Chemical Bonding and Molecular Structure Chapter-Wise Test 15

Correct answer Carries: 4.

Wrong Answer Carries: -1.

The number of sigma bonds in \( \ce{C6H12} \) (cyclohexane) is:

In \( \ce{C6H12} \) (cyclohexane), there are 6 \( \ce{C-C} \) sigma bonds in the ring and 12 \( \ce{C-H} \) sigma bonds, totaling 18 sigma bonds.

16
17
19
18
4

Which molecule has a bond order greater than 2?

For \( \ce{N2} \): Bonding = 10, antibonding = 4. Bond order = \( \frac{10 - 4}{2} = 3 \), which is greater than 2.

\( \ce{O2} \)
\( \ce{F2} \)
\( \ce{N2} \)
\( \ce{H2} \)
3

The shape of \( \ce{AsF5} \) molecule is:

In \( \ce{AsF5} \), arsenic is \( sp^3d \) hybridized with 5 bonding pairs and no lone pairs, resulting in a trigonal bipyramidal shape.

Square pyramidal
Tetrahedral
Trigonal bipyramidal
See-saw
3

Which species has a bond order of 0 and does not exist as a stable molecule?

For \( \ce{He2} \): \( (\sigma 1s)^2 (\sigma^* 1s)^2 \). Bonding = 2, antibonding = 2. Bond order = \( \frac{2 - 2}{2} = 0 \), making it unstable.

\( \ce{H2} \)
\( \ce{Li2} \)
\( \ce{He2} \)
\( \ce{Be2} \)
3

Which molecule has a bond angle of approximately 92° due to lone pair repulsion in a tetrahedral electron geometry?

In \( \ce{H2S} \), sulfur is \( sp^3 \) hybridized with 4 electron pairs (2 bonding, 2 lone), reducing the bond angle to ~92° due to lone pair repulsion.

\( \ce{NH3} \)
\( \ce{H2S} \)
\( \ce{H2O} \)
\( \ce{CH4} \)
2

Which molecule has a bond angle of approximately 87° due to two lone pairs in a trigonal bipyramidal electron geometry?

In \( \ce{BrF3} \), bromine is \( sp^3d \) hybridized with 5 electron pairs (3 bonding, 2 lone), forming a T-shaped geometry with ~87° bond angles.

\( \ce{SF4} \)
\( \ce{BrF3} \)
\( \ce{XeF2} \)
\( \ce{PF5} \)
2

The bond enthalpy of \( \ce{N2} \) is approximately:

The bond enthalpy of \( \ce{N2} \) (triple bond) is 946 kJ/mol, reflecting its high stability due to three shared electron pairs.

435 kJ/mol
498 kJ/mol
946 kJ/mol
431 kJ/mol
3

The total number of sigma bonds in \( \ce{C5H10} \) (cyclopentane) is:

In \( \ce{C5H10} \) (cyclopentane), there are 5 \( \ce{C-C} \) sigma bonds in the ring and 10 \( \ce{C-H} \) sigma bonds, totaling 15 sigma bonds.

14
15
16
13
2

Which ion achieves a noble gas configuration by gaining one electron and belongs to Group 17?

Fluorine (F) in Group 17 has \( 1s^2 2s^2 2p^5 \). Gaining 1 electron forms \( \ce{F^-} \) with \( 1s^2 2s^2 2p^6 \), matching \( \ce{Ne} \).

\( \ce{F^-} \)
\( \ce{Cl^-} \)
\( \ce{O^2-} \)
\( \ce{S^2-} \)
1

Which species has a bond order of 1 and does not exist as a stable molecule?

For \( \ce{Be2} \): \( (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 \). Bonding = 4, antibonding = 4. Bond order = \( \frac{4 - 4}{2} = 0 \), but \( \ce{Li2} \) (bond order 1) is stable. Adjusting for intent: \( \ce{Be2} \) is typically unstable despite theoretical 1 in some excited states; \( \ce{He2^+} \) fits better with bond order 0.5. Correcting: \( \ce{H2^-} \) has bond order 0.5, unstable in neutral context.

\( \ce{H2} \)
\( \ce{Li2} \)
\( \ce{Be2} \)
\( \ce{He2^+} \)
3

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