Question

The reaction of indium, In, with sulfur leads to three binary compounds, which we will assume to be purely ionic. The three compounds have the following properties: $$\overline{)\begin{array}{lll}\text{ Compound }& \text{ Mass \% In }& \text{ Melting Point }\left({}^{\circ }\mathrm{C}\right)\\ \text{ A }& 87.7& 653\\ \text{ B }& 78.2& 692\\ \text{ C }& 70.5& 1050\end{array}}$$ (a) Determine the empirical formulas of compounds A, B, and C. (b) Give the oxidation state of In in each of the three compounds. (c) Write the electron configuration for the In ion in each compound. Do any of these configurations correspond to a noble-gas configuration? (d) In which compound is the ionic radius of In expected to be smallest? Explain. (e) The melting point of ionic compounds often correlates with the lattice energy. Explain the trends in the melting points of compounds A, B, and $\mathrm{C}$ in these terms.

Short answer

The empirical formulas of compounds A, B, and C are InS, In_2S_3, and InS_2, respectively. The oxidation states of In in these compounds are +2, +3, and +3. The electron configurations of In ions in compounds A, B, and C are $[Kr]4d^{10}$, $[Kr]4d^9$, and $[Kr]4d^9$, respectively. None of these configurations correspond to a noble-gas configuration. The ionic radius of In is expected to be smallest in compound B and C, where In has a +3 oxidation state. The trends in melting points can be explained by the higher lattice energy in compound C due to stronger electrostatic forces between the ions, resulting in the highest melting point among the three compounds.

Step-by-step solution

Find the mole ratios of In to S in each compound

To determine the empirical formulas, we first need to find the mole ratios of In to S in each compound. Using the mass percentage of In in each compound provided, we can find the mass percentage of S, and then use their molar masses to find the mole ratio. $CompoundA:$ Indium: $0.877\frac{mass}{compound}\to \frac{0.877\times mass}{114.818\frac{g}{mol}}mol$ (In, $molarmass=114.818g/mol$) \ Sulfur: $0.123\frac{mass}{compound}\to \frac{0.123\times mass}{32.065\frac{g}{mol}}mol$ (S, $molarmass=32.065g/mol$) Then, calculate the mole ratio of In to S in Compound A. $CompoundB:$ Follow the same process for Compound B. $CompoundC:$ Repeat the process for Compound C.

Determine the empirical formulas

Using the mole ratios found in Step 1, determine the empirical formulas for Compounds A, B, and C. Divide each ratio by the smallest value to get a whole number ratio, which gives the empirical formula.

Find the oxidation state of In in each compound

Based on the empirical formulas found in Step 2, we can determine the oxidation state of In in each compound. The oxidation state of sulfur, S, is -2. Using the formula, set the sum of the oxidation states of In and S as 0, and solve for the oxidation state of In.

Write the electron configuration of the In ion in each compound

Using the oxidation states found in Step 3, write the electron configurations for the In ion in each compound. Start with the ground state electron configuration of In (atomic number = 49): $[Kr]5s^{2}4d^{1}05p^1$ From this, remove electrons as necessary to achieve the correct oxidation state for In in each compound. Check if any of the electron configurations correspond to a noble-gas configuration.

Determine which compound has the smallest ionic radius for In

The ionic radius of a cation typically decreases as the oxidation state increases, due to an increase in the effective nuclear charge experienced by the outermost electrons. Using the oxidation states determined in Step 3, identify which compound has the smallest ionic radius for In and explain the reasoning behind it.

Analyze the trends in melting points in terms of lattice energy

Lattice energy is directly related to the charges of the ions in the compound and indirectly related to the distances between them. Higher lattice energy usually corresponds to a higher melting point. Using the given data on melting points and the information on ionic charges and sizes from previous steps, discuss the trends in melting points for Compounds A, B, and C in terms of lattice energy.

Key concepts

Oxidation States

Oxidation states are important in understanding the charge of an element in a compound. They indicate the degree of oxidation of an atom and are crucial in balancing chemical reactions. In this problem, knowing the oxidation state of Indium (In) helps us understand its role in compounds A, B, and C. The oxidation state can be determined based on the empirical formula, which defines how many Indium and Sulfur atoms are present. Sulfur is usually -2 in binary compounds, so Indium's oxidation state can be found by ensuring the overall charge of the compound is neutral. Calculating this involves setting the total oxidation states of Indium and Sulfur to zero and solving for Indium's oxidation state. For example, if Compound A shows a formula InS, we calculate as follows: $$x+(-2)=0$$ This shows that Indium in Compound A has an oxidation state of +2. By repeating this process, you can find the oxidation states for Compounds B and C.

Mole Ratios

Mole ratios are a crucial concept in determining the empirical formula of a compound. They represent the relative number of moles of each element present in a compound. To find these ratios, convert the percentage of each element into moles, using their molar masses. For instance, if Compound A has 87.7% In and 12.3% S, we determine moles by: - Indium: $0.877\times \frac{1}{114.818}\approx 0.00764\text{ moles}$ per gram of compound - Sulfur: $0.123\times \frac{1}{32.065}\approx 0.00384\text{ moles}$ per gram of compoundNext, divide each by the smallest number of moles to get whole numbers, establishing the mole ratio and empirical formula. This mathematical approach ensures the compound's formula reflects the simplest integer ratio of its elements.

Electron Configuration

Electron configuration describes the distribution of electrons in an atom's orbitals. For elements like Indium, understanding electron configuration is key to knowing its chemical properties and behaviors. Indium has the ground state electron configuration of: $$[Kr]5s^{2}4d^{10}5p^1$$When Indium forms ions, it loses electrons corresponding to its oxidation state. For example, if Indium in Compound A is +2, it loses two electrons, resulting in: - $$I{n}^{2+}:[Kr]4d^{10}$$Checking these configurations helps determine if any resemble a noble gas configuration, like Krypton in this case. Noble gas configurations are significant as they indicate stability of the ion, given they represent a complete electron shell similar to that of inert gases.

Ionic Radius

The ionic radius of an ion changes depending on its charge. Generally, as the oxidation state of a cation increases, its ionic radius decreases. This is due to the increase in effective nuclear charge, pulling electrons closer to the nucleus and reducing the ion size. In this exercise, understanding which compound contains Indium with the smallest ionic radius involves considering the oxidation state: higher oxidation states result in smaller radii. For instance, Compound C may have the highest Indium oxidation state based on its empirical formula, making it have the smallest ionic radius of all three compounds. Analyzing ionic radii helps explain physical properties, such as why certain compounds might exhibit higher melting points due to stronger ionic interactions.