Given : $$XNa_2HAsO_3 + YNaBrO_3 + ZHCl \rightarrow NaBr + H_3AsO_4 + NaCl$$
The values of X, Y and Z in the above redox reaction are respectively :

The given redox reaction is:

$$XNa_2HAsO_3 + YNaBrO_3 + ZHCl \rightarrow NaBr + H_3AsO_4 + NaCl$$

We need to find the values of X, Y, and Z by balancing the reaction using the oxidation number method. First, we determine the oxidation states of the elements involved.

In the reactant Na₂HAsO₃:

• Sodium (Na) has an oxidation state of +1.
• Hydrogen (H) has an oxidation state of +1.
• Oxygen (O) has an oxidation state of -2.
• Let the oxidation state of arsenic (As) be a. The sum of oxidation states in Na₂HAsO₃ is zero: 2*(+1) + (+1) + a + 3*(-2) = 0 → 2 + 1 + a - 6 = 0 → a - 3 = 0 → a = +3.

In the reactant NaBrO₃:

• Sodium (Na) has an oxidation state of +1.
• Oxygen (O) has an oxidation state of -2.
• Let the oxidation state of bromine (Br) be b. The sum is: +1 + b + 3*(-2) = 0 → 1 + b - 6 = 0 → b - 5 = 0 → b = +5.

In the product NaBr:

• Sodium (Na) has an oxidation state of +1, so bromine (Br) must be -1.

In the product H₃AsO₄:

• Hydrogen (H) has an oxidation state of +1.
• Oxygen (O) has an oxidation state of -2.
• Let the oxidation state of arsenic (As) be c. The sum is: 3*(+1) + c + 4*(-2) = 0 → 3 + c - 8 = 0 → c - 5 = 0 → c = +5.

Arsenic changes from +3 to +5, losing 2 electrons per atom. Bromine changes from +5 to -1, gaining 6 electrons per atom (since +5 to -1 is a decrease of 6).

Let X be the number of Na₂HAsO₃ molecules and Y be the number of NaBrO₃ molecules. The total electrons lost by arsenic are 2X, and the total electrons gained by bromine are 6Y. For the reaction to be balanced, electrons lost must equal electrons gained:

$$2X = 6Y \rightarrow X = 3Y$$

So, X must be 3 times Y. The smallest integer solution is Y = 1 and X = 3.

Now, substitute X = 3 and Y = 1 into the reaction:

$$3Na_2HAsO_3 + NaBrO_3 + ZHCl \rightarrow NaBr + H_3AsO_4 + NaCl$$

Balance the atoms. On the reactant side:

• Sodium (Na): From 3Na₂HAsO₃ → 6 Na, from NaBrO₃ → 1 Na, total Na = 7.
• Hydrogen (H): From 3Na₂HAsO₃ → 3 H (one per molecule), from ZHCl → Z H, total H = 3 + Z.
• Arsenic (As): From 3Na₂HAsO₃ → 3 As.
• Bromine (Br): From NaBrO₃ → 1 Br.
• Oxygen (O): From 3Na₂HAsO₃ → 9 O, from NaBrO₃ → 3 O, total O = 12.
• Chlorine (Cl): From ZHCl → Z Cl.

On the product side, as written, we have NaBr, H₃AsO₄, and NaCl. However, with 3 As atoms, we need 3 H₃AsO₄. With 1 Br atom, we need 1 NaBr. Sodium atoms on the left are 7, but NaBr provides 1 Na and NaCl provides 1 Na, totaling 2 Na, which is insufficient. Therefore, we adjust the products to include coefficients:

Products should be: NaBr (for Br), 3H₃AsO₄ (for As), and NaCl for the remaining Na and Cl. Sodium required: total Na from reactants is 7. NaBr provides 1 Na, so the remaining 6 Na must come from NaCl. Thus, we need 6 NaCl.

So, the balanced products are: NaBr + 3H₃AsO₄ + 6NaCl.

Now, the product side has:

• Sodium (Na): NaBr → 1 Na, 6NaCl → 6 Na, total Na = 7.
• Hydrogen (H): 3H₃AsO₄ → 9 H.
• Arsenic (As): 3H₃AsO₄ → 3 As.
• Bromine (Br): NaBr → 1 Br.
• Oxygen (O): 3H₃AsO₄ → 12 O.
• Chlorine (Cl): 6NaCl → 6 Cl.

Set reactant and product atoms equal:

• Na: 7 = 7 → balanced.
• H: 3 + Z = 9 → Z = 6.
• As: 3 = 3 → balanced.
• Br: 1 = 1 → balanced.
• O: 12 = 12 → balanced.
• Cl: Z = 6 → balanced.

Thus, X = 3, Y = 1, Z = 6. The balanced equation is:

$$3Na_2HAsO_3 + NaBrO_3 + 6HCl \rightarrow NaBr + 3H_3AsO_4 + 6NaCl$$

Comparing with the options:

• A. 2, 1, 2
• B. 2, 1, 3
• C. 3, 1, 6
• D. 3, 1, 4

Hence, the correct answer is Option C.