The bond order (BO) in Molecular Orbital Theory is calculated using the formula:

$\text{Bond Order}=\frac{1}{2}(\text{Number of bonding electrons}-\text{Number of antibonding electrons})$

Let's analyze the options:

Why (D) is correct:

• Bond order can be a positive integer (e.g., 1 for H\u2082, 2 for O\u2082, 3 for N\u2082).
• Bond order can be a positive fractional value (e.g., 0.5 for H\u2082\u207a, 1.5 for O\u2082\u207a, 2.5 for N\u2082\u207a).
• Bond order can be zero if the number of bonding electrons equals the number of antibonding electrons (e.g., He\u2082, Be\u2082, Ne\u2082). A zero bond order indicates that the molecule is unstable and does not exist.
• Therefore, bond order can assume any positive value (integral or fractional) or zero.

Option Analysis:

• A) can have a -ve value: This is incorrect. The number of bonding electrons cannot be less than the number of antibonding electrons to such an extent that the bond order becomes negative. A negative bond order would imply repulsion stronger than attraction, which is not observed for stable species.
• B) has always an integral value: This is incorrect. As seen with species like H\u2082\u207a (BO = 0.5) or O\u2082\u207a (BO = 1.5), bond order can be fractional.
• C) is a non-zero quantity: This is incorrect. For molecules like He\u2082 or Be\u2082, the bond order is zero, indicating that these molecules are unstable and do not exist.

Correct Answer: (D)