Which of the following statements is / are true ?
(i) `PH_5 and BiCl_5` do not exist
(ii) `p pi - d pi` bond is present in `SO_2`
(iii) `I_3^(+)` has bent geometry
(iv) `SeF_4 and CH_4` have same shape

Options

A. (i) , (ii) , (iii)
B. (i) , (iii)
C. (i) , (iii) , (iv)
D. (i) , (ii) , (iv)

Answer

To determine which statements are true, we will analyze each statement one by one.

### Statement (i): `PH5 and BiCl5` do not exist.

1. **Analysis of PH5**:
- Phosphorus (P) is in the third period and has the electronic configuration of [Ne] 3s² 3p³.
- For PH5 to exist, phosphorus would need to use sp³d hybridization, which involves the mixing of 3s, 3p, and 3d orbitals. However, the energy difference between these orbitals is too large for effective hybridization to occur.
- Therefore, PH5 does not exist.

2. **Analysis of BiCl5**:
- Bismuth (Bi) is also in group 15 and can exhibit +5 oxidation state. However, due to the inert pair effect, Bi prefers to exist in the +3 oxidation state.
- The bond strength between Bi and Cl is weak due to the larger size of Bi compared to Cl, making BiCl5 unstable.
- Therefore, BiCl5 does not exist.

**Conclusion**: Statement (i) is true.

### Statement (ii): `p pi - d pi` bond is present in `SO2`.

1. **Analysis of SO2**:
- The structure of SO2 involves a sulfur atom (S) double-bonded to two oxygen atoms (O).
- Sulfur has vacant d orbitals available for bonding, while oxygen can form p pi bonds.
- In SO2, there are two p pi bonds formed between sulfur and oxygen, and sulfur can also utilize its d orbitals to form a p pi - d pi bond.

**Conclusion**: Statement (ii) is true.

### Statement (iii): `I3+` has bent geometry.

1. **Analysis of I3+**:
- The central atom in I3+ is iodine (I) with a +1 charge, meaning it has lost one electron. The valence shell of iodine originally has 7 electrons, so I3+ has 6 electrons remaining.
- In I3+, there are 2 bond pairs and 2 lone pairs on the central iodine atom.
- The steric number is calculated as the sum of bond pairs and lone pairs: 2 (bond pairs) + 2 (lone pairs) = 4.
- For a steric number of 4, the hybridization is sp³, which results in a tetrahedral arrangement. However, the presence of two lone pairs distorts this arrangement, leading to a bent geometry.

**Conclusion**: Statement (iii) is true.

### Statement (iv): `SeF4 and CH4` have the same shape.

1. **Analysis of SeF4**:
- Selenium (Se) has 6 valence electrons. In SeF4, there are 4 bond pairs (Se-F) and 1 lone pair.
- The steric number is 4 (bond pairs) + 1 (lone pair) = 5, leading to sp³d hybridization. The molecular geometry is trigonal bipyramidal, but the presence of a lone pair results in a seesaw shape.

2. **Analysis of CH4**:
- Carbon (C) has 4 valence electrons and forms 4 C-H bonds with no lone pairs.
- The steric number is 4, leading to sp³ hybridization, and the molecular geometry is tetrahedral.

**Conclusion**: Statement (iv) is false since SeF4 has a seesaw shape, while CH4 has a tetrahedral shape.

### Final Conclusion:
The true statements are (i), (ii), and (iii). Therefore, the correct answer is that statements (i), (ii), and (iii) are true.

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