Question

Give the formula of a molecule that you would expect to have the same Lewis structure as (a) \(\mathrm{OH}^{-}\) (b) \(\mathrm{O}_{2}^{2-}\) (c) \(\mathrm{CN}^{-}\) (d) \(\mathrm{SO}_{4}^{2-}\)

Short answer

Question: Identify a molecule/ion with the same Lewis structure as: (a) OH⁻ Answer: F⁻ (b) O₂²⁻ Answer: S₂²⁻ (c) CN⁻ Answer: N₂⁺ (d) SO₄²⁻ Answer: CrO₄²⁻

Step-by-step solution

(a) Finding Lewis structure of \(\mathrm{OH}^{-}\) and a similar structure

First, let's find the Lewis structure of the hydroxide ion \(\mathrm{OH}^{-}\). It has one oxygen atom and one hydrogen atom, with a -1 charge. Oxygen has 6 valence electrons, and hydrogen has 1. Due to the negative charge, we have one additional electron, bringing the total count to 8 electrons. To draw the Lewis structure, we arrange the atoms with the least electronegative atom in the center (O in this case). Then we form single bonds between O and H. This consumes 2 of our 8 electrons, leaving us 6 electrons. These remaining electrons will form 3 lone pairs around oxygen. Now, let's find a molecule/ion with the same Lewis structure. We need an atom with similar electronegativity as oxygen that can form a single bond with hydrogen and have 3 lone pairs. A matching example is the fluoride ion, \(\mathrm{F}^{-}\). It forms a single bond with hydrogen and has 3 lone pairs with a -1 charge. So, the answer for (a) is \(\mathrm{F}^{-}\).

(b) Finding Lewis structure of \(\mathrm{O}_{2}^{2-}\) and a similar structure

First, let's find the Lewis structure of the peroxide ion \(\mathrm{O}_{2}^{2-}\). Here, we have two oxygen atoms, with a -2 charge. Each oxygen atom has 6 valence electrons, so our overall count is 14 electrons, including the extra 2 electrons from the -2 charge. Placing both oxygen atoms side by side, we form a single bond between them, leaving us with 12 electrons. We then distribute the remaining electrons around the oxygen atoms as lone pairs, with 6 electrons on each oxygen atom, leading to 3 lone pairs on each. A molecule/ion with a similar Lewis structure will have two atoms with similar electronegativity forming a single bond and having 3 lone pairs each. Let's consider the sulfur: \(\mathrm{S}_{2}^{2-}\) is a suitable candidate, as it satisfies these conditions. So, the answer for (b) is \(\mathrm{S}_{2}^{2-}\).

(c) Finding Lewis structure of \(\mathrm{CN}^{-}\) and a similar structure

First, find the Lewis structure of the cyanide ion \(\mathrm{CN}^{-}\). Carbon has 4 valence electrons, and nitrogen has 5. Due to the -1 charge, we have one extra electron, bringing the total count to 10 electrons. Put the less electronegative carbon in the center and form a triple bond between carbon and nitrogen. This uses up 6 of the 10 electrons. The remaining 4 electrons will form 2 lone pairs around nitrogen. Now, let's find a molecule/ion with the same Lewis structure. We need two elements with a combined electron count of 10 and able to form a triple bond. A suitable candidate is \(\mathrm{N}_{2}^{+}\) which has a triple bond and a +1 charge, leaving one lone pair on each of the nitrogen atoms. So, the answer for (c) is \(\mathrm{N}_{2}^{+}\).

(d) Finding Lewis structure of \(\mathrm{SO}_{4}^{2-}\) and a similar structure

First, let's find the Lewis structure of the sulfate ion \(\mathrm{SO}_{4}^{2-}\). Sulfur has 6 valence electrons, and oxygen has 6 valence electrons for each of the four atoms. The -2 charge provides us with 2 additional electrons, giving us a total of 32 electrons. Place sulfur in the center and form double bonds between sulfur and each oxygen atom. This utilizes 16 electrons. Then, place two lone pairs around each oxygen atom, consuming the remaining 16 electrons. Now, let's find a molecule/ion with the same Lewis structure. We need a central atom that can form double bonds with four identical outer atoms. A suitable candidate is the chromate ion \(\mathrm{CrO}_{4}^{2-}\). This ion contains one chromium atom and four oxygen atoms, forming a structure equivalent to that of the sulfate ion and sharing the same -2 charge. So, the answer for (d) is \(\mathrm{CrO}_{4}^{2-}\).

Key concepts

Chemical Bonding

Chemical bonding is a fundamental concept in chemistry where atoms combine to form molecules or compounds. Bonds are formed due to attractions between atoms, ions, or molecules that enable the formation of chemical compounds. There are three main types of chemical bonds: ionic, covalent, and metallic.

Ionic Bonds

Ionic bonds occur when there is a complete transfer of valence electrons between atoms, typically between a metal and a non-metal. This transfer creates ions, with one becoming positively charged (cation) and the other negatively charged (anion). The electrostatic forces between these oppositely charged ions hold them together. For example, in the exercise, we mentioned the fluoride ion \(\mathrm{F}^{-}\), which forms an ionic bond with hydrogen to create hydrofluoric acid.

Covalent Bonds

Covalent bonds involve the sharing of valence electron pairs between atoms. This type of bond usually occurs between non-metal atoms with similar electronegativities. In our practice problems, the peroxide ion \(\mathrm{O}_{2}^{2-}\) and the cyanide ion \(\mathrm{CN}^{-}\) are examples of species with covalent bonds. The peroxide ion has a single covalent bond and cyanide ion has a triple covalent bond.

Metallic Bonds

Metallic bonds are a type of chemical bond found in metals where electrons are shared over many atoms, allowing them to be delocalized. This bond is responsible for some of the properties of metals, such as conductivity and malleability.

Understanding the type of bond formed is crucial for predicting the properties of a compound and for knowing how atoms can be grouped together to form stable chemical structures.

Valence Electrons

Valence electrons play a critical role in chemical bonding because they are the outermost electrons of an atom and are involved in the formation of bonds. The number of valence electrons an atom has determines how it will bond with other atoms.

In the Lewis structure practice problems, analyzing the valence electrons is the first step in predicting the type of bond and the structure of the molecule or ion. For example, oxygen has 6 valence electrons, hydrogen has 1, carbon has 4, nitrogen has 5, and sulfur has 6. These valence electrons are either shared or transferred to achieve the stable electronic configuration of a noble gas, often known as the 'octet rule', where atoms tend to have 8 electrons in their valence shell.

The exercise mentioned ions like \(\mathrm{OH}^{-}\) and \(\mathrm{O}_{2}^{2-}\), each having a negative charge, indicating an excess of valence electrons. \(\mathrm{CN}^{-}\)'s Lewis structure, with a triple bond, shows it has 10 valence electrons, whereas \(\mathrm{SO}_{4}^{2-}\) has 32 valence electrons accounting for double bonds and lone pairs on the oxygen atoms. These practices illustrate how valence electrons are arranged around atoms to form stable structures.

Molecular Geometry

Molecular geometry, also known as molecular structure, is the three-dimensional arrangement of atoms within a molecule. Understanding the geometry is essential for predicting the physical and chemical properties of a molecule, such as reactivity, polarity, phase of matter, color, magnetism, and biological activity.

The VSEPR (Valence Shell Electron Pair Repulsion) theory is a common method used to predict molecular geometry. It states that because electron pairs repel each other, they tend to arrange themselves as far apart as possible to minimize repulsion. In the practice problems, the molecular geometry for the hydroxide ion \(\mathrm{OH}^{-}\) can be inferred from its Lewis structure, which shows two electron domains (a single bond and three lone pairs), resulting in a bent shape.

Common Geometries

On the other hand, \(\mathrm{O}_{2}^{2-}\) with two bonding domains (a single bond between two oxygens) and several lone pairs, reflects a linear geometry. A more complex ion like \(\mathrm{SO}_{4}^{2-}\), with four double bonds, exhibits a tetrahedral geometry. The geometry of these molecules and ions is pivotal in understanding the angles between bonds and the overall shape of the compound, influencing how they interact with each other and with other substances.