The isoelectronic pair is 

Options

A. `Cl_2 O, ICl_(2)^(-)`
B. `Cl_(2)^(-), ClO_2`
C. `IF_(2)^(+), I_(3)^(-)`
D. `ClO_(2)^(-), ClF_(2)^(+)`

Answer

To determine the isoelectronic pair from the given options, we need to find pairs of molecules or ions that have the same number of electrons. Let's analyze each option step by step.

### Step 1: Understand the definition of isoelectronic pairs
Isoelectronic pairs are molecules or ions that have the same number of electrons.

### Step 2: Analyze each option
We will calculate the total number of electrons for each given molecule or ion in the options.

#### Option 1: Cl2O and ICl2^-
- **Cl2O**:
- Chlorine (Cl) has 17 electrons, so for 2 Cl: 2 × 17 = 34 electrons.
- Oxygen (O) has 8 electrons.
- Total for Cl2O = 34 + 8 = 42 electrons.

- **ICl2^-**:
- Iodine (I) has 53 electrons, and Chlorine (Cl) has 17 electrons.
- Total for ICl2^- = 53 + (2 × 17) + 1 (for the negative charge) = 53 + 34 + 1 = 88 electrons.

Since 42 ≠ 88, Cl2O and ICl2^- are not isoelectronic.

#### Option 2: Cl2^- and ClO2
- **Cl2^-**:
- Chlorine has 17 electrons, and with a negative charge, it gains 1 electron.
- Total for Cl2^- = (2 × 17) + 1 = 34 + 1 = 35 electrons.

- **ClO2**:
- Chlorine has 17 electrons, and Oxygen has 8 electrons.
- Total for ClO2 = 17 + (2 × 8) = 17 + 16 = 33 electrons.

Since 35 ≠ 33, Cl2^- and ClO2 are not isoelectronic.

#### Option 3: IF2^+ and I3^-
- **IF2^+**:
- Iodine has 53 electrons, and Fluorine has 9 electrons.
- Total for IF2^+ = 53 + (2 × 9) - 1 (for the positive charge) = 53 + 18 - 1 = 70 electrons.

- **I3^-**:
- Iodine has 53 electrons.
- Total for I3^- = (3 × 53) + 1 (for the negative charge) = 159 + 1 = 160 electrons.

Since 70 ≠ 160, IF2^+ and I3^- are not isoelectronic.

#### Option 4: ClO2^- and ClF2^+
- **ClO2^-**:
- Chlorine has 17 electrons, and Oxygen has 8 electrons.
- Total for ClO2^- = 17 + (2 × 8) + 1 (for the negative charge) = 17 + 16 + 1 = 34 electrons.

- **ClF2^+**:
- Chlorine has 17 electrons, and Fluorine has 9 electrons.
- Total for ClF2^+ = 17 + (2 × 9) - 1 (for the positive charge) = 17 + 18 - 1 = 34 electrons.

Since 34 = 34, ClO2^- and ClF2^+ are isoelectronic.

### Conclusion
The isoelectronic pair is **ClO2^- and ClF2^+**.