Question

the valence of aluminum is +3, and the valence of chlorine is −1. the formula for aluminum chloride is correctly written as a) al₃cl. b) clal₃. c) cl₃al.

Explanation:

Step1: Recall the criss - cross method for writing chemical formulas

In the criss - cross method, the valency of the cation (aluminum, \(Al^{3 +}\)) becomes the subscript of the anion (chlorine, \(Cl^{-}\)) and the valency of the anion becomes the subscript of the cation. But we need to simplify if possible.
The valency of \(Al\) is \(+ 3\) and the valency of \(Cl\) is \(-1\). So, when we apply the criss - cross method, the formula should be \(Al_{1}Cl_{3}\) (since we take the absolute value of the valency of each ion and cross - multiply). We can write \(Al_{1}Cl_{3}\) as \(AlCl_{3}\), but in the given options, we need to check the order and subscripts. Also, the correct order of writing a chemical formula is that the metal (aluminum) comes first and then the non - metal (chlorine). But let's check the subscripts again. The charge of \(Al\) is \(+3\) and \(Cl\) is \(- 1\). To balance the charges, we need 3 \(Cl^{-}\) ions to balance the charge of 1 \(Al^{3+}\) ion. So the formula is \(AlCl_{3}\), but in the options, we have to check the format. Wait, the options are in a different order? Wait no, maybe a typo in the options. Wait, the correct formula is \(AlCl_{3}\), but let's check the options. Wait, option C is \(Cl_{3}Al\), which is the same as \(AlCl_{3}\) (just reversed order, but in some notations, but actually the correct formula has metal first. Wait, no, maybe I made a mistake. Wait, the valency of \(Al\) is \(+3\), \(Cl\) is \(- 1\). So the number of \(Al\) atoms \(n_{Al}\) and \(Cl\) atoms \(n_{Cl}\) should satisfy \(n_{Al}\times(+3)+n_{Cl}\times(- 1)=0\). Let \(n_{Al} = 1\), then \(3 + n_{Cl}\times(-1)=0\), so \(n_{Cl}=3\). So the formula is \(AlCl_{3}\), which can also be written as \(Cl_{3}Al\) (though the standard is metal first). Now let's check the options:

• Option A: \(Al_{3}Cl\) would mean \(3\times(+3)+1\times(-1)=9 - 1 = 8

eq0\), so incorrect.

• Option B: \(ClAl_{3}\) would mean \(1\times(-1)+3\times(+3)=- 1 + 9 = 8

eq0\), so incorrect.

• Option C: \(Cl_{3}Al\) (or \(AlCl_{3}\)): \(1\times(+3)+3\times(-1)=3 - 3 = 0\), which is correct.

Answer:

C) \(Cl_{3}Al\)