NEET Chemical Bonding and Molecular Structure — Set 10

Question 1

Which molecule has a see-saw shape with an axial-equatorial bond angle of approximately 86°?

Accepted Answer

In $ \ce{SF4} $, sulfur is $ sp^3d $ hybridized with 5 electron pairs (4 bonding, 1 lone), forming a see-saw shape. The axial-equatorial F-S-F bond angle is ~86°, and the equatorial bond angle is ~101°, both reduced due to lone pair repulsion.

Question 2

The hybridization of nitrogen in $ \ce{N2H4} $ (hydrazine) is:

Accepted Answer

In $ \ce{N2H4} $, each nitrogen is bonded to two hydrogens and one nitrogen, with one lone pair, requiring four $ sp^3 $ hybrid orbitals.

Question 3

Which species has a bond order of 2.5 and is paramagnetic?

Accepted Answer

For $ \ce{NO} $: $ (\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)^1 $. Bonding = 10, antibonding = 5. Bond order = $ \frac{10 - 5}{2} = 2.5 $, with 1 unpaired electron (paramagnetic).

NEET Chemistry: Chemical Bonding and Molecular Structure — Practice Set 10

Q1. Which molecule has a see-saw shape with an axial-equatorial bond angle of approximately 86°?

Q2. The hybridization of nitrogen in \( \ce{N2H4} \) (hydrazine) is:

Q3. Which species has a bond order of 2.5 and is paramagnetic?

Q4. Which molecule has an octahedral electron pair geometry but a square planar molecular shape?

Q5. The shape of \( \ce{PCl3} \) molecule is:

Q6. Which ion achieves a noble gas configuration by losing two electrons and belongs to Group 2?

Q7. Which molecule has a see-saw shape with an equatorial bond angle of approximately 101°?

Q8. The geometry of \( \ce{SF6} \) molecule is:

Q9. In which molecule does the central atom have an odd number of electrons, violating the octet rule?

Q10. Which of the following molecules has a single covalent bond formed by the sharing of one electron pair between two atoms?