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 of the following pairs of species have identical shapes, we will analyze each pair using the concept of steric number and hybridization. The steric number can be calculated using the formula:

\[ \text{Steric Number} = \frac{1}{2} \left( V + M + \text{(charge)} \right) \]

Where:
- \( V \) = number of valence electrons of the central atom
- \( M \) = number of monovalent atoms attached to the central atom
- Add the charge if it’s an anion, or subtract the charge if it’s a cation.

### Step-by-Step Solution:

**Step 1: Analyze the first pair (NO2⁺ and NO2⁻)**

- **For NO2⁺:**
- Central atom: Nitrogen (N)
- Valence electrons (V) = 5
- Charge = +1 (cation)
- So, \( \text{Steric Number} = \frac{1}{2} (5 + 2 - 1) = \frac{1}{2} (6) = 3 \)
- Hybridization = sp² (since 3 is the steric number)
- Shape = Linear.

- **For NO2⁻:**
- Valence electrons (V) = 5
- Charge = -1 (anion)
- So, \( \text{Steric Number} = \frac{1}{2} (5 + 2 + 1) = \frac{1}{2} (8) = 4 \)
- Hybridization = sp³ (since 4 is the steric number)
- Shape = Bent (due to one lone pair).

**Conclusion for Pair 1:** NO2⁺ is linear and NO2⁻ is bent. They do not have identical shapes.

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**Step 2: Analyze the second pair (PCl5 and BrF5)**

- **For PCl5:**
- Central atom: Phosphorus (P)
- Valence electrons (V) = 5
- Number of monovalent atoms (M) = 5
- So, \( \text{Steric Number} = \frac{1}{2} (5 + 5) = 5 \)
- Hybridization = sp³d
- Shape = Trigonal bipyramidal.

- **For BrF5:**
- Central atom: Bromine (Br)
- Valence electrons (V) = 7
- Number of monovalent atoms (M) = 5
- So, \( \text{Steric Number} = \frac{1}{2} (7 + 5) = 6 \)
- Hybridization = sp³d²
- Shape = Square pyramidal (due to one lone pair).

**Conclusion for Pair 2:** PCl5 is trigonal bipyramidal and BrF5 is square pyramidal. They do not have identical shapes.

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**Step 3: Analyze the third pair (XeF4 and ICl4⁻)**

- **For XeF4:**
- Central atom: Xenon (Xe)
- Valence electrons (V) = 8
- Number of monovalent atoms (M) = 4
- So, \( \text{Steric Number} = \frac{1}{2} (8 + 4) = 6 \)
- Hybridization = sp³d²
- Shape = Square planar (due to two lone pairs).

- **For ICl4⁻:**
- Central atom: Iodine (I)
- Valence electrons (V) = 7
- Number of monovalent atoms (M) = 4
- Charge = -1 (anion)
- So, \( \text{Steric Number} = \frac{1}{2} (7 + 4 + 1) = 6 \)
- Hybridization = sp³d²
- Shape = Square planar (due to two lone pairs).

**Conclusion for Pair 3:** Both XeF4 and ICl4⁻ are square planar. They have identical shapes.

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**Step 4: Analyze the fourth pair (TeCl4 and HCO4)**

- **For TeCl4:**
- Central atom: Tellurium (Te)
- Valence electrons (V) = 6
- Number of monovalent atoms (M) = 4
- So, \( \text{Steric Number} = \frac{1}{2} (6 + 4) = 5 \)
- Hybridization = sp³d
- Shape = Seesaw (due to one lone pair).

- **For HCO4:**
- Central atom: Carbon (C)
- Valence electrons (V) = 4
- Number of monovalent atoms (M) = 4 (Oxygen is divalent)
- So, \( \text{Steric Number} = \frac{1}{2} (4 + 4) = 4 \)
- Hybridization = sp³
- Shape = Tetrahedral.

**Conclusion for Pair 4:** TeCl4 is seesaw and HCO4 is tetrahedral. They do not have identical shapes.

### Final Answer:
The only pair of species that have identical shapes is **XeF4 and ICl4⁻**, both of which are square planar.

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