Question

Draw the Lewis structures for the following species to assist you in answering the questions: \(\mathrm{SeCl}_{2}, \mathrm{ICl}, \mathrm{BH}_{3}, \mathrm{NO}_{2}^{-}\). a. Which of the species is(are) polar? b. Which of the species exhibits hydrogen-bonding interactions?

Short answer

a. The polar species are \(\mathrm{SeCl}_{2}, \mathrm{ICl},\) and \(\mathrm{NO}_{2}^{-}\). b. None of the given species exhibits hydrogen-bonding interactions.

Step-by-step solution

Drawing Lewis structures

Draw the Lewis structures for the given species: 1. \(\mathrm{SeCl}_{2}\): Selenium has 6 valence electrons and each chlorine atom has 7 valence electrons. The Lewis structure can be drawn as: \[ \mathrm{Se} ( :_\mathrm{Cl} -_\mathrm{Se} -_\mathrm{Cl} :) \] 2. \(\mathrm{ICl}\): Iodine has 7 valence electrons, and chlorine has 7 valence electrons. The Lewis structure can be drawn as: \[ \mathrm{I} ( :_\mathrm{Cl} -_\mathrm{I} : : : ) \] 3. \(\mathrm{BH}_{3}\): Boron has 3 valence electrons and each hydrogen atom has 1 valence electron. The Lewis structure can be drawn as: \[ \mathrm{B} ( -_\mathrm{H} -_\mathrm{B} -_\mathrm{H} ) \] 4. \(\mathrm{NO}_{2}^{-}\): Nitrogen has 5 valence electrons, each oxygen atom has 6 valence electrons, and there is 1 extra electron due to the negative charge. The Lewis structure can be drawn as: \[ \mathrm{N} ( -_\mathrm{O} =_\mathrm{N} -_\mathrm{O} :) \]

Determine polarity of the species

Now, analyze the Lewis structure of each species to determine its polarity. 1. \(\mathrm{SeCl}_{2}\): The molecular geometry of \(\mathrm{SeCl}_{2}\) is bent, which results in a net dipole moment. Therefore, \(\mathrm{SeCl}_{2}\) is polar. 2. \(\mathrm{ICl}\): The molecular geometry of \(\mathrm{ICl}\) is linear, but due to the difference in electronegativity between Iodine and Chlorine, there is a net dipole moment. Therefore, \(\mathrm{ICl}\) is polar. 3. \(\mathrm{BH}_{3}\): The molecular geometry of \(\mathrm{BH}_{3}\) is trigonal planar, which results in a zero dipole moment. Therefore, \(\mathrm{BH}_{3}\) is nonpolar. 4. \(\mathrm{NO}_{2}^{-}\): The molecular geometry of \(\mathrm{NO}_{2}^{-}\) is bent, which results in a net dipole moment. Therefore, \(\mathrm{NO}_{2}^{-}\) is polar. a. The polar species are \(\mathrm{SeCl}_{2}, \mathrm{ICl},\) and \(\mathrm{NO}_{2}^{-}\).

Determine species with hydrogen-bonding interactions

To exhibit hydrogen-bonding interactions, a species must have an H atom bonded to a highly electronegative atom (N, O, or F). 1. \(\mathrm{SeCl}_{2}\): There are no hydrogen atoms bonded to a highly electronegative atom, so it cannot exhibit hydrogen-bonding interactions. 2. \(\mathrm{ICl}\): There are no hydrogen atoms bonded to a highly electronegative atom, so it cannot exhibit hydrogen-bonding interactions. 3. \(\mathrm{BH}_{3}\): Hydrogen atoms are bonded to boron, which is not a highly electronegative atom, so it cannot exhibit hydrogen-bonding interactions. 4. \(\mathrm{NO}_{2}^{-}\): There are no hydrogen atoms in this species, so it cannot exhibit hydrogen-bonding interactions. b. None of the given species exhibits hydrogen-bonding interactions.

Key concepts

Polarity of Molecules

Polarity in molecules is determined by the arrangement of the atoms and the difference in electronegativity between them. When a molecule has an uneven distribution of electrical charge, it is considered polar. In simple terms, polar molecules have a positive end and a negative end.
For instance, let's analyze the examples from our exercise:

• **SeCl extsubscript{2}**: This molecule has a "bent" shape because of the electron pairs. The geometry causes an unequal sharing of electrons, leading to a net dipole moment, making it polar.
• **ICl**: Even though this molecule is linear, the difference in electronegativity between iodine and chlorine creates a net dipole moment, resulting in a polar molecule.
• **BH extsubscript{3}**: This molecule has a "trigonal planar" shape, which leads to an even distribution of charge. Therefore, it is nonpolar.
• **NO extsubscript{2} extsuperscript{-}**: This molecule has a "bent" shape, much like SeCl extsubscript{2}, resulting in a polar molecule as well.

Understanding polarity is crucial as it affects how molecules interact with each other and their environment, playing vital roles in solubility, boiling and melting points, and chemical reactivity.

Hydrogen Bonding

Hydrogen bonding is a special type of dipole-dipole interaction that occurs when hydrogen is bonded to highly electronegative atoms such as nitrogen (N), oxygen (O), or fluorine (F). This creates a situation where the hydrogen atoms have a partial positive charge, enabling them to interact with negative charges from electronegative atoms nearby.
In our exercise, none of the molecules exhibit hydrogen bonding. Here's why:

• **SeCl extsubscript{2} and ICl**: Although these are polar molecules, they lack hydrogen atoms bonded to electronegative atoms like N, O, or F, which are necessary for hydrogen bonding.
• **BH extsubscript{3}**: Hydrogen atoms here are bonded to boron, an element that isn't highly electronegative, hence not supporting hydrogen bonding.
• **NO extsubscript{2} extsuperscript{-}**: This ion doesn’t have hydrogen atoms at all, so hydrogen bonding is impossible.

Despite the name, hydrogen bonds are not actual bonds, but rather strong intermolecular forces. They significantly affect a molecule's physical properties such as boiling point and solubility.

Valence Electrons

Valence electrons are the outermost electrons of an atom. They play a critical role in the formation of chemical bonds, as they are the electrons involved in bonding with other atoms. By counting and arranging valence electrons using Lewis structures, we can predict the shape and behavior of molecules.
Consider how valence electrons are used in the Lewis structures discussed:

• **SeCl extsubscript{2}**: Selenium has 6 valence electrons, while each chlorine adds 7. These electrons arrange themselves to form bonds and lone pairs in a "bent" structure.
• **ICl**: Iodine and chlorine each contribute 7 valence electrons, resulting in a single bond between them with lone pairs around each atom.
• **BH extsubscript{3}**: Boron’s 3 valence electrons pair with each hydrogen's single electron to form covalent bonds in a "trigonal planar" structure.
• **NO extsubscript{2} extsuperscript{-}**: Here, the presence of an extra electron due to the negative charge needs to be considered in addition to nitrogen's 5 and each oxygen's 6 valence electrons, leading to a "bent" structure.

Understanding the arrangement of valence electrons is essential for exploring molecular geometry, chemical reactivity, and bonding characteristics.