Which of the following species contains equal number of sigma and pi bonds ?

Options

A. `CH_(2)(CN)_(2)`
B. `HCO_(3)^(-)`
C. `XeO_(4)`
D. `(CN)_(2)`

Explanation

To determine which of the given species contains an equal number of sigma and pi bonds, we will analyze each species one by one. The key concepts to remember are:

1. A single bond consists of 1 sigma bond and 0 pi bonds.
2. A double bond consists of 1 sigma bond and 1 pi bond.
3. A triple bond consists of 1 sigma bond and 2 pi bonds.

Now, let's analyze the species step by step.

### Step 1: Analyze the first species (CH2=C=C=CH2)
1. **Structure**: Draw the structure of the species.
- CH2=C=C=CH2 has two double bonds.

2. **Count Sigma Bonds**:
- Each double bond contributes 1 sigma bond.
- There are 4 carbon-hydrogen single bonds (2 from each CH2).
- Total sigma bonds = 4 (C-H) + 2 (C=C) = 6 sigma bonds.

3. **Count Pi Bonds**:
- Each double bond contributes 1 pi bond.
- Total pi bonds = 2 (from two C=C double bonds) = 2 pi bonds.

4. **Conclusion**:
- Sigma bonds = 6, Pi bonds = 2. Not equal.

### Step 2: Analyze the second species (HCO3^-)
1. **Structure**: Draw the structure of the bicarbonate ion.
- HCO3^- has one double bond (C=O) and two single bonds (C-O and C-H).

2. **Count Sigma Bonds**:
- 1 (C-H) + 1 (C-O) + 1 (C=O) = 3 sigma bonds.

3. **Count Pi Bonds**:
- 1 (from the C=O double bond) = 1 pi bond.

4. **Conclusion**:
- Sigma bonds = 3, Pi bonds = 1. Not equal.

### Step 3: Analyze the third species (XeF4)
1. **Structure**: Draw the structure of xenon tetrafluoride.
- XeF4 has four single bonds (Xe-F) and no double or triple bonds.

2. **Count Sigma Bonds**:
- 4 (Xe-F) = 4 sigma bonds.

3. **Count Pi Bonds**:
- There are no pi bonds in this molecule.

4. **Conclusion**:
- Sigma bonds = 4, Pi bonds = 0. Not equal.

### Step 4: Analyze the fourth species (C2N2)
1. **Structure**: Draw the structure of dicyano.
- C2N2 has a triple bond between the two carbons and single bonds to nitrogen.

2. **Count Sigma Bonds**:
- 1 (C-C) + 2 (C-N) = 3 sigma bonds.

3. **Count Pi Bonds**:
- 2 (from the C≡C triple bond) + 1 (from C-N) = 3 pi bonds.

4. **Conclusion**:
- Sigma bonds = 3, Pi bonds = 3. Equal.

### Final Conclusion
After analyzing all the species, we find that **C2N2** is the species that contains an equal number of sigma and pi bonds.