Which of the following possess two lone pair of electrons on the central atom and square planar in shape ?
(I) `SF_(4) " " (II)XeO_(4) " " (III)XeF_(4) " " (IV)IC l_(4)^(-)`

Options

A. I,III
B. II,IV
C. III, IV
D. All

Answer

To solve the problem of identifying which of the given compounds possess two lone pairs of electrons on the central atom and are square planar in shape, we will analyze each option step by step.

### Step 1: Analyze SF₄
- **Valence Electrons**: Sulfur (S) has 6 valence electrons, and each fluorine (F) has 7 valence electrons. Thus, for SF₄:
- Total valence electrons = 6 (S) + 4 × 7 (F) = 6 + 28 = 34 electrons.
- **Lone Pairs**: Sulfur will form 4 bonds with fluorine, using 8 electrons (4 bonds). This leaves 26 electrons, which means there is 1 lone pair on sulfur.
- **Steric Number**: The steric number (number of bonded atoms + lone pairs) = 4 (bonded to F) + 1 (lone pair) = 5.
- **Hybridization**: The hybridization is sp³d.
- **Shape**: The shape is see-saw due to the presence of one lone pair.

### Step 2: Analyze XeO₄
- **Valence Electrons**: Xenon (Xe) has 8 valence electrons, and each oxygen (O) has 6 valence electrons. Thus, for XeO₄:
- Total valence electrons = 8 (Xe) + 4 × 6 (O) = 8 + 24 = 32 electrons.
- **Lone Pairs**: Xenon forms 4 bonds with oxygen, using 8 electrons. This leaves 24 electrons, which means there are 0 lone pairs on xenon.
- **Steric Number**: The steric number = 4 (bonded to O) + 0 (lone pairs) = 4.
- **Hybridization**: The hybridization is sp³.
- **Shape**: The shape is tetrahedral.

### Step 3: Analyze XeF₄
- **Valence Electrons**: Xenon (Xe) has 8 valence electrons, and each fluorine (F) has 7 valence electrons. Thus, for XeF₄:
- Total valence electrons = 8 (Xe) + 4 × 7 (F) = 8 + 28 = 36 electrons.
- **Lone Pairs**: Xenon forms 4 bonds with fluorine, using 8 electrons. This leaves 28 electrons, which means there are 2 lone pairs on xenon.
- **Steric Number**: The steric number = 4 (bonded to F) + 2 (lone pairs) = 6.
- **Hybridization**: The hybridization is sp³d².
- **Shape**: The shape is square planar due to the presence of 2 lone pairs.

### Step 4: Analyze ICl₄⁻
- **Valence Electrons**: Iodine (I) has 7 valence electrons, and each chlorine (Cl) has 7 valence electrons. The negative charge adds 1 more electron. Thus, for ICl₄⁻:
- Total valence electrons = 7 (I) + 4 × 7 (Cl) + 1 (negative charge) = 7 + 28 + 1 = 36 electrons.
- **Lone Pairs**: Iodine forms 4 bonds with chlorine, using 8 electrons. This leaves 28 electrons, which means there are 2 lone pairs on iodine.
- **Steric Number**: The steric number = 4 (bonded to Cl) + 2 (lone pairs) = 6.
- **Hybridization**: The hybridization is sp³d².
- **Shape**: The shape is square planar due to the presence of 2 lone pairs.

### Conclusion
The compounds that possess two lone pairs of electrons on the central atom and are square planar in shape are:
- **(III) XeF₄**
- **(IV) ICl₄⁻**

### Final Answer
**(III) XeF₄ and (IV) ICl₄⁻**