Total number of unpaired electrons(s) present in both cationic and anionic part of compound `O_(2)[PtF_(6)]`.

Explanation

To determine the total number of unpaired electrons present in both the cationic and anionic parts of the compound \( O_2[PtF_6] \), we will analyze each part separately.

### Step 1: Analyze the cation \( O_2^{+} \)

1. **Determine the total number of electrons in \( O_2 \)**:
- Each oxygen atom has 8 electrons. Therefore, for \( O_2 \), the total number of electrons is:
\[ 2 \times 8 = 16 \text{ electrons} \]

2. **Account for the positive charge**:
- The \( O_2^{+} \) cation has lost one electron, so:
\[ 16 - 1 = 15 \text{ electrons} \]

3. **Determine the electron configuration**:
- The electron configuration for \( O_2 \) is \( (1s^2)(2s^2)(2p^4) \).
- For \( O_2^{+} \), we remove one electron from the \( 2p \) orbital. Thus, the configuration becomes:
\[ (1s^2)(2s^2)(2p^3) \]

4. **Count unpaired electrons**:
- In the \( 2p^3 \) configuration, there are 3 electrons in the \( p \) orbitals, which can be represented as:
\[ \uparrow \, \uparrow \, \uparrow \]
- All three electrons are unpaired. Therefore, \( O_2^{+} \) has:
\[ 1 \text{ unpaired electron} \]

### Step 2: Analyze the anion \( PtF_6^{-} \)

1. **Determine the oxidation state of platinum (Pt)**:
- The oxidation state of \( Pt \) in \( PtF_6^{-} \) can be calculated as follows:
\[ x + 6(-1) = -1 \implies x - 6 = -1 \implies x = +5 \]

2. **Determine the electron configuration of \( Pt \)**:
- The electron configuration of neutral platinum (Pt) is:
\[ [Xe] 4f^{14} 5d^9 6s^1 \]
- In the +5 oxidation state, we remove 5 electrons (2 from \( 6s \) and 3 from \( 5d \)):
\[ [Xe] 4f^{14} 5d^8 \]

3. **Count unpaired electrons in \( 5d^8 \)**:
- The \( 5d^8 \) configuration can be visualized as:
\[ \uparrow \, \uparrow \, \uparrow \, \uparrow \, \uparrow \, \downarrow \, \downarrow \, \downarrow \]
- In this arrangement, there are 2 paired electrons and 6 unpaired electrons. Therefore, \( PtF_6^{-} \) has:
\[ 5 \text{ unpaired electrons} \]

### Step 3: Calculate the total number of unpaired electrons

- From \( O_2^{+} \): 1 unpaired electron
- From \( PtF_6^{-} \): 5 unpaired electrons

Thus, the total number of unpaired electrons in the compound \( O_2[PtF_6] \) is:
\[ 1 + 5 = 6 \text{ unpaired electrons} \]

### Final Answer:
The total number of unpaired electrons present in both the cationic and anionic parts of the compound \( O_2[PtF_6] \) is **6**.

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