Which is the following pairs of species have identical shapes ?

Options

A. `NO_(2)^(+) and NO_(2)^(-)`
B. `PCl_(5) and BrF_(5)`
C. `XeF_(4) and IC l_(4)^(-)`
D. `TeCl_(4) and XeO_(4)`

Answer

To determine which pairs of species have identical shapes, we will analyze the molecular geometry of each species based on their valence electrons, bonding pairs, and lone pairs. Here is a step-by-step breakdown:

### Step 1: Analyze NO2+
- **Valence Electrons**: Nitrogen (N) has 5 valence electrons, and each Oxygen (O) has 6. Since NO2+ has a positive charge, we remove one electron from the total.
- **Total Valence Electrons**: 5 (N) + 2 * 6 (O) - 1 (positive charge) = 16 electrons.
- **Lewis Structure**: N is bonded to two O atoms with one double bond and one single bond.
- **Bonding Pairs**: 2 (one for each bond).
- **Lone Pairs**: 0.
- **Steric Number**: 2 (bonding pairs + lone pairs).
- **Hybridization**: sp.
- **Shape**: Linear.

### Step 2: Analyze Na2-
- **Valence Electrons**: Nitrogen has 5, and each O has 6. With a negative charge, we add one electron.
- **Total Valence Electrons**: 5 (N) + 6 (O) + 1 (negative charge) = 12 electrons.
- **Lewis Structure**: N is bonded to one O atom with three bonds.
- **Bonding Pairs**: 3.
- **Lone Pairs**: 0.
- **Steric Number**: 3.
- **Hybridization**: sp².
- **Shape**: Trigonal planar.

### Step 3: Analyze PCl5
- **Valence Electrons**: Phosphorus (P) has 5, and each Cl has 7.
- **Total Valence Electrons**: 5 (P) + 5 * 7 (Cl) = 40 electrons.
- **Lewis Structure**: P is bonded to five Cl atoms.
- **Bonding Pairs**: 5.
- **Lone Pairs**: 0.
- **Steric Number**: 5.
- **Hybridization**: sp³d.
- **Shape**: Trigonal bipyramidal.

### Step 4: Analyze BrF5-
- **Valence Electrons**: Bromine (Br) has 7, and each F has 7. With a negative charge, we add one electron.
- **Total Valence Electrons**: 7 (Br) + 5 * 7 (F) + 1 (negative charge) = 43 electrons.
- **Lewis Structure**: Br is bonded to five F atoms with one lone pair.
- **Bonding Pairs**: 5.
- **Lone Pairs**: 1.
- **Steric Number**: 6.
- **Hybridization**: sp³d².
- **Shape**: Square pyramidal.

### Step 5: Analyze XeF4
- **Valence Electrons**: Xenon (Xe) has 8, and each F has 7.
- **Total Valence Electrons**: 8 (Xe) + 4 * 7 (F) = 36 electrons.
- **Lewis Structure**: Xe is bonded to four F atoms with two lone pairs.
- **Bonding Pairs**: 4.
- **Lone Pairs**: 2.
- **Steric Number**: 6.
- **Hybridization**: sp³d².
- **Shape**: Square planar.

### Step 6: Analyze ICl4-
- **Valence Electrons**: Iodine (I) has 7, and each Cl has 7. With a negative charge, we add one electron.
- **Total Valence Electrons**: 7 (I) + 4 * 7 (Cl) + 1 (negative charge) = 43 electrons.
- **Lewis Structure**: I is bonded to four Cl atoms with two lone pairs.
- **Bonding Pairs**: 4.
- **Lone Pairs**: 2.
- **Steric Number**: 6.
- **Hybridization**: sp³d².
- **Shape**: Square planar.

### Step 7: Analyze TeCl4
- **Valence Electrons**: Tellurium (Te) has 6, and each Cl has 7.
- **Total Valence Electrons**: 6 (Te) + 4 * 7 (Cl) = 34 electrons.
- **Lewis Structure**: Te is bonded to four Cl atoms with one lone pair.
- **Bonding Pairs**: 4.
- **Lone Pairs**: 1.
- **Steric Number**: 5.
- **Hybridization**: sp³d.
- **Shape**: Seesaw.

### Step 8: Analyze XeO4
- **Valence Electrons**: Xenon has 8, and each O has 6.
- **Total Valence Electrons**: 8 (Xe) + 4 * 6 (O) = 32 electrons.
- **Lewis Structure**: Xe is bonded to four O atoms with no lone pairs.
- **Bonding Pairs**: 4.
- **Lone Pairs**: 0.
- **Steric Number**: 4.
- **Hybridization**: sp³.
- **Shape**: Tetrahedral.

### Conclusion
The species that have identical shapes are **XeF4** and **ICl4-**, both of which have a square planar shape.

### Final Answer
**The correct answer is: XeF4 and ICl4- have identical shapes.**