Chemical Bonding and Molecular Structure Chapter-Wise Test 14

Correct answer Carries: 4.

Wrong Answer Carries: -1.

The bond order of \( \ce{F2^-} \) ion, based on molecular orbital theory, is:

For \( \ce{F2^-} \) (19 electrons): \( (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^2 (\pi^* 2p_y)^2 (\sigma^* 2p_z)^1 \). Bonding = 10, antibonding = 9. Bond order = \( \frac{10 - 9}{2} = 0.5 \).

0.5
1
1.5
2
1

What is the bond order of \( \ce{N2} \) molecule?

For \( \ce{N2} \): \( (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^2 \). Bonding = 10, antibonding = 4. Bond order = \( \frac{10 - 4}{2} = 3 \).

1
2
3
4
3

Which molecule has the highest bond enthalpy among the halogens?

\( \ce{Cl2} \) has a bond enthalpy of ~243 kJ/mol, higher than \( \ce{F2} \) (159 kJ/mol), \( \ce{Br2} \) (193 kJ/mol), or \( \ce{I2} \) (151 kJ/mol) due to optimal bond strength.

\( \ce{F2} \)
\( \ce{Br2} \)
\( \ce{I2} \)
\( \ce{Cl2} \)
4

The number of lone pairs on xenon in \( \ce{XeF2} \) is:

In \( \ce{XeF2} \), xenon has 8 valence electrons, forms 2 bonds (4 electrons used), leaving 6 electrons as 3 lone pairs in a \( sp^3d \) hybridized state.

2
1
3
4
3

Which molecule has the highest bond enthalpy among those with a single bond?

\( \ce{HF} \) has a single bond with a bond enthalpy of ~567 kJ/mol, higher than \( \ce{HCl} \) (431 kJ/mol), \( \ce{HBr} \) (366 kJ/mol), or \( \ce{HI} \) (299 kJ/mol) due to strong electronegativity difference.

\( \ce{HCl} \)
\( \ce{HBr} \)
\( \ce{HI} \)
\( \ce{HF} \)
4

Which molecule has the highest bond enthalpy?

\( \ce{N2} \) has a triple bond with a bond enthalpy of ~946 kJ/mol, higher than \( \ce{O2} \) (498 kJ/mol), \( \ce{F2} \) (159 kJ/mol), or \( \ce{Cl2} \) (243 kJ/mol).

\( \ce{O2} \)
\( \ce{F2} \)
\( \ce{N2} \)
\( \ce{Cl2} \)
3

The hybridization of oxygen in \( \ce{OF2} \) is:

In \( \ce{OF2} \), oxygen has 4 electron pairs (2 bonding, 2 lone), requiring \( sp^3 \) hybridization, forming a bent shape.

\( sp^2 \)
\( sp \)
\( sp^3d \)
\( sp^3 \)
4

The shape of \( \ce{SF4} \) molecule is:

In \( \ce{SF4} \), sulfur is \( sp^3d \) hybridized with 5 electron pairs (4 bonding, 1 lone), resulting in a see-saw shape.

Tetrahedral
Trigonal bipyramidal
Square planar
See-saw
4

Which molecule has an expanded octet with 10 electrons around the central atom?

In \( \ce{PCl5} \), phosphorus has 5 bonds (10 electrons), exceeding the octet due to d-orbital availability.

\( \ce{BF3} \)
\( \ce{NH3} \)
\( \ce{PCl5} \)
\( \ce{CH4} \)
3

Which molecule has the shortest bond length?

\( \ce{N2} \) has a triple bond (bond order 3), resulting in the shortest bond length (~110 pm) due to greater electron density pulling the nuclei closer.

\( \ce{O2} \)
\( \ce{F2} \)
\( \ce{Cl2} \)
\( \ce{N2} \)
4

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