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: Compound Mass % In Melting Point (ºC) A 87.7 653 B 78.2 692 C 70.5 1050 a. Determine the empirical formulas of compounds A, B, and C. b. Give the charge of In in each of the three compounds. c. Write the electron configurations (shorthand) for the In 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 the smallest? Explain. e. The melting point of ionic compounds often correlates with the lattice energy (see Chapter 9 in Tro). Explain the trends in the melting points of compounds A, B, and C in these terms.

Answer 1

(a): Let's consider 100 g of each binary compound. Since we have considered 100 g, mass percent will be equal to mass.

Compound-A: Mass of In = 87.7 g

=> Mass of S = 100 - 87.7 = 12.3 g

=> Moles of In = Mass / atomic mass = 87.7 g / 114.818 g/mol = 0.7444 mol In

Moles of S = 12.3 g / 32.065 g/mol = 0.3836 mol S

Moles of In : Moles of S

= 0.7444 mol In : 0.3836 mol S

Dividing the above ratio by the smaller mol (0.3836)

= (0.7444/0.3836) mol In : (0.3836 / 0.3836) mol S

= 2 mol In : 1 mol S

=> Empirical formula of A = In₂S

Compound-B: Mass of In = 78.2 g

=> Mass of S = 100 - 78.2 = 21.8 g

=> Moles of In = Mass / atomic mass = 78.2 g / 114.818 g/mol = 0.681 mol In

Moles of S = 21.8 g / 32.065 g/mol = 0.680 mol S

Moles of In : Moles of S

= 0.681 mol In : 0.680 mol S

Dividing the above ratio by the smaller mol (0.680)

= (0.681/0.680) mol In : (0.680 / 0.680) mol S

= 1 mol In : 1 mol S

=> Empirical formula of B = InS

Compound-C: Mass of In = 70.5 g

=> Mass of S = 100 - 70.5 = 29.5 g

=> Moles of In = Mass / atomic mass = 70.5 g / 114.818 g/mol = 0.614 mol In

Moles of S = 29.5 g / 32.065 g/mol = 0.920 mol S

Moles of In : Moles of S

= 0.614 mol In : 0.920 mol S

Dividing the above ratio by the smaller mol (0.614)

= (0.614/0.614) mol In : (0.920 / 0.614) mol S

= 1 mol In : 1.5 mol S

To round up the above ratio, we have to multiply them by 2

= 2*[1 mol In : 1.5 mol S]

= 2 mol In : 3 mol S

=> Empirical formula of C = In₂S₃.

(b): Charge of compound-A, In₂S = +1

Charge of compound-B, InS = +2

Charge of compound-A, In₂S₃ = +3

(c): Electronic configuration of In is: [Kr] 4d¹⁰ 5s² 5p¹

=> Electronic configuration of In⁺ =  [Kr] 4d¹⁰ 5s²

Electronic configuration of In²⁺ = [Kr] 4d¹⁰ 5s¹

Electronic configuration of In³⁺ = [Kr] 4d¹⁰

Due the presence of extra  4d¹⁰, non of them have noble gas configuration.

(d): Higher the positive charge of cation, smaller is the radius. Hence In³⁺ has smallest radius.