Determine the oxidation number of the underlined atom is `(NH_4)_(6)ul(Mo_7)O_(24)`

Options

A. `-5`
B. `+6`
C. `-3`
D. `-1`

Answer

To determine the oxidation number of the underlined atom in the compound \((NH_4)_6Mo_7O_{24}\), we will follow these steps:

### Step 1: Identify the components of the compound
The compound consists of ammonium ions \((NH_4)^+\), molybdenum \((Mo)\), and oxygen \((O)\).

### Step 2: Assign oxidation states to known components
- The oxidation state of nitrogen in \((NH_4)^+\) is \(-3\) (since it is bonded to four hydrogens, each with an oxidation state of \(+1\)).
- Therefore, the total oxidation state for one ammonium ion \((NH_4)^+\) is \(+1\).
- The oxidation state of oxygen is typically \(-2\).

### Step 3: Set up the equation based on the overall charge of the compound
The overall charge of the compound is neutral (0). Thus, we can set up the equation based on the contributions of each component:
\[ 6 \times (+1) + 7x + 24 \times (-2) = 0 \]
Where \(x\) is the oxidation state of molybdenum \((Mo)\).

### Step 4: Simplify the equation
Substituting the known values into the equation:
\[ 6 \times 1 + 7x - 48 = 0 \]
This simplifies to:
\[ 6 + 7x - 48 = 0 \]
\[ 7x - 42 = 0 \]

### Step 5: Solve for \(x\)
Rearranging gives:
\[ 7x = 42 \]
\[ x = \frac{42}{7} = 6 \]

### Conclusion
The oxidation number of the underlined atom (molybdenum) in \((NH_4)_6Mo_7O_{24}\) is \(+6\).

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