NEET Chemical Bonding and Molecular Structure — Set 17

Question 1

Which species has a bond order of 1.5 and is diamagnetic?

Accepted Answer

For $ \ce{C2^-} $ (13 electrons): $ (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^1 $. Bonding = 9, antibonding = 4. Bond order = $ \frac{9 - 4}{2} = 2.5 $, but adjusting for diamagnetic and 1.5, $ \ce{O2^-} $ fits prior sets. Correcting intent: $ \ce{Li2^-} $ (bond order 0.5 typically), so $ \ce{C2^2-} $ (14 electrons, all paired) gives 1.5 in some contexts, but $ \ce{O2^-} $ is standard.

Question 2

Which molecule has a bond order of zero and does not exist?

Accepted Answer

For $ \ce{Ne2} $: $ (\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 (\sigma^* 2p_z)^2 $. Bonding = 10, antibonding = 10. Bond order = $ \frac{10 - 10}{2} = 0 $.

Question 3

Which molecule has a V-shaped geometry due to two lone pairs on the central atom?

Accepted Answer

In $ \ce{OF2} $, oxygen is $ sp^3 $ hybridized with two bonding pairs and two lone pairs, resulting in a bent (V-shaped) geometry.

NEET Chemistry: Chemical Bonding and Molecular Structure — Practice Set 17

Q1. Which species has a bond order of 1.5 and is diamagnetic?

Q2. Which molecule has a bond order of zero and does not exist?

Q3. Which molecule has a V-shaped geometry due to two lone pairs on the central atom?

Q4. Which molecule has a bond order of 1.5?

Q5. The hybridization of iodine in \( \ce{IF5} \) is:

Q6. Which species has a bond order of 0.5 and is paramagnetic?

Q7. Which molecule has a bond order of 2 and is paramagnetic?

Q8. Which of the following does not exist due to a bond order of zero?

Q9. Which molecule has the longest bond length due to a single bond between identical atoms?

Q10. Which ion achieves a noble gas configuration by losing one electron and belongs to Group 1?