Question

An element Z has valency equal to 3, what will be its formula with sulphate ions :

• A. \(\text{Z}_2\text{SO}_4\)
• B. \(\text{ZSO}_4\)
• C. \(\text{Z}_2\text{(SO}_4\text{)}_3\)
• D. \(\text{Z(SO}_4\text{)}_3\)

Hint:

1. Identify ions and their valencies/charges: Element Z: Valency 3 (so, \(Z^{3+}\)) Sulphate ion: \(\text{SO}_4\), Valency 2 (so, \(\text{SO}_4^{2-}\)) 2. Criss-cross the numerical values of the valencies to get the subscripts: The '3' from Z goes to \(\text{SO}_4\). The '2' from \(\text{SO}_4\) goes to Z. 3. Write the formula: \(\text{Z}_2(\text{SO}_4)_3\). (Use parentheses for polyatomic ions if their subscript is greater than 1).

Answer 1

The Correct Option is C

Concept: To write the chemical formula of an ionic compound, we use the "criss-cross" method, where the numerical value of the valency (or charge) of one ion becomes the subscript of the other ion. The formula should represent the simplest whole number ratio of ions that results in a neutral compound. Step 1: Identify the valencies (or charges) of the element Z and the sulphate ion
Valency of element Z is given as 3. This means the ion formed by Z will likely be \(Z^{3+}\) (assuming it's a metal or forms a cation).
The sulphate ion is a polyatomic ion with the formula \(\text{SO}_4^{2-}\). Its charge (and thus its valency in ionic compounds) is 2-. Step 2: Apply the criss-cross method Write the symbols of the ions with their valencies (ignoring the sign for now for subscript determination): Z with valency 3 \(\text{SO}_4\) with valency 2 Criss-cross the valencies:
The valency of Z (3) becomes the subscript for the \(\text{SO}_4\) group.
The valency of \(\text{SO}_4\) (2) becomes the subscript for Z. This gives the formula: \(\text{Z}_2 (\text{SO}_4)_3\). The parentheses around \(\text{SO}_4\) are important because the subscript 3 applies to the entire sulphate group. Step 3: Check if the formula is in its simplest ratio The subscripts are 2 and 3. These are already in their simplest whole number ratio (their GCD is 1). Step 4: Verify neutrality (optional but good practice) Charge from \(2 \times Z^{3+}\) ions = \(2 \times (+3) = +6\). Charge from \(3 \times SO_4^{2-}\) ions = \(3 \times (-2) = -6\). Total charge = \(+6 + (-6) = 0\). The compound is neutral. The formula for the compound formed by element Z (valency 3) and sulphate ions (\(\text{SO}_4^{2-}\)) is \(\text{Z}_2\text{(SO}_4\text{)}_3\). This matches option (3). The handwritten criss-cross on the image also indicates this.