Question

Among the following compounds, the one that is polar and has the central atom with sp \(^{2}\) hybridization is (a) \(\mathrm{SiF}_{4}\) (b) \(\mathrm{BF}_{3}\) (c) \(\mathrm{HClO}_{2}\) (d) \(\mathrm{H}_{2} \mathrm{CO}_{3}\)

Short answer

The compound is \(\mathrm{H}_2\mathrm{CO}_3\).

Step-by-step solution

Identify the Central Atom in Each Compound

First, let's identify the central atom in each of the given compounds. In \(\mathrm{SiF}_4\), the central atom is silicon (Si). In \(\mathrm{BF}_3\), it is boron (B). In \(\mathrm{HClO}_2\), chlorine (Cl) is the central atom. Lastly, in \(\mathrm{H}_2 \mathrm{CO}_3\), the central atom is carbon (C).

Determine the Hybridization of the Central Atom

To determine the hybridization of the central atom, count the number of sigma bonds and lone pairs around the central atom:- \(\mathrm{SiF}_4\): Si forms four sigma bonds with F atoms, giving \(\mathrm{sp}^3\) hybridization.- \(\mathrm{BF}_3\): B forms three sigma bonds with F atoms, resulting in \(\mathrm{sp}^2\) hybridization.- \(\mathrm{HClO}_2\): Cl forms one sigma bond with each of the two oxygen atoms and has a lone pair, which results in \(\mathrm{sp}^3\) hybridization.- \(\mathrm{H}_2\mathrm{CO}_3\): C forms three sigma bonds (two to oxygen, one to the H-O group) and one double bond, leading to \(\mathrm{sp}^2\) hybridization.

Evaluate the Polarity of Each Compound

For a compound to be polar, it must have an uneven distribution of electron density. This usually happens when there are electronegative atoms creating dipoles and the molecular geometry is asymmetric:- \(\mathrm{SiF}_4\): The molecule is \(\mathrm{sp}^3\) hybridized, tetrahedral, and symmetrical, so it is nonpolar.- \(\mathrm{BF}_3\): Despite being \(\mathrm{sp}^2\) hybridized, it is trigonal planar and symmetric, making it nonpolar.- \(\mathrm{HClO}_2\): With \(\mathrm{sp}^3\) hybridization and an asymmetric bent/angular shape, it can be polar.- \(\mathrm{H}_2\mathrm{CO}_3\): It is \(\mathrm{sp}^2\) hybridized and polar due to its molecular geometry, with an uneven distribution of electron density.

Conclusion

From the analysis, the compounds with the central atom having \(\mathrm{sp}^2\) hybridization are \(\mathrm{BF}_3\) and \(\mathrm{H}_2\mathrm{CO}_3\). However, \(\mathrm{BF}_3\) is nonpolar due to its symmetry, whereas \(\mathrm{H}_2\mathrm{CO}_3\) is polar due to its asymmetric molecular structure. Therefore, \(\mathrm{H}_2\mathrm{CO}_3\) is the compound that is both polar and has the central atom with \(\mathrm{sp}^2\) hybridization.

Key concepts

Hybridization

When atoms form molecules, their atomic orbitals mix to create new hybrid orbitals that are energetically favorable for bonding. This is called hybridization. It provides a useful way to explain the shapes of molecules and the types of bonds they form.

The type of hybridization depends on the number and arrangement of electron pairs around the central atom. For example:

• \( sp \) hybridization occurs with two regions of electron density, like in linear structures.
• \( sp^2 \) hybridization involves three regions of electron density, forming trigonal planar structures.
• \( sp^3 \) hybridization involves four regions, leading to tetrahedral structures.

Each specific hybridization type accounts for a different geometry, affecting molecular polarity and properties. In the exercise, \( H_2CO_3 \) has \( sp^2 \) hybridization, meaning it has a trigonal planar arrangement which gives it certain chemical characteristics.

Molecular Geometry

Molecular geometry refers to the three-dimensional arrangement of atoms in a molecule. This shape affects both the molecule's physical and chemical properties, including its polarity. Understanding molecular geometry is essential for predicting how molecules interact with each other.

To determine the molecular geometry, consider both the bonding pairs of electrons as well as lone pairs around the central atom:

• Tetrahedral geometry, found in \( SiF_4 \), is symmetrical, usually resulting in nonpolar molecules.
• Trigonal planar geometry, which is typical of \( BF_3 \), also results in nonpolarity if there is symmetry.
• Bent or angular shapes, like in \( HClO_2 \), lead to polar molecules due to asymmetric electron distribution.

In the case of \( H_2CO_3 \), even though it's \( sp^2 \) hybridized leading to trigonal planar geometry, the asymmetry caused by the presence of different atoms results in a polar molecule. This highlights how vital it is to consider molecular geometry for understanding the chemical behavior.

Sigma Bonds

Sigma bonds are the strongest and most common type of covalent bonds between two atoms. They occur when two orbitals directly overlap along the axis connecting the two nuclei, allowing for efficient electron sharing.

In compounds like those in the exercise, each sigma bond corresponds to a single line in the molecule’s structural drawing:

• \( SiF_4 \) has four sigma bonds, one for each Si-F bond, contributing to its \( sp^3 \) hybridization.
• \( BF_3 \) features three sigma bonds, one for each B-F bond, leading to \( sp^2 \) hybridization.
• \( HClO_2 \), with its single bonds between chlorine and oxygen, forms sigma bonds as well within \( sp^3 \) hybridized surroundings.
• \( H_2CO_3 \) features three sigma bonds around carbon with oxygen and hydrogen, plus pi bonds in one case, giving \( sp^2 \) hybridization.

Each sigma bond aids in determining not just hybridization but also the molecular shape and polarity. This concept underpins why \( H_2CO_3 \) is polar: the types and connections of these bonds create an uneven distribution of electron density.