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.

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

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.

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.

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).

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.