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 from which of the given reactions we can calculate the second ionization energy of \( M(g) \), we need to analyze each reaction step by step.

### Step 1: Understanding Second Ionization Energy
The second ionization energy of an element \( M \) is defined as the energy required to remove the second electron from a gaseous ion \( M^+ \) to form \( M^{2+} \):
\[ M^+(g) \rightarrow M^{2+}(g) + e^- \]
This means we need to find a combination of the provided reactions that will yield this equation.

### Step 2: Analyzing the Given Reactions
Let's analyze each of the provided reactions to see if they can be combined to yield the second ionization energy equation.

1. **Reaction (a)**:
\[ M(s) \rightarrow M(g) \]
This reaction converts solid \( M \) to gaseous \( M \).

2. **Reaction (b)**:
\[ M(s) \rightarrow M^{2+}(g) + 2e^- \]
This reaction indicates that solid \( M \) is converted to gaseous \( M^{2+} \) while releasing two electrons.

3. **Reaction (c)**:
\[ M(g) \rightarrow M^+(g) + e^- \]
This reaction shows the ionization of gaseous \( M \) to form \( M^+ \) and one electron.

4. **Reaction (d)**:
\[ M^+(g) \rightarrow M^{2+}(g) + e^- \]
This reaction represents the second ionization process, where \( M^+ \) loses another electron to form \( M^{2+} \).

5. **Reaction (e)**:
\[ M(g) \rightarrow M^{2+}(g) + 2e^- \]
This reaction indicates that gaseous \( M \) directly converts to \( M^{2+} \) while releasing two electrons.

### Step 3: Combining Reactions
Now, we need to check combinations of these reactions to derive the second ionization energy equation.

- **Option 1**: \( (a) + (c) + (d) \)
- This results in:
\[ M(s) \rightarrow M(g) \rightarrow M^+(g) + e^- \rightarrow M^{2+}(g) + e^- \]
This does not yield the desired equation for second ionization energy.

- **Option 2**: \( (b) - (a) + (c) \)
- This results in:
\[ M(s) \rightarrow M^{2+}(g) + 2e^- - M(g) + M^+(g) + e^- \]
This does not yield the desired equation for second ionization energy.

- **Option 3**: \( (a) + (e) \)
- This results in:
\[ M(s) \rightarrow M(g) \rightarrow M^{2+}(g) + 2e^- \]
This does not yield the desired equation for second ionization energy.

- **Option 4**: \( (e) - (c) \)
- This results in:
\[ M(g) \rightarrow M^{2+}(g) + 2e^- - (M(g) \rightarrow M^+(g) + e^-) \]
This simplifies to:
\[ M^+(g) \rightarrow M^{2+}(g) + e^- \]
This is exactly the equation for the second ionization energy of \( M \).

### Conclusion
The correct option from which we can calculate the second ionization energy of \( M(g) \) is **Option 4**.