Sum of `sigma and pi` bonds in `NH_(4)^(+)` cation is ..

Answer

To determine the sum of sigma and pi bonds in the \( NH_4^+ \) cation, we can follow these steps:

### Step 1: Understand the structure of \( NH_4^+ \)
The ammonium ion (\( NH_4^+ \)) consists of one nitrogen atom (N) and four hydrogen atoms (H). The nitrogen atom is the central atom in this cation.

### Step 2: Determine the hybridization of nitrogen in \( NH_4^+ \)
In the \( NH_4^+ \) ion, nitrogen has four valence electrons. When it forms bonds with four hydrogen atoms, it undergoes hybridization to form \( sp^3 \) hybrid orbitals. This is because nitrogen uses its four valence electrons to form four equivalent bonds.

### Step 3: Identify the types of bonds present
In \( NH_4^+ \):
- Each N-H bond is a single bond, which is a sigma bond.
- There are no double or triple bonds present in \( NH_4^+ \), which means there are no pi bonds.

### Step 4: Count the sigma and pi bonds
- **Sigma bonds**: There are four N-H sigma bonds in \( NH_4^+ \).
- **Pi bonds**: There are no pi bonds in \( NH_4^+ \).

### Step 5: Calculate the total number of bonds
The total number of bonds in \( NH_4^+ \) is the sum of sigma and pi bonds:
- Total sigma bonds = 4
- Total pi bonds = 0

Thus, the sum of sigma and pi bonds in \( NH_4^+ \) is:
\[ \text{Total bonds} = \text{Sigma bonds} + \text{Pi bonds} = 4 + 0 = 4 \]

### Final Answer:
The sum of sigma and pi bonds in \( NH_4^+ \) cation is **4**.

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