Question

For the following collection of nonmetallic elements, \(\mathrm{O}, \mathrm{P},\) Te, \(I, B,(a)\) which two would form the most polar single bond? (b) Which two would form the longest single bond? (c) Which two would be likely to form a compound of formula \(\mathrm{XY}_{2} ?\) (d) Which combinations of elements would likely yield a compound of empirical formula \(\mathrm{X}_{2} \mathrm{Y}_{3} ?\) In each case explain your answer.

Short answer

(a) The most polar single bond would be formed between Oxygen (O) and Boron (B), as their electronegativity difference is the highest, at 1.4. (b) The longest single bond would be formed between Tellurium (Te) and Iodine (I), as their combined atomic sizes have the highest sum, at 273 pm. (c) The compound with the formula XY2 would be OI2 (Oxygen Diiodide), as Oxygen has a valency of 2 and Iodine has a valency of 1. (d) Four possible combinations of elements would yield a compound of empirical formula X2Y3: P2O3 (Diphosphorus Trioxide), P2Te3 (Diphosphorus Trtelluride), B2O3 (Diboron Trioxide), and B2Te3 (Diboron Trtelluride).

Step-by-step solution

1. Identify the given elements and their properties

Before proceeding to determine the properties of the possible bonds and compounds, let's analyze the given nonmetallic elements and list their electronegativity values, atomic size, and valency. - O (Oxygen): Electronegativity: 3.44, Atomic size: 48 pm, Valency: 2 - P (Phosphorus): Electronegativity: 2.19, Atomic size: 98 pm, Valency: 3 - Te (Tellurium): Electronegativity: 2.1, Atomic size: 140 pm, Valency: 2 - I (Iodine): Electronegativity: 2.66, Atomic size: 133 pm, Valency: 1 - B (Boron): Electronegativity: 2.04, Atomic size: 85 pm, Valency: 3

2. Determine the most polar single bond (a)

To find the most polar single bond, we need to compare the electronegativity difference (ΔEN) between each pair of elements. The highest electronegativity difference would indicate the most polar bond. - O-P: \( |\Delta EN|=|3.44 - 2.19|=1.25 \) - O-Te: \( |\Delta EN|=|3.44 - 2.1|=1.34 \) - O-I: \( |\Delta EN|=|3.44 - 2.66|=0.78 \) - O-B: \( |\Delta EN|=|3.44 - 2.04|=1.4 \) - P-Te: \( |\Delta EN|=|2.19 - 2.1|=0.09 \) - P-I: \( |\Delta EN|=|2.19 - 2.66|=0.47 \) - P-B: \( |\Delta EN|=|2.19 - 2.04|=0.15 \) - Te-I: \( |\Delta EN|=|2.1 - 2.66|=0.56 \) - Te-B: \( |\Delta EN|=|2.1 - 2.04|=0.06 \) - I-B: \( |\Delta EN|=|2.66 - 2.04|=0.62 \) The highest electronegativity difference is 1.4 for the O-B bond. Therefore, the most polar single bond would be formed between Oxygen and Boron.

3. Determine the longest single bond (b)

The length of a single bond is primarily dependent on the atomic size of each element. To find the longest single bond, we need to compare the sum of atomic sizes for each pair of elements. The highest sum would indicate the longest bond. - O-P: Length: 48 pm + 98 pm = 146 pm - O-Te: Length: 48 pm + 140 pm = 188 pm - O-I: Length: 48 pm + 133 pm = 181 pm - O-B: Length: 48 pm + 85 pm = 133 pm - P-Te: Length: 98 pm + 140 pm = 238 pm - P-I: Length: 98 pm + 133 pm = 231 pm - P-B: Length: 98 pm + 85 pm = 183 pm - Te-I: Length: 140 pm + 133 pm = 273 pm - Te-B: Length: 140 pm + 85 pm = 225 pm - I-B: Length: 133 pm + 85 pm = 218 pm The highest sum of atomic sizes is 273 pm for the Te-I bond. Therefore, the longest single bond would be formed between Tellurium and Iodine.

4. Determine the compound with the formula XY2 (c)

The formula XY2 indicates that element X has a valency of 2, while element Y has a valency of 1. From the given elements, Oxygen (O) has a valency of 2, and Iodine (I) has a valency of 1. Therefore, the compound with the formula XY2 would be OI2 (Oxygen Diiodide).

5. Determine the compound with the empirical formula X2Y3 (d)

The empirical formula X2Y3 indicates that element X has a valency of 3, while element Y has a valency of 2. From the given elements, Phosphorus (P) and Boron (B) have a valency of 3, and Oxygen (O) and Tellurium (Te) have a valency of 2. The possible combinations that yield a compound of empirical formula X2Y3 are as follows: - P2O3 (Diphosphorus Trioxide) - P2Te3 (Diphosphorus Trtelluride) - B2O3 (Diboron Trioxide) - B2Te3 (Diboron Trtelluride) Therefore, four possible combinations of elements would yield a compound of empirical formula X2Y3.

Key concepts

Electronegativity

Electronegativity is a chemical property that describes the ability of an atom to attract electrons towards itself when forming a chemical bond. It is a crucial concept in understanding the polarity of molecules.

- **High Electronegativity:** Atoms with high electronegativity values tend to attract shared electrons strongly. Oxygen, with an electronegativity of 3.44, is an excellent example of a highly electronegative element.
- **Electronegativity Difference:** The difference in electronegativity between two bonded atoms determines the polarity of the bond. A large difference indicates a polar bond, while a small difference indicates a nonpolar bond.

In the exercise, the bond between Oxygen and Boron is identified as the most polar due to their electronegativity difference of 1.4, which is the highest among the given pairs. Understanding this concept helps explain why some substances have such strong or weak intermolecular forces.

Valency

Valency refers to the combining power of an element, mainly determined by the number of electrons an atom can gain, lose, or share to achieve a stable configuration. Typically, the valency of an element corresponds to the number of electrons in its outer shell.

- **Valency 2 and 3 Examples:** In the provided exercise, Oxygen (O) and Tellurium (Te) have a valency of 2, while Phosphorus (P) and Boron (B) have a valency of 3.
- **Formula Interpretation:** The valency of elements helps predict the ratio of atoms in compounds. For instance, XY2 implies element X contributes 2 valency electrons, and Y contributes 1, thus forming a two-to-one bond ratio.

Understanding valency is pivotal for constructing chemical formulas and understanding the types of bonds elements are likely to form with one another. It's a foundational aspect of predicting molecular structures and reactivity.

Empirical Formula

An empirical formula represents the simplest whole-number ratio of atoms in a compound. It provides insight into the elemental composition but not the actual numbers a molecule might contain. The empirical formula is essential in chemistry as it offers a straightforward means of describing the stoichiometry of a substance.

- **Determining Empirical Formulas:** In our exercise, the empirical formula concept helps deduce compounds like P2O3, which defines the simplest stoichiometric ratio of phosphorus to oxygen.
- **Difference from Molecular Formulas:** While empirical formulas show the simplest ratio, molecular formulas depict the actual number of atoms in a molecule. For example, glucose has an empirical formula of CH2O, indicating a 1:2:1 ratio of carbon, hydrogen, and oxygen, but its molecular formula is C6H12O6.

Empirical formulas are crucial in guiding chemists toward understanding a compound's basic structure and eventual behavior in reactions.

Atomic Size

Atomic size, also known as atomic radius, refers to the distance from the nucleus to the outermost boundary of the electron cloud surrounding an atom. This property varies among elements and influences many chemical behaviors, including bond length and reactivity.

- **Trends in Atomic Size:** Generally, atomic size increases down a group and decreases across a period in the periodic table. For instance, Tellurium has a larger atomic size compared to Oxygen due to its position further down the periodic table.
- **Influence on Bond Length:** In chemical bonding, larger atoms usually form longer bonds. The exercise highlights this by identifying the longest single bond between Tellurium and Iodine, with an atomic size sum of 273 pm, indicating larger atomic sizes lead to longer bond distances.

Understanding atomic size is fundamental to predicting the physical properties of elements and their compounds, including the ability to form stable or metastable molecular structures.