Which of the following arrangements is correct on the basis of the increasing `p-` character of the hybrid orbitals of the central atoms in the followings `:`
`(I)ClO_(2)^(-)" "(II)CS_(2)" "(III)SnCl_(2)`

Options

A. Potential energy of electron `prop(Z^(2))/(n^(2))`
B. The product of velocity of electron and principle quantum number `(n) prop Z^(2)`
C. Frequency of revolution of electron in an orbit `prop (Z^(2))/(n^(3))`
D. Coulombic force of attraction on the electron `prop (Z^(2))/(n^(2))`

Explanation

To determine the increasing p-character of the hybrid orbitals of the central atoms in the given compounds (I) ClO2^(-), (II) CS2, and (III) SnCl2, we will follow these steps:

### Step 1: Calculate the hybridization of ClO2^(-)
1. **Identify the valence electrons**: Chlorine (Cl) has 7 valence electrons (3s² 3p⁵), and each oxygen (O) has 6 valence electrons (2s² 2p⁴). Since there are two oxygen atoms, that gives us a total of 12 valence electrons from oxygen.
2. **Account for the charge**: The overall charge of the ion is -1, which means we add one more electron.
3. **Use the hybridization formula**:
\[ \text{Hybridization} = \frac{(\text{Valence Electrons}) + (\text{Monovalent Atoms}) - (\text{Charge})}{2} \]
Here, we have:
\[ \text{Valence Electrons} = 7 + 12 + 1 = 20 \]
There are no monovalent atoms, so:
\[ \text{Hybridization} = \frac{20}{2} = 10 \]
This indicates a hybridization of sp³.

### Step 2: Calculate the hybridization of CS2
1. **Identify the valence electrons**: Carbon (C) has 4 valence electrons (2s² 2p²), and sulfur (S) has 6 valence electrons. Since there are two sulfur atoms, that gives us a total of 12 valence electrons from sulfur.
2. **Account for the charge**: The overall charge is 0.
3. **Use the hybridization formula**:
\[ \text{Hybridization} = \frac{(4 + 12 + 0)}{2} = \frac{16}{2} = 8 \]
This indicates a hybridization of sp.

### Step 3: Calculate the hybridization of SnCl2
1. **Identify the valence electrons**: Tin (Sn) has 4 valence electrons (4s² 4p²), and chlorine (Cl) has 7 valence electrons. Since there are two chlorine atoms, that gives us a total of 14 valence electrons from chlorine.
2. **Account for the charge**: The overall charge is 0.
3. **Use the hybridization formula**:
\[ \text{Hybridization} = \frac{(4 + 14 + 0)}{2} = \frac{18}{2} = 9 \]
This indicates a hybridization of sp².

### Step 4: Determine the increasing p-character order
- **Hybridization and p-character**:
- sp³ has 3 p orbitals (maximum p-character)
- sp² has 2 p orbitals
- sp has 1 p orbital (minimum p-character)

### Final Order
Based on the hybridizations calculated:
1. CS2: sp (least p-character)
2. SnCl2: sp²
3. ClO2^(-): sp³ (most p-character)

Thus, the correct order based on increasing p-character is:
**CS2 < SnCl2 < ClO2^(-)**