Which of the following ions show a bonds order of `1.75:`

Options

A. `NO_(3)^(-)`
B. `CO_(3)^(2-)`
C. `ClO_(4)^(-)`
D. None of these

Answer

To determine which of the given ions has a bond order of 1.75, we need to calculate the bond order for each ion using the formula:

**Bond Order = 1 + (Number of π bonds) / (Total number of surrounding atoms)**

Let's analyze each ion step by step.

### Step 1: Calculate the bond order for NO3⁻ (Nitrate Ion)

1. **Structure**: The nitrate ion (NO3⁻) has one nitrogen atom bonded to three oxygen atoms. It contains one double bond (N=O) and two single bonds (N-O).
2. **Count π bonds**: There is 1 π bond (from the double bond).
3. **Count surrounding atoms**: There are 3 surrounding atoms (the three oxygen atoms).
4. **Apply the formula**:
\[ \text{Bond Order} = 1 + \frac{1 \text{ (π bond)}}{3 \text{ (surrounding atoms)}} = 1 + \frac{1}{3} = \frac{4}{3} \approx 1.33 \]

### Step 2: Calculate the bond order for CO3²⁻ (Carbonate Ion)

1. **Structure**: The carbonate ion (CO3²⁻) has one carbon atom bonded to three oxygen atoms. It contains one double bond (C=O) and two single bonds (C-O).
2. **Count π bonds**: There is 1 π bond (from the double bond).
3. **Count surrounding atoms**: There are 3 surrounding atoms (the three oxygen atoms).
4. **Apply the formula**:
\[ \text{Bond Order} = 1 + \frac{1 \text{ (π bond)}}{3 \text{ (surrounding atoms)}} = 1 + \frac{1}{3} = \frac{4}{3} \approx 1.33 \]

### Step 3: Calculate the bond order for ClO4⁻ (Perchlorate Ion)

1. **Structure**: The perchlorate ion (ClO4⁻) has one chlorine atom bonded to four oxygen atoms. It contains three double bonds (Cl=O) and one single bond (Cl-O).
2. **Count π bonds**: There are 3 π bonds (from the three double bonds).
3. **Count surrounding atoms**: There are 4 surrounding atoms (the four oxygen atoms).
4. **Apply the formula**:
\[ \text{Bond Order} = 1 + \frac{3 \text{ (π bonds)}}{4 \text{ (surrounding atoms)}} = 1 + \frac{3}{4} = \frac{7}{4} = 1.75 \]

### Conclusion

After calculating the bond orders for all three ions, we find:
- NO3⁻: Bond order = 1.33
- CO3²⁻: Bond order = 1.33
- ClO4⁻: Bond order = 1.75

Thus, the ion that shows a bond order of 1.75 is **ClO4⁻**.

### Final Answer:
**ClO4⁻**

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