In the balanced chemical reaction
`IO_(3)^(ө)+al^(ө)+bH^(ө)rarrcH_(2)O+dI_(2)`
`a, b,c`, and `d`, respectively, correspond to

Options

A. `a = 5`
B. `c = 3`
C. `d = 5`
D. `b = 6`

Explanation

To solve the balanced chemical reaction given by:

\[ IO_3^{-} + Al + bH \rightarrow cH_2O + dI_2 \]

we need to balance the equation step by step.

### Step 1: Identify the oxidation states
First, we need to identify the oxidation states of the elements involved in the reaction. In \( IO_3^{-} \), iodine is in the +5 oxidation state. The aluminum (Al) is in the 0 oxidation state, and hydrogen (H) is also in the 0 oxidation state.

### Step 2: Determine the products
The products of the reaction include water (\( H_2O \)) and iodine (\( I_2 \)). In \( I_2 \), iodine is in the 0 oxidation state.

### Step 3: Write the half-reactions
Next, we can write the half-reactions for the oxidation and reduction processes.

1. **Reduction half-reaction**:
\[ IO_3^{-} + 6H^{+} + 5e^{-} \rightarrow I_2 + 3H_2O \]

2. **Oxidation half-reaction**:
\[ Al \rightarrow Al^{3+} + 3e^{-} \]

### Step 4: Balance the electrons
To balance the electrons transferred, we need to multiply the oxidation half-reaction by 5 to match the 5 electrons in the reduction half-reaction:

\[ 5(Al \rightarrow Al^{3+} + 3e^{-}) \rightarrow 5Al \rightarrow 5Al^{3+} + 15e^{-} \]

### Step 5: Combine the half-reactions
Now, we can combine the half-reactions:

\[ IO_3^{-} + 5Al + 6H^{+} \rightarrow I_2 + 3H_2O + 5Al^{3+} \]

### Step 6: Identify coefficients
Now we can identify the coefficients \( a, b, c, \) and \( d \) in the original equation:

- \( a \) (for \( Al \)) = 5
- \( b \) (for \( H \)) = 6
- \( c \) (for \( H_2O \)) = 3
- \( d \) (for \( I_2 \)) = 3

### Final Answer
Thus, the values are:
- \( a = 5 \)
- \( b = 6 \)
- \( c = 3 \)
- \( d = 3 \)

### Summary
The balanced equation gives us the values:
- \( A = 5 \)
- \( B = 6 \)
- \( C = 3 \)
- \( D = 3 \)