Question 9: How many bonding pair are present in carbon(IV) oxide molecule? Carbon(IV) oxide is commonly known as carbon dioxide ($\text{CO}_2$). To determine the number of bonding pairs, we draw its Lewis structure: 1. Count total valence electrons: Carbon (C) has 4 valence electrons. Oxygen (O) has 6 valence electrons. Total valence electrons = $4 + (2 \times 6) = 4 + 12 = 16$ electrons. 2. Arrange atoms: Carbon is the central atom, bonded to two oxygen atoms: O - C - O. 3. Form single bonds: Place a single bond between the central carbon and each oxygen atom. This uses $2 \times 2 = 4$ electrons. Remaining electrons = $16 - 4 = 12$ electrons. 4. Distribute remaining electrons as lone pairs to outer atoms: Each oxygen needs 6 more electrons to complete its octet. $\ddot{\text{O}} - \text{C} - \ddot{\text{O}}$ This uses $2 \times 6 = 12$ electrons. Remaining electrons = $12 - 12 = 0$ electrons. 5. Check octets: At this point, each oxygen has an octet (2 from bond + 6 lone pair), but carbon only has 4 electrons (2 from each single bond). 6. Form multiple bonds: To satisfy carbon's octet, move one lone pair from each oxygen atom to form a double bond with the carbon atom. $\ddot{\text{O}} = \text{C} = \ddot{\text{O}}$ 7. Final check: Each oxygen atom now has 4 electrons from the double bond and 4 lone pair electrons, totaling 8 electrons. The carbon atom has 4 electrons from each double bond, totaling 8 electrons. All atoms have complete octets. 8. Count bonding pairs: There is a double bond between the left oxygen and carbon. A double bond consists of 2 bonding pairs. There is a double bond between the right oxygen and carbon. A double bond consists of 2 bonding pairs. Total bonding pairs = $2 + 2 = 4$. Therefore, there are 4 bonding pairs in a carbon(IV) oxide molecule. The correct option is B.