The first four successive ionization energies for an element are 6.113, 11.871, 50.908 and 6701 (in eV) respectively. The number of valence shell electrons is________.

Answer

To determine the number of valence shell electrons based on the given ionization energies, we can analyze the provided values step by step.

### Step-by-Step Solution:

1. **List the Ionization Energies**:
- The first four successive ionization energies are given as:
- \( IE_1 = 6.113 \, \text{eV} \)
- \( IE_2 = 11.871 \, \text{eV} \)
- \( IE_3 = 50.908 \, \text{eV} \)
- \( IE_4 = 6701 \, \text{eV} \)

2. **Identify the Large Increase in Ionization Energy**:
- Look for the significant jump in the ionization energies.
- The increase from \( IE_2 \) to \( IE_3 \) is substantial (from 11.871 eV to 50.908 eV).
- Similarly, the jump from \( IE_3 \) to \( IE_4 \) is even more pronounced (from 50.908 eV to 6701 eV).

3. **Interpret the Large Increase**:
- A large increase in ionization energy typically indicates that the electron being removed is from a more stable, inner electron shell (i.e., a core electron).
- Since the first two ionization energies (IE_1 and IE_2) are relatively close, it suggests that the first two electrons are valence electrons.

4. **Conclusion**:
- Given that the significant jump occurs after the removal of the second electron, we can conclude that the element has **2 valence shell electrons**.

### Final Answer:
The number of valence shell electrons is **2**.

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