Consider the following changes `:-`
`M_((s)rarrM_((g))" ""....."(a)`
`M_((s))rarrM_((g))^(+2)+2e^(ө)" ""....."(b)`
`M_((g))rarrM^(+)+e^(ө)" ""....."(c)`
`M_((g))^(+)rarrM_((g))^(+2)+e^(ө)" ""....."(d)`
`M_((g))rarrM_((g))^(+2)+2e^(ө)" ""....."(e)`
The second ionisation energy of `M_((g))` could be calculated from which of the above given reactions`:`

Options

A. 1 + 3 + 4
B. 2 – 1 + 3
C. 1 + 5
D. 5 – 3

Explanation

To determine the second ionization energy of the element \( M \) in the gaseous state, we need to analyze the provided reactions and see which ones can help us derive the necessary equation for the second ionization energy.

### Step-by-Step Solution:

1. **Understanding Ionization Energy**:
- The first ionization energy is the energy required to remove one electron from a neutral atom in the gaseous state to form a cation.
- The second ionization energy is the energy required to remove a second electron from the cation formed after the first ionization.

2. **Identifying Relevant Reactions**:
- We have the following reactions:
- (a) \( M_{(s)} \rightarrow M_{(g)} \)
- (b) \( M_{(s)} \rightarrow M_{(g)}^{+2} + 2e^{-} \)
- (c) \( M_{(g)} \rightarrow M^{+} + e^{-} \)
- (d) \( M_{(g)}^{+} \rightarrow M_{(g)}^{+2} + e^{-} \)
- (e) \( M_{(g)} \rightarrow M_{(g)}^{+2} + 2e^{-} \)

3. **Finding the Second Ionization Energy**:
- The second ionization energy can be represented as:
\[ M^{+} \rightarrow M^{+2} + e^{-} \]
- This corresponds to reaction (d), where the gaseous ion \( M^{+} \) loses an electron to form \( M^{+2} \).

4. **Combining Reactions**:
- To derive the second ionization energy using the provided reactions, we can combine reactions (c) and (d):
- From (c): \( M_{(g)} \rightarrow M^{+} + e^{-} \) (First ionization)
- From (d): \( M^{+} \rightarrow M^{+2} + e^{-} \) (Second ionization)
- By combining these two reactions, we can derive the overall process of removing two electrons from the neutral atom \( M \).

5. **Conclusion**:
- The second ionization energy of \( M_{(g)} \) can be calculated from reaction (d) \( M_{(g)}^{+} \rightarrow M_{(g)}^{+2} + e^{-} \).

### Final Answer:
The second ionization energy of \( M_{(g)} \) could be calculated from reaction (d).

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