Chemical Bonding and Molecular Structure Chapter-Wise Test 10

Correct answer Carries: 4.

Wrong Answer Carries: -1.

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).

\( \ce{N2} \)
\( \ce{NO} \)
\( \ce{O2} \)
\( \ce{F2} \)
2

Which molecule has an octahedral electron pair geometry but a square planar molecular shape?

In \( \ce{XeF4} \), xenon has 6 electron pairs (4 bonding, 2 lone) with \( sp^3d^2 \) hybridization, giving an octahedral electron geometry and a square planar shape.

\( \ce{XeF4} \)
\( \ce{SF6} \)
\( \ce{BrF5} \)
\( \ce{ClF5} \)
1

Which molecule has a see-saw shape with an equatorial bond angle of approximately 101°?

In \( \ce{SF4} \), sulfur is \( sp^3d \) hybridized with 5 electron pairs (4 bonding, 1 lone), forming a see-saw shape. The equatorial F-S-F bond angle is ~101°, reduced due to lone pair repulsion.

\( \ce{SF4} \)
\( \ce{ClF3} \)
\( \ce{XeF4} \)
\( \ce{PF5} \)
1

Which ion achieves a noble gas configuration by losing two electrons and belongs to Group 2?

Magnesium (Mg) in Group 2 has \( [\ce{Ne}] 3s^2 \). Losing 2 electrons forms \( \ce{Mg^2+} \) with \( [\ce{Ne}] \), a noble gas configuration.

\( \ce{Mg^2+} \)
\( \ce{Ca^2+} \)
\( \ce{Na^+} \)
\( \ce{Al^3+} \)
1

The geometry of \( \ce{SF6} \) molecule is:

In \( \ce{SF6} \), sulfur undergoes \( sp^3d^2 \) hybridization, forming an octahedral geometry with 6 fluorine atoms.

Tetrahedral
Trigonal bipyramidal
Square planar
Octahedral
4

Which of the following molecules has a single covalent bond formed by the sharing of one electron pair between two atoms?

A single covalent bond involves sharing one electron pair. In \( \ce{H2} \), two hydrogen atoms share one pair of electrons, forming a single bond.

\( \ce{H2} \)
\( \ce{O2} \)
\( \ce{N2} \)
\( \ce{CO} \)
1

In which molecule does the central atom have an odd number of electrons, violating the octet rule?

In \( \ce{NO} \), nitrogen has 11 electrons total, leading to an odd-electron system that cannot satisfy the octet rule for all atoms.

\( \ce{CO2} \)
\( \ce{NO} \)
\( \ce{H2O} \)
\( \ce{N2} \)
2

Which molecule has a see-saw shape with an axial-equatorial bond angle of approximately 86°?

In \( \ce{SF4} \), sulfur is \( sp^3d \) hybridized with 5 electron pairs (4 bonding, 1 lone), forming a see-saw shape. The axial-equatorial F-S-F bond angle is ~86°, and the equatorial bond angle is ~101°, both reduced due to lone pair repulsion.

\( \ce{ClF3} \)
\( \ce{SF4} \)
\( \ce{XeF2} \)
\( \ce{PF5} \)
2

The hybridization of nitrogen in \( \ce{N2H4} \) (hydrazine) is:

In \( \ce{N2H4} \), each nitrogen is bonded to two hydrogens and one nitrogen, with one lone pair, requiring four \( sp^3 \) hybrid orbitals.

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

The shape of \( \ce{PCl3} \) molecule is:

In \( \ce{PCl3} \), phosphorus is \( sp^3 \) hybridized with 4 electron pairs (3 bonding, 1 lone), resulting in a trigonal pyramidal shape.

Trigonal planar
Bent
Tetrahedral
Trigonal pyramidal
4

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