The number of a bonds in `ClO_(4)^(-)` ion is 

Options

A. 2
B. 3
C. 7
D. 1

Answer

To determine the number of bonds in the \( \text{ClO}_4^{-} \) ion, we can follow these steps:

### Step 1: Identify the central atom and surrounding atoms
In the \( \text{ClO}_4^{-} \) ion, chlorine (Cl) is the central atom, and there are four oxygen (O) atoms surrounding it.

**Hint:** Remember that in polyatomic ions, the central atom is usually the least electronegative element.

### Step 2: Determine the total number of valence electrons
Chlorine has 7 valence electrons, and each oxygen has 6 valence electrons. Since there are four oxygen atoms, the total number of valence electrons is:
\[ 7 + (4 \times 6) + 1 = 7 + 24 + 1 = 32 \]
(The additional 1 is due to the negative charge of the ion.)

**Hint:** Count the valence electrons carefully, including the charge of the ion.

### Step 3: Draw the Lewis structure
In the Lewis structure of \( \text{ClO}_4^{-} \), chlorine is the central atom with four oxygen atoms bonded to it. To minimize formal charges, three of the oxygen atoms will form double bonds with chlorine, while one oxygen will have a single bond and carry the negative charge.

**Hint:** Use double bonds to satisfy the octet rule for both chlorine and oxygen atoms.

### Step 4: Count the bonds
In the Lewis structure:
- There are three double bonds (Cl=O) and one single bond (Cl-O).
- Each double bond counts as 2 bonds, and the single bond counts as 1 bond.

Thus, the total number of bonds is:
\[ 3 \times 2 + 1 = 6 + 1 = 7 \]

**Hint:** Remember to count each bond type correctly (single vs. double).

### Conclusion
The total number of bonds in the \( \text{ClO}_4^{-} \) ion is **7**.

**Final Answer:** 7