The number of unpaired electrons in the valence shell of the members of oxygen family is ____.

Explanation

To determine the number of unpaired electrons in the valence shell of the members of the oxygen family (Group 16 elements), we will analyze the electronic configurations of each member. The members of the oxygen family include oxygen (O), sulfur (S), selenium (Se), tellurium (Te), and polonium (Po).

### Step-by-Step Solution:

1. **Identify the Group 16 Elements**:
The elements in the oxygen family are:
- Oxygen (O)
- Sulfur (S)
- Selenium (Se)
- Tellurium (Te)
- Polonium (Po)

2. **Determine the Atomic Numbers**:
- Oxygen (O): Atomic number 8
- Sulfur (S): Atomic number 16
- Selenium (Se): Atomic number 34
- Tellurium (Te): Atomic number 52
- Polonium (Po): Atomic number 84

3. **Write the Electron Configurations**:
- Oxygen: \(1s^2 2s^2 2p^4\)
- Sulfur: \([Ne] 3s^2 3p^4\)
- Selenium: \([Ar] 4s^2 4p^4\)
- Tellurium: \([Kr] 5s^2 5p^4\)
- Polonium: \([Xe] 6s^2 6p^4\)

4. **Identify the Valence Electrons**:
The valence shell configuration for all these elements can be summarized as \(ns^2 np^4\). Here, \(n\) represents the principal quantum number for the outermost shell.

5. **Count the Unpaired Electrons**:
In the \(np^4\) configuration, we can visualize the distribution of electrons in the p-orbitals:
- The p-orbitals can hold a maximum of 6 electrons (3 orbitals: \(p_x\), \(p_y\), \(p_z\)).
- For \(np^4\), we can represent the filling as follows:
- The first two electrons fill one orbital (let's say \(p_x\)).
- The next two electrons will occupy the next two orbitals (\(p_y\) and \(p_z\)).
- This results in two unpaired electrons since the last two electrons will occupy separate orbitals.

6. **Conclusion**:
Therefore, the number of unpaired electrons in the valence shell of the members of the oxygen family is **2**.

### Final Answer:
The number of unpaired electrons in the valence shell of the members of the oxygen family is **2**.