Question

Which one of the following is the correct set with reference to molecular formula, hybridization of central atom and shape of the molecule? (a) \(\mathrm{CO}_{2}, \mathrm{sp}^{2}\), bent (b) \(\mathrm{H}_{2} \mathrm{O}, \mathrm{sp}^{2}\), bent (c) \(\mathrm{BeCl}_{2}\), sp, linear (d) \(\mathrm{H}_{2} \mathrm{O}, \mathrm{sp}^{3}\), linear

Short answer

Option (c) is correct: \(\mathrm{BeCl}_{2}, \mathrm{sp}, \text{linear}\).

Step-by-step solution

Analyze Option (a)

Option (a) is \( \mathrm{CO}_{2}, \mathrm{sp}^{2}, \text{bent} \). \(\mathrm{CO}_2\) has carbon as the central atom with two double bonds with oxygen. The carbon atom adopts \(\mathrm{sp}\) hybridization because it forms two sigma bonds and has no lone pairs. Therefore, the shape is linear, not bent. So, this option is incorrect.

Analyze Option (b)

Option (b) is \(9 \mathrm{H}_{2} \mathrm{O}, \mathrm{sp}^{2}, \text{bent} \).Water molecule, \(\mathrm{H}_2\mathrm{O}\), has oxygen as the central atom. It comprises 2 sigma bonds and 2 lone pairs. This gives it a \(\mathrm{sp}^{3}\) hybridization, resulting in a bent shape with a bond angle of approximately 104.5°. Thus, this option is incorrect regarding hybridization.

Analyze Option (c)

Option (c) is \(9 \mathrm{BeCl}_{2}, \mathrm{sp}, \text{linear} \). \(\mathrm{BeCl}_2\) has beryllium as the central atom. It forms two sigma bonds with chlorine atoms and has no lone pairs. Hence, the hybridization is \(\mathrm{sp}\), and the molecule is indeed linear. This option is correct.

Analyze Option (d)

Option (d) is \(9 \mathrm{H}_{2} \mathrm{O}, \mathrm{sp}^{3}, \text{linear} \).As established from analyzing option (b), water has \(\mathrm{sp}^{3}\) hybridization but features a bent shape instead of linear because of its 2 lone pairs of electrons. Hence, this option is incorrect regarding the shape.

Key concepts

Hybridization

Hybridization is a concept used to describe the mixing of atomic orbitals to form new hybrid orbitals. These hybrid orbitals tend to have different energies, shapes, and orientations compared to the original atomic orbitals. The type of hybridization in a molecule depends on the number of sigma bonds and lone pairs around the central atom.

For instance, in the case of \(\small{\mathrm{CO}_2}\), the central atom, carbon, forms two double bonds with oxygen. Despite each double bond containing one sigma and one pi-bond, only the sigma bonds count towards hybridization. With no lone pairs on carbon, the hybridization is \(\mathrm{sp}\), indicating a linear shape due to the 180° bond angle.

In contrast, water (\(\small{\mathrm{H}_2\mathrm{O}}\)) has an oxygen central atom with two sigma bonds and two lone pairs. This arrangement results in \(\mathrm{sp}^3\) hybridization despite having only two bonded atoms. This type of hybrid orbitals makes the molecular shape bent, deviating slightly from 109.5° to about 104.5° due to lone pair-bond pair repulsion.

Hybridization is crucial in determining both the geometry and properties of molecules, thus providing insightful predictions about molecular behavior.

Molecular Shape

The molecular shape of a molecule determines how the atoms in a molecule are arranged in space, which can influence the physical and chemical properties of the substance. VSEPR (Valence Shell Electron Pair Repulsion) theory helps predict the shape by considering electron pairs around the central atom.

For example, \(\small{\mathrm{CO}_2}\) has a linear shape because of its \(\mathrm{sp}\) hybridization and symmetrical distribution of bonding pairs around carbon. Each oxygen atom forms a double bond with carbon, leading to a 180° angle between the atoms.

• Linear Geometries: Seen when there are two regions of electron density, often leading to a 180° bond angle.

Alternatively, the water molecule, \(\small{\mathrm{H}_2\mathrm{O}}\), exhibits a bent shape because of the \(\mathrm{sp}^3\) hybridization and the presence of two lone pairs on the oxygen. The repulsion caused by the lone pairs compresses the bond angle to approximately 104.5°.

• Bent Structures: Occur when there are lone pairs exerting repulsive forces, decreasing bond angles.

Understanding molecular shapes is crucial because it affects reactivity, polarity, phase of matter, color, magnetism, biological activity, and many other properties.

Sigma Bonds

Sigma bonds are one of the primary types of covalent bonds in chemistry, formed by the head-on overlap of atomic orbitals. A sigma bond (c) is characterized by its cylindrical symmetry around the bond axis, providing a single bond that is strong and stable.

In the molecule \(\small{\mathrm{CO}_2}\), carbon forms two sigma bonds, one with each oxygen. Despite each being part of a double bond, only the sigma component is considered in hybridization, leading to the \(\mathrm{sp}\) hybrid arrangement;
while in \(\small{\mathrm{H}_2\mathrm{O}}\), the oxygen atom is involved in two sigma bonds with the hydrogen atoms. Alongside \sp^3 hybrid orbitals,
the sigma bonds ensure the bent shape of the molecule.

• Characteristics of Sigma Bonds:
• Formed by direct orbital overlap.
• Allows free rotation of bonded atoms.
• Strong and stable, usually forming the base for more complex interactions like pi-bonds.

Sigma bonds are fundamental in constructing the basic frameworks of molecules, influencing reactivity, and helping to define molecular shapes.