Question

What is the total number of electrons present in bonding and antibonding orbitals respectively in \(\text{F}_2\) molecule according to MO theory ?

• A. Bonding - 8, Antibonding - 10
• B. Bonding - 6, Antibonding - 12
• C. Bonding - 10, Antibonding - 8
• D. Bonding - 12, Antibonding - 6

Hint:

Fill 18 electrons into the MO energy order and count those in bonding and antibonding orbitals.

Answer 1

The Correct Option is C

Step 1: Count electrons
Each fluorine atom has \(9\) electrons, so \(F_2\) has \(18\) electrons.
For \(F_2\) the MO energy order is \(\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < \pi 2p_x = \pi 2p_y < \pi^* 2p_x = \pi^* 2p_y < \sigma^* 2p_z\).

Step 2: Fill the orbitals
\(\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 \pi 2p_y^2, \pi^* 2p_x^2 \pi^* 2p_y^2\).

Step 3: Count
Bonding: \(2+2+2+4 = 10\) electrons. Antibonding: \(2+2+4 = 8\) electrons.
Check: \(10+8 = 18\).

Step 4: Bond order
\(\frac{10-8}{2} = 1\), which agrees with the single bond in fluorine. Option (C) is correct. The other options do not add to the correct pattern or do not give bond order 1.

Final Answer:
Bonding electrons are 10 and antibonding electrons are 8. \[ \boxed{\text{(C)}\ \text{Bonding }10,\ \text{Antibonding }8} \]