Equivalent weight of `Br_(2)` in the following reaction is `Br_(2)+HgO+H_(2)OtoHgBr_(2).HgO+HBrO`
(given `Br=80`)

Options

A. `(160)/(3)`
B. `80`
C. `160`
D. `160xx3`

Answer

To find the equivalent weight of \( \text{Br}_2 \) in the reaction:

\[ \text{Br}_2 + \text{HgO} + \text{H}_2\text{O} \rightarrow \text{HgBr}_2 \cdot \text{HgO} + \text{HBrO} \]

we will follow these steps:

### Step 1: Identify the molecular weight of \( \text{Br}_2 \)

The molecular weight of bromine (\( \text{Br} \)) is given as 80. Therefore, the molecular weight of \( \text{Br}_2 \) is:

\[ \text{Molecular weight of } \text{Br}_2 = 2 \times 80 = 160 \]

### Step 2: Determine the change in oxidation state

In this reaction, \( \text{Br}_2 \) is being converted into \( \text{HgBr}_2 \) and \( \text{HBrO} \). We need to find out how many electrons are involved in the change of oxidation states of bromine.

- In \( \text{Br}_2 \), bromine has an oxidation state of 0.
- In \( \text{HgBr}_2 \), bromine has an oxidation state of -1.
- In \( \text{HBrO} \), bromine has an oxidation state of +1.

Thus, the change in oxidation state for each bromine atom can be summarized as follows:
- From 0 to -1 (in \( \text{HgBr}_2 \)): this is a reduction of 1 electron.
- From 0 to +1 (in \( \text{HBrO} \)): this is an oxidation of 1 electron.

### Step 3: Calculate the total change in electrons

Since \( \text{Br}_2 \) consists of 2 bromine atoms, the total change in oxidation state for \( \text{Br}_2 \) can be calculated as follows:

- Each bromine atom undergoes a change of 1 electron (one is reduced and one is oxidized).
- Therefore, for \( \text{Br}_2 \), the total change is \( 2 \) electrons.

### Step 4: Calculate the valence factor

The valence factor is the total number of electrons transferred per molecule of \( \text{Br}_2 \). Since we have determined that 2 electrons are transferred, the valence factor for \( \text{Br}_2 \) is 2.

### Step 5: Calculate the equivalent weight

The equivalent weight is calculated using the formula:

\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{Valence factor}} \]

Substituting the values we have:

\[ \text{Equivalent weight of } \text{Br}_2 = \frac{160}{2} = 80 \]

### Final Answer

The equivalent weight of \( \text{Br}_2 \) in the given reaction is **80**.

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