Which of the following statement is correct about `I_(3)^(+) and I_(3)^(-)` molecular ions ?

Options

A. Number of lone pairs at central atoms are same in both molecular ions
B. Hybridization of central atoms in both ions are same
C. Both are polar species
D. Both are planar species

Explanation

To determine which statement is correct about the molecular ions \( I_3^+ \) and \( I_3^- \), we will analyze their structures, hybridization, and other properties step by step.

### Step 1: Determine the Valence Electrons
- **For \( I_3^+ \)**: Each iodine (I) atom has 7 valence electrons. Since there are 3 iodine atoms, that gives us \( 3 \times 7 = 21 \) electrons. However, since it has a positive charge, we subtract 1 electron, giving us \( 21 - 1 = 20 \) valence electrons.
- **For \( I_3^- \)**: Again, we have 3 iodine atoms contributing 21 electrons. Since it has a negative charge, we add 1 electron, resulting in \( 21 + 1 = 22 \) valence electrons.

### Step 2: Draw the Lewis Structures
- **For \( I_3^+ \)**:
- Place one iodine atom in the center and the other two iodine atoms on either side.
- Connect the iodine atoms with single bonds. Each bond uses 2 electrons, so 2 bonds use 4 electrons.
- This leaves \( 20 - 4 = 16 \) electrons. Distributing these as lone pairs around the outer iodine atoms, we find that the central iodine will have 2 lone pairs.

- **For \( I_3^- \)**:
- Similarly, place one iodine atom in the center and the other two iodine atoms on either side.
- Connect the iodine atoms with single bonds, using 4 electrons for 2 bonds.
- This leaves \( 22 - 4 = 18 \) electrons. Distributing these, we find that the central iodine will have 3 lone pairs.

### Step 3: Determine the Hybridization
- **For \( I_3^+ \)**: The steric number is calculated as the number of sigma bonds plus the number of lone pairs. Here, we have 2 sigma bonds and 2 lone pairs, giving a steric number of 4. Hence, the hybridization is \( sp^3 \).

- **For \( I_3^- \)**: The steric number is 2 sigma bonds and 3 lone pairs, giving a steric number of 5. Therefore, the hybridization is \( sp^3d \).

### Step 4: Analyze the Molecular Geometry
- **For \( I_3^+ \)**: With a steric number of 4 and 2 lone pairs, the molecular shape is bent or V-shaped.

- **For \( I_3^- \)**: With a steric number of 5 and 3 lone pairs, the molecular shape is linear.

### Step 5: Determine Polarity
- **Polarity**: Both ions consist of the same type of atom (iodine), which means there is no significant difference in electronegativity between the atoms. Therefore, both \( I_3^+ \) and \( I_3^- \) are non-polar.

### Step 6: Planarity
- **Planarity**: Both \( I_3^+ \) and \( I_3^- \) are planar species, meaning all atoms lie in the same plane.

### Conclusion
From the analysis, we can conclude that:
- The number of lone pairs at the central atom is not the same.
- The hybridization of the central atom is not the same.
- Both are non-polar species.
- Both are planar species.

Thus, the correct statement is that **both are planar species**.