To determine the correct statements for the given reaction, we first need to balance the chemical equation and analyze the oxidation states of the elements involved.

The unbalanced reaction is: ${I^ - } + ClO_3^ - + {H_2}S{O_4} \to C{l^ - } + HSO_4^ - + {I_2}$

Step 1: Identify Oxidation and Reduction

1. Assign oxidation states to the elements that change their state:

   • Iodine (I): The oxidation state changes from -1 in $I^-$ to 0 in $I_2$. An increase in oxidation state is oxidation. Therefore, iodide is oxidised.
   • Chlorine (Cl): In $ClO_3^-$, the oxidation state of Cl is $x + 3(-2) = -1 \implies x = +5$. The oxidation state changes from +5 in $ClO_3^-$ to -1 in $Cl^-$. A decrease in oxidation state is reduction.
   • Sulphur (S): In $H_2SO_4$, the oxidation state of S is $2(+1) + S + 4(-2) = 0 \implies S = +6$. In $HSO_4^-$, the oxidation state of S is $+1 + S + 4(-2) = -1 \implies S = +6$. The oxidation state of Sulphur does not change.

2. Evaluate statements B and C based on this analysis:

   • Statement B: iodide is oxidised. This is correct.
   • Statement C: sulphur is reduced. This is incorrect as its oxidation state doesn't change.

Step 2: Balance the Redox Reaction

We will use the oxidation number method to balance the equation.

1. Write the half-reactions based on the redox pairs and determine the change in oxidation number:

   • Oxidation: $2I^- \to I_2$. The total change in oxidation number is from $2 \times (-1)$ to $0$, which is an increase of 2. ($+2$)
   • Reduction: $ClO_3^- \to Cl^-$. The change in oxidation number is from +5 to -1, which is a decrease of 6. ($-6$)

2. Equalize the change in oxidation numbers:

   • To make the total increase equal to the total decrease, we multiply the oxidation half-reaction by 3.
   • $3 \times (2I^- \to I_2)$ gives a total change of $3 \times (+2) = +6$.
   • The reduction half-reaction already has a change of $-6$.

3. Determine the stoichiometric coefficients for the redox species:

   • This gives us the ratio: $6I^-$ to $3I_2$ and $1ClO_3^-$ to $1Cl^-$.
   • The partially balanced equation is: $6{I^ - } + {ClO_3^ - } + {H_2}S{O_4} \to {Cl^ - } + 3{I_2} + HSO_4^ -$

4. Balance the remaining atoms ($S, H, O$) and charges.

   • Let the coefficients of $H_2SO_4$, $HSO_4^-$, and $H_2O$ be $a$, $b$, and $c$ respectively. $6{I^ - } + {ClO_3^ - } + a{H_2}S{O_4} \to {Cl^ - } + 3{I_2} + b{HSO_4^ - } + c{H_2}O$
   • Balance S atoms: The number of S atoms on both sides must be equal. Thus, $a = b$.
   • Balance O atoms: Reactants: 3 (from $ClO_3^-$) + $4a$ (from $H_2SO_4$) Products: $4b$ (from $HSO_4^-$) + $c$ (from $H_2O$) $3 + 4a = 4b + c$. Since $a=b$, this simplifies to $3 + 4a = 4a + c$, which gives $c = 3$. So, $H_2O$ is a product.
   • Balance H atoms: Reactants: $2a$ (from $H_2SO_4$) Products: $b$ (from $HSO_4^-$) + $2c$ (from $H_2O$) $2a = b + 2c$. Substituting $a=b$ and $c=3$, we get: $2a = a + 2(3) \implies 2a = a + 6 \implies a = 6$.
   • Since $a=b$, we have $b=6$.

Step 3: Write the Final Balanced Equation

Substituting the coefficients $a=6, b=6, c=3$ into the equation: $6{I^ - } + ClO_3^ - + 6{H_2}S{O_4} \to C{l^ - } + 6HSO_4^ - + 3{I_2} + 3{H_2}O$

Step 4: Evaluate the Remaining Statements

• A: stoichiometric coefficient of HSO${}_4^ -$ is 6
  • In the balanced equation, the coefficient for $HSO_4^-$ is 6. This statement is correct.
• D: $H_2O$ is one of the products
  • In the balanced equation, 3 moles of $H_2O$ are formed on the product side. This statement is correct.

Conclusion

The correct statements are A, B, and D.