$H$_2$ (Ethyne): Structure: $HC \equiv CH$. There is one $C \equiv C$ triple bond and two $C-H$ single bonds. Total: $(1+2) = 3$ $\sigma$ bonds and $2$ $\pi$ bonds. B matches with I.
C. CH$_4$ (Methane): Structure: A central carbon bonded to four hydrogen atoms by single bonds. Total: 4 $\sigma$ bonds. C matches with III.
D. NH$_3$ (Ammonia): Structure: Nitrogen has 5 valence electrons. 3 are used for single bonds with Hydrogen, leaving 2 electrons as a lone pair. Total: 3 $\sigma$ bonds and 1 lone pair. D matches with II.
Step 4: Final Answer:
The matching sequence is A-IV, B-I, C-III, D-II, which is option (B).