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 molybdenum (Mo) in the compound \((NH_4)_6Mo_7O_{24}\), we can follow these steps:

### Step 1: Identify the oxidation states of known elements
- The oxidation state of ammonium \((NH_4^+)\) is \(+1\).
- The oxidation state of oxygen \((O)\) is \(-2\).

### Step 2: Set up the equation
Let the oxidation state of molybdenum (Mo) be \(x\). The compound has:
- 6 ammonium ions, contributing \(6 \times (+1) = +6\).
- 7 molybdenum atoms, contributing \(7 \times x\).
- 24 oxygen atoms, contributing \(24 \times (-2) = -48\).

### Step 3: Write the overall charge equation
Since the overall charge of the compound is neutral (0), we can set up the equation:
\[ 6(+1) + 7x + 24(-2) = 0 \]

### Step 4: Simplify the equation
Substituting the known values into the equation:
\[ 6 + 7x - 48 = 0 \]

### Step 5: Solve for \(x\)
Rearranging the equation gives:
\[ 7x - 42 = 0 \]
\[ 7x = 42 \]
\[ x = \frac{42}{7} = 6 \]

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

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