Consider the following unbalanced redox reaction:
`H_(2)O+AX+BYrarrHA+OY+X_(2)B`
The oxidation number of `X` is `-2` and niether `X` nor water is involved in the redox process.
If the above reaction is balanced with smallest whole number coefficients, the sum of the stoichiometric coefficients of all the compound is

Options

A. `9`
B. `8`
C. `7`
D. `6`

Explanation

To solve the given unbalanced redox reaction:

**Reaction:**
\[ H_2O + AX + BY \rightarrow HA + OY + X_2B \]

**Step 1: Identify the components of the reaction.**
- Reactants: \( H_2O, AX, BY \)
- Products: \( HA, OY, X_2B \)

**Step 2: Write down the number of atoms for each element on both sides.**
- **Left Side:**
- H: 2 (from \( H_2O \))
- A: 1 (from \( AX \))
- X: 1 (from \( AX \))
- B: 1 (from \( BY \))
- Y: 1 (from \( BY \))
- O: 1 (from \( H_2O \))

- **Right Side:**
- H: 1 (from \( HA \))
- A: 1 (from \( HA \))
- X: 2 (from \( X_2B \))
- B: 2 (from \( X_2B \))
- Y: 1 (from \( OY \))
- O: 1 (from \( OY \))

**Step 3: Balance the hydrogen (H) atoms.**
- Left Side: 2 H
- Right Side: 1 H
- To balance, we place a coefficient of 2 in front of \( HA \):
\[ H_2O + AX + BY \rightarrow 2HA + OY + X_2B \]

**Step 4: Update the atom counts after balancing H.**
- **Left Side:**
- H: 2
- A: 1
- X: 1
- B: 1
- Y: 1
- O: 1

- **Right Side:**
- H: 2 (from \( 2HA \))
- A: 2 (from \( 2HA \))
- X: 2 (from \( X_2B \))
- B: 2 (from \( X_2B \))
- Y: 1 (from \( OY \))
- O: 1 (from \( OY \))

**Step 5: Balance the A atoms.**
- Left Side: 1 A
- Right Side: 2 A
- To balance, we place a coefficient of 2 in front of \( AX \):
\[ H_2O + 2AX + BY \rightarrow 2HA + OY + X_2B \]

**Step 6: Update the atom counts after balancing A.**
- **Left Side:**
- H: 2
- A: 2 (from \( 2AX \))
- X: 2 (from \( 2AX \))
- B: 1
- Y: 1
- O: 1

- **Right Side:**
- H: 2 (from \( 2HA \))
- A: 2 (from \( 2HA \))
- X: 2 (from \( X_2B \))
- B: 2 (from \( X_2B \))
- Y: 1 (from \( OY \))
- O: 1 (from \( OY \))

**Step 7: Balance the B atoms.**
- Left Side: 1 B
- Right Side: 2 B
- To balance, we place a coefficient of 2 in front of \( BY \):
\[ H_2O + 2AX + 2BY \rightarrow 2HA + OY + X_2B \]

**Step 8: Update the atom counts after balancing B.**
- **Left Side:**
- H: 2
- A: 2
- X: 2
- B: 2 (from \( 2BY \))
- Y: 2 (from \( 2BY \))
- O: 1

- **Right Side:**
- H: 2 (from \( 2HA \))
- A: 2 (from \( 2HA \))
- X: 2 (from \( X_2B \))
- B: 2 (from \( X_2B \))
- Y: 1 (from \( OY \))
- O: 1 (from \( OY \))

**Step 9: Final check for balance.**
- All elements are balanced:
- H: 2 = 2
- A: 2 = 2
- X: 2 = 2
- B: 2 = 2
- Y: 2 = 1 (not balanced)
- O: 1 = 1

**Step 10: Final adjustments.**
- To balance Y, we need to adjust \( OY \). Since we have 2 Y in the left side, we need to place a coefficient of 2 in front of \( OY \):
\[ H_2O + 2AX + 2BY \rightarrow 2HA + 2OY + X_2B \]

**Step 11: Calculate the sum of stoichiometric coefficients.**
- Coefficients:
- \( H_2O \): 1
- \( 2AX \): 2
- \( 2BY \): 2
- \( 2HA \): 2
- \( 2OY \): 2
- \( X_2B \): 1

**Sum:**
\[ 1 + 2 + 2 + 2 + 2 + 1 = 10 \]

**Final Answer:** The sum of the stoichiometric coefficients is **10**.

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