Chemical Bonding and Molecular Structure Chapter-Wise Test 4

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Which molecule has a bond angle of approximately 102° due to lone pair repulsion?

In \(\ce{NF3}\), nitrogen is \(sp^3\) hybridized with 4 electron pairs (3 bonding, 1 lone), forming a trigonal pyramidal shape. The F-N-F bond angle is ~102°, reduced from 107° in \(\ce{NH3}\) due to greater lone pair repulsion from electronegative fluorine atoms.

\(\ce{NH3}\)
\(\ce{NF3}\)
\(\ce{H2O}\)
\(\ce{CH4}\)
2

The hybridization of boron in \( \ce{B2H6} \) (diborane) is:

In \( \ce{B2H6} \), each boron is \( sp^3 \) hybridized, forming two normal \( \ce{B-H} \) bonds and two 3-center-2-electron bridge bonds.

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

Which molecule has an intramolecular hydrogen bond?

In o-nitrophenol, the hydrogen of the \( \ce{OH} \) group forms a hydrogen bond with the oxygen of the \( \ce{NO2} \) group within the same molecule.

\( \ce{HF} \)
\( \ce{H2O} \)
\( \ce{NH3} \)
o-nitrophenol
4

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

In \( \ce{HCN} \), carbon has 2 bonding domains (1 triple bond, 1 single bond), requiring \( sp \) hybridization, forming a linear shape.

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

The number of lone pairs on chlorine in \( \ce{ClF5} \) is:

In \( \ce{ClF5} \), chlorine 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
0
1
3
3

Which ion achieves a stable noble gas configuration by gaining two electrons?

Oxygen (O) has \( 1s^2 2s^2 2p^4 \). Gaining 2 electrons forms \( \ce{O^2-} \) with \( 1s^2 2s^2 2p^6 \), matching \( \ce{Ne} \).

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

Which species is paramagnetic and has a bond order of 1.5?

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 = \( \frac{10 - 7}{2} = 1.5 \), with 1 unpaired electron (paramagnetic).

\( \ce{N2} \)
\( \ce{F2} \)
\( \ce{O2^-} \)
\( \ce{C2} \)
3

Which molecule has a coordinate bond formed by donation of a lone pair?

In \( \ce{NH4^+} \), nitrogen donates a lone pair to \( \ce{H^+} \), forming a coordinate bond.

\( \ce{NH4^+} \)
\( \ce{CH4} \)
\( \ce{H2O} \)
\( \ce{N2} \)
1

Which molecule is paramagnetic due to unpaired electrons?

\( \ce{O2} \) has two unpaired electrons in \( \pi^* 2p \) orbitals, making it paramagnetic.

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

The number of sigma bonds in \( \ce{C2H4} \) is:

In \( \ce{C2H4} \) (ethene), there is one \( \ce{C=C} \) (1 sigma + 1 pi) and four \( \ce{C-H} \) bonds (4 sigma), totaling 5 sigma bonds.

4
5
6
3
2

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