Question

A compound \(\mathrm{XF}_{5}\) is \(42.81 \%\) fluorine by mass. Identify the element \(\mathrm{X}\) and draw the Lewis structure for the compound. What is the molecular structure of \(\mathrm{XF}_{5} ?\)

Short answer

The element X is Iodine, and the compound is IF5. Its Lewis structure has an expanded octet around Iodine with five single bonds to Fluorine atoms and a lone pair. The molecular structure of IF5 is square pyramidal.

Step-by-step solution

Calculate the mass percentage of element X

Since the mass percentage of fluorine is given, we know that it is \(42.81 \%\). We can find the mass percentage of element X by subtracting the mass percentage of fluorine from \(100 \%\): Mass percentage of element X = \(100 \% - 42.81 \% = 57.19 \%\)

Identify element X using the molar ratio

Let's denote the molar mass of element X as M(X). For the compound XF5, the molar mass can be calculated as: M(XF5) = M(X) + 5 * M(F) Here, M(F) is the molar mass of Fluorine, which is approximately \(19 \mathrm{g/mol}\). Now, we can set up a proportion using mass percentages and molar masses: \(\frac{57.19 \times 100}{M(X)} = \frac{42.81 \times 100}{5 \times 19}\) Solving for M(X): M(X) = \(\frac{57.19 \times 100 \times 5 \times 19}{42.81 \times 100} = 127\) Using the periodic table, we can see that an element with molar mass 127 is Iodine (I). So, element X is Iodine and the compound is \(\mathrm{IF}_{5}\).

Draw the Lewis structure of IF5

To draw the Lewis structure of IF5, we first calculate the total number of valence electrons: - Iodine (I) has 7 valence electrons. - Each Fluorine (F) atom has 7 valence electrons, and there are 5 F atoms in the compound, giving a total of 5 x 7 = 35 valence electrons. Thus, there are a total of 7 (Iodine) + 35 (Fluorine) = 42 valence electrons in \(\mathrm{IF}_{5}\). Next, we arrange the atoms around the central Iodine atom and use the valence electrons to form bonds: - Place Iodine (I) at the center, and arrange the 5 Fluorine (F) atoms around it. - First, form single bonds between Iodine (I) and each Fluorine (F) atom, using 2 electrons for each bond (a total of 10 electrons). - The remaining 32 electrons are distributed as lone pairs on the Fluorine atoms, with each Fluorine receiving 3 lone pairs (2 electrons in each pair, a total of 30 electrons). - Note that Iodine has an expanded octet, with 5 single bonds and 2 lone pairs, for a total of 12 electrons in its valence shell.

Determine the molecular structure of IF5

The molecular structure of IF5 can be determined using VSEPR (Valence Shell Electron Pair Repulsion) theory. IF5 has five bonding pairs and one lone pair on Iodine. According to VSEPR theory, the molecular structure of IF5 is square pyramidal because five electron domains are in a trigonal bipyramidal arrangement with one equatorial position occupied by the lone pair. In conclusion, the element X is Iodine, the Lewis structure of IF5 shows an expanded octet around Iodine with five single bonds and a lone pair, and the molecular structure is square pyramidal.

Key concepts

Lewis Structure

The Lewis structure is a symbolic representation of a molecule showing how valence electrons are arranged among atoms to form bonds. This diagram helps in predicting the number and types of bonds that can be formed around an atom. To create a Lewis structure, you should follow these steps:

• Determine the total number of valence electrons in the molecule.
• Choose the central atom, generally the least electronegative that's not hydrogen.
• Place single bonds between the central atom and surrounding atoms to begin forming the molecule's skeleton.
• Satisfy the octet rule for atoms bonded to the central atom using remaining electrons.
• For atoms that can have an expanded octet, like Iodine in our example, add additional electrons to satisfy their bonding requirements.
• If any atoms have less than an octet, form double or triple bonds as necessary to satisfy the octet rule, moving lone pair electrons to form multiple bonds.

For the compound \( \mathrm{IF}_{5} \), the Lewis structure shows Iodine at the center with five single bonds connecting to five Fluorine atoms, and one lone pair on Iodine.

VSEPR Theory

The Valence Shell Electron Pair Repulsion (VSEPR) theory is a model used to predict the geometry of individual molecules based on the repulsion between the electron pairs in the valence shell of the central atom. VSEPR theory posits that electron pairs will organize themselves in such a way as to minimize repulsion, leading to specific molecular shapes. Key guidelines of VSEPR theory include:

• Electron domains around the central atom, including both bonding domains (such as single, double, or triple bonds) and non-bonding domains (lone pairs), should be counted.
• Shapes are adjusted to keep electron domains as far apart as possible, resulting in specific molecular geometries.
• Certain geometries are associated with certain numbers of electron domains.

A molecule such as \( \mathrm{IF}_{5} \) with five bonding pairs and one lone pair will have a square pyramidal structure. This structure is due to the lone pair occupying an equatorial position in a trigonal bipyramidal domain arrangement, causing the other bonds to shift into a square pyramidal shape.

Molar Mass Calculation

Calculating molar mass is a fundamental skill in chemistry that allows us to convert between the mass of a substance and the number of moles. The molar mass is the weight of one mole of a chemical element or chemical compound in grams. To calculate the molar mass of a compound:

• Add the atomic masses of each element in the compound, multiplied by the number of atoms of that element in the formula.
• Use the atomic masses from the periodic table where the atomic weight of each element is listed in atomic mass units (amu) or grams per mole (g/mol).

For example, in the compound \( \mathrm{IF}_{5} \), we have one iodine atom and five fluorine atoms. The molar mass would be the sum of the molar mass of one iodine atom and the molar mass of five fluorine atoms, which amounts to the molar mass of the compound.

Percent Composition by Mass

Percent composition by mass expresses the percentage by mass of each element within a compound. It's a critical concept for stoichiometry, as it allows chemists to determine the relative amounts of elements in a substance. To calculate percent composition, follow these steps:

• Calculate the molar mass of the compound by summing the molar masses of all the atoms in the compound.
• Find the mass of each element in the compound by multiplying the atomic mass of the element by the number of atoms of that element in the compound.
• Divide the mass of each element by the total molar mass of the compound and multiply by 100 to get the percentage.

In our \( \mathrm{XF}_{5} \) exercise, given \( \mathrm{XF}_{5} \) is \(42.81 \%\) fluorine by mass, we calculated that element \( X \) (which turned out to be Iodine) must therefore be \(57.19 \%\) of the total mass, using the iterative process of identifying the molar mass of the unknown element and the known element, fluorine.