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

Options

A. `5,6,3,3`
B. `5,3,6,3`
C. `3,5,3,6`
D. `5,6,5,5`

Explanation

To solve the given redox reaction and find the values of \( a \), \( b \), \( c \), and \( d \) in the balanced equation:

\[ IO_3^- + aI^- + bH^+ \rightarrow cH_2O + dI_2 \]

we will follow these steps:

### Step 1: Identify the oxidation states
First, we need to identify the oxidation states of iodine in the reactants and products. In \( IO_3^- \), iodine has an oxidation state of +5, while in \( I^- \) it is -1, and in \( I_2 \) it is 0.

### Step 2: Write the half-reactions
Next, we can write the half-reactions for the oxidation and reduction processes. The reduction half-reaction involves \( IO_3^- \) being reduced to \( I_2 \), and the oxidation half-reaction involves \( I^- \) being oxidized to \( I_2 \).

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

2. Oxidation half-reaction:
\[ 2I^- \rightarrow I_2 + 2e^- \]

### Step 3: Balance the electrons
To balance the electrons, we need to ensure that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction. The least common multiple of 5 and 2 is 10.

- Multiply the reduction half-reaction by 2:
\[ 2IO_3^- + 12H^+ + 10e^- \rightarrow 2I_2 + 6H_2O \]

- Multiply the oxidation half-reaction by 5:
\[ 10I^- \rightarrow 5I_2 + 10e^- \]

### Step 4: Combine the half-reactions
Now, we can combine the balanced half-reactions:

\[ 2IO_3^- + 10I^- + 12H^+ \rightarrow 6H_2O + 5I_2 \]

### Step 5: Identify coefficients
From the balanced equation, we can identify the coefficients corresponding to \( a \), \( b \), \( c \), and \( d \):

- \( a = 10 \) (for \( I^- \))
- \( b = 12 \) (for \( H^+ \))
- \( c = 6 \) (for \( H_2O \))
- \( d = 5 \) (for \( I_2 \))

### Final Answer
Thus, the values of \( a \), \( b \), \( c \), and \( d \) are:

- \( a = 10 \)
- \( b = 12 \)
- \( c = 6 \)
- \( d = 5 \)