Q.Write formulas for the following compounds:

The key idea is Stock notation: Roman numerals in parentheses indicate the oxidation state of the metal, which determines the ratio of ions needed for a neutral compound.

Reasoning steps:

1. Identify the cation (metal) with its given oxidation state and the anion (non-metal or polyatomic ion) with its usual charge.
2. Write the symbols with charges: $M^{n+}$ and $X^{m-}$.
3. Cross-multiply charges to find the smallest whole-number ratio of ions, then write the formula.

1. $\ce{Hg^{2+}}$ and $\ce{Cl-}$ → 1:2 ratio → $\ce{HgCl2}$
2. $\ce{Ni^{2+}}$ and $\ce{SO4^{2-}}$ → 1:1 ratio → $\ce{NiSO4}$
3. $\ce{Sn^{4+}}$ and $\ce{O^{2-}}$ → 1:2 ratio → $\ce{SnO2}$ …

The Core Idea: Stock Notation

The Roman numeral in parentheses — like (II), (III), (IV) — is the Stock notation for the oxidation state of the metal. It tells you exactly how many electrons the metal atom has lost (its positive charge). The trick is that you don't need to memorise "mercury is +2 here" — the name gives you the charge. Your only job is to pair that charged cation with the right number of anions so the compound is neutral.

Let's walk through each one.

1. (a) Mercury(II) chloride

• Cation: Mercury(II) means $\text{Hg}^{2+}$.
• Anion: Chloride is $\text{Cl}^-$ (always -1).
• Balancing: One $\text{Hg}^{2+}$ needs two $\text{Cl}^-$ ions to cancel the +2 charge.
• Formula: $\text{HgCl}_2$.

2. (b) Nickel(II) sulphate

• Cation: Nickel(II) is $\text{Ni}^{2+}$.
• Anion: Sulphate is $\text{SO}_4^{2-}$ (a polyatomic ion with charge -2).
• Balancing: One $\text{Ni}^{2+}$ and one $\text{SO}_4^{2-}$ already cancel: $+2 + (-2) = 0$.
• Formula: $\text{NiSO}_4$. No parentheses needed because there's only one sulphate.

3. (c) Tin(IV) oxide

• Cation: Tin(IV) is $\text{Sn}^{4+}$.
• Anion: Oxide is $\text{O}^{2-}$.
• Balancing: To get a total of +4 from the cation, you need two $\text{O}^{2-}$ ions: $2 \times (-2) = -4$.
• Formula: $\text{SnO}_2$.

4. (d) Thallium(I) sulphate

• Cation: Thallium(I) is $\text{Tl}^{+}$.
• Anion: Sulphate is $\text{SO}_4^{2-}$.
• Balancing: You need two $\text{Tl}^{+}$ ions to balance one $\text{SO}_4^{2-}$: $2 \times (+1) = +2$ cancels $-2$.
• Formula: $\text{Tl}_2\text{SO}_4$. The subscript 2 goes on thallium, and the sulphate stays as a single unit.

5. (e) Iron(III) sulphate

• Cation: Iron(III) is $\text{Fe}^{3+}$.
• Anion: Sulphate is $\text{SO}_4^{2-}$.
• Balancing: This is a classic cross-multiplication. The lowest common multiple of 3 and 2 is 6. You need two $\text{Fe}^{3+}$ (total +6) and three $\text{SO}_4^{2-}$ (total -6). …