Chemical Bonding and Molecular Structure Chapter-Wise Test 7

Correct answer Carries: 4.

Wrong Answer Carries: -1.

The number of pi bonds in \( \ce{C2H2} \) is:

In \( \ce{C2H2} \) (ethyne), the triple bond between carbons consists of 1 sigma and 2 pi bonds, so there are 2 pi bonds total.

1
3
4
2
4

Which molecule has a triple bond and exhibits linear geometry?

In \( \ce{HCN} \), carbon is \( sp \) hybridized, forming a triple bond with nitrogen and a single bond with hydrogen, resulting in a linear shape with a 180° bond angle. No lone pairs on carbon affect the geometry.

\( \ce{HCN} \)
\( \ce{C2H2} \)
\( \ce{N2} \)
\( \ce{CO} \)
1

The hybridization of carbon in \( \ce{CS2} \) is:

In \( \ce{CS2} \), carbon has 2 double bonds (2 bonding domains), requiring \( sp \) hybridization for a linear shape.

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

In which molecule does hydrogen bonding significantly influence its properties?

In \( \ce{HF} \), hydrogen bonding between H and F atoms leads to higher boiling points due to strong intermolecular forces.

\( \ce{CH4} \)
\( \ce{N2} \)
\( \ce{HF} \)
\( \ce{Cl2} \)
3

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

In \( \ce{XeF2} \), xenon has 8 valence electrons and forms 2 bonds (4 electrons used), leaving 4 electrons as 3 lone pairs. With \( sp^3d \) hybridization, the 5 electron pairs (2 bonding, 3 lone) result in a trigonal bipyramidal electron geometry and a linear molecular shape.

1
4
2
3
4

The hybridization of carbon in \( \ce{C2H6} \) is:

In \( \ce{C2H6} \) (ethane), each carbon is bonded to three hydrogens and one carbon, using four \( sp^3 \) hybrid orbitals.

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

The bond order of \( \ce{B2^-} \) ion, based on molecular orbital theory, is:

For \( \ce{B2^-} \) (11 electrons): \( (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^1 \). Bonding = 7, antibonding = 4. Bond order = \( \frac{7 - 4}{2} = 1.5 \).

1.5
1
2
0.5
1

The stability of an ionic compound is primarily due to:

The stability of ionic compounds like \( \ce{NaCl} \) arises from the lattice enthalpy, which compensates for the energy required to form ions.

Lattice enthalpy
Covalent bonding
Hydrogen bonding
Van der Waals forces
1

The number of lone pairs on iodine in \( \ce{IF5} \) is:

In \( \ce{IF5} \), iodine has 7 valence electrons, forms 5 bonds (10 electrons used), leaving 2 electrons as 1 lone pair in a \( sp^3d^2 \) hybridized state.

2
3
1
0
3

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

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).

\( \ce{N2^+} \)
\( \ce{NO} \)
\( \ce{O2^+} \)
\( \ce{C2} \)
2

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