Chemical Bonding and Molecular Structure Chapter-Wise Test 1

\( \ce{N2} \) has a triple bond with a bond length of ~110 pm, shorter than \( \ce{O2} \)’s double bond (121 pm) due to higher bond order (3 vs. 2).

In \( \ce{SiCl4} \), silicon is \( sp^3 \) hybridized with no lone pairs, resulting in a tetrahedral shape with a bond angle of 109.5°.

In \( \ce{XeF6} \), xenon has 8 valence electrons, forms 6 bonds (12 electrons used), leaving 2 electrons as 1 lone pair in a \( sp^3d^3 \) hybridized state.

In \( \ce{[Cu(NH3)4]^2+} \), each \( \ce{NH3} \) donates a lone pair from nitrogen to \( \ce{Cu^2+} \), forming coordinate bonds.

For \( \ce{F2} \): \( (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^2 (\pi^* 2p_y)^2 \). All electrons are paired, so it’s diamagnetic.

In \( \ce{TeF6} \), tellurium is \( sp^3d^2 \) hybridized with 6 bonding pairs and no lone pairs, resulting in an octahedral shape.

Carbon has 4 valence electrons, and oxygen has 6. Total = 4 + 6 = 10 valence electrons in \( \ce{CO} \).

In \( \ce{NH3} \), hydrogen bonding occurs between the hydrogen of one molecule and the nitrogen lone pair of another, stronger than van der Waals forces in others.

For \( \ce{O2} \): \( (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^1 \). Bonding = 10, antibonding = 6. Bond order = \( \frac{10 - 6}{2} = 2 \), with 2 unpaired electrons (paramagnetic).

In \( \ce{BrF3} \), bromine is \( sp^3d \) hybridized with 5 electron pairs (3 bonding, 2 lone), resulting in a T-shaped molecular geometry.