Chemical Bonding and Molecular Structure Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Which molecule violates the octet rule by having fewer than 8 electrons around the central atom?

In \( \ce{BH3} \), boron has 3 bonds (6 electrons), resulting in an incomplete octet due to only 3 valence electrons.

\( \ce{BH3} \)
\( \ce{PCl5} \)
\( \ce{SF6} \)
\( \ce{NH3} \)
1

Which species has a bond order of 1.5 due to the addition of an electron to a diatomic molecule?

For \( \ce{O2^-} \): \( (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^2 (\pi^* 2p_y)^1 \). Bonding = 10, antibonding = 7. Bond order = \( \frac{10 - 7}{2} = 1.5 \).

\( \ce{N2^+} \)
\( \ce{O2} \)
\( \ce{F2^+} \)
\( \ce{O2^-} \)
4

Which molecule has a bond angle of approximately 90° in a square pyramidal shape?

In \(\ce{IF5}\), iodine is \(sp^3d^2\) hybridized with 6 electron pairs (5 bonding, 1 lone), forming a square pyramidal shape. The axial-equatorial F-I-F bond angles are ~90°, slightly reduced by lone pair repulsion.

\(\ce{BrF5}\)
\(\ce{IF5}\)
\(\ce{SF6}\)
\(\ce{XeF4}\)
2

The number of sigma bonds in \( \ce{C4H10} \) (butane) is:

In \( \ce{CH3-CH2-CH2-CH3} \): 3 \( \ce{C-C} \) sigma bonds, 10 \( \ce{C-H} \) sigma bonds. Total = 13 sigma bonds.

12
13
14
11
2

Which species is diamagnetic and has a bond order of 3?

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 = \( \frac{10 - 4}{2} = 3 \), with all electrons paired (diamagnetic).

\( \ce{O2} \)
\( \ce{B2} \)
\( \ce{N2} \)
\( \ce{NO} \)
3

Which molecule has a bond angle of approximately 107° due to lone pair repulsion?

In \( \ce{NH3} \), nitrogen is \( sp^3 \) hybridized with 4 electron pairs (3 bonding, 1 lone), reducing the bond angle to ~107° due to lone pair repulsion.

\( \ce{H2O} \)
\( \ce{NH3} \)
\( \ce{CH4} \)
\( \ce{BF3} \)
2

The hybridization of phosphorus in \( \ce{PCl5} \) is:

In \( \ce{PCl5} \), phosphorus forms 5 bonds using one s, three p, and one d orbital, resulting in \( sp^3d \) hybridization.

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

The hybridization of sulfur in \( \ce{SO3} \) is:

In \( \ce{SO3} \), sulfur has 3 double bonds (3 bonding domains), requiring \( sp^2 \) hybridization for a trigonal planar shape.

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

The number of lone pairs on sulfur in \( \ce{SF4} \) is:

In \( \ce{SF4} \), sulfur has 6 valence electrons, forms 4 bonds (8 electrons used), leaving 2 electrons as 1 lone pair in a \( sp^3d \) hybridized state.

2
3
1
0
3

The shape of \( \ce{ClF5} \) molecule is:

In \( \ce{ClF5} \), chlorine is \( sp^3d^2 \) hybridized with 6 electron pairs (5 bonding, 1 lone), resulting in a square pyramidal shape.

Trigonal bipyramidal
Octahedral
Square pyramidal
See-saw
3

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