Which one of the following is a correct set?

Options

A. `H_(2)O, sp^(3)` angular
B. `BCl_(3), sp^(3)`, angular
C. `NH_(4)^(+), dsp^(2)`, square planar
D. `CH_(4), dsp^(2)`, tetrahedral

Explanation

To determine which of the given sets is correct, we need to analyze the molecular geometry and hybridization of each compound mentioned in the question. Let's break down the analysis step by step.

### Step 1: Analyze H2O (Water)
1. **Valence Electrons**: Oxygen has 6 valence electrons, and each hydrogen has 1 valence electron. Therefore, total valence electrons = 6 (O) + 1 (H) + 1 (H) = 8.
2. **Bonding**: Oxygen forms 2 single bonds with 2 hydrogen atoms, using 2 of its 6 valence electrons.
3. **Lone Pairs**: After forming 2 bonds, oxygen has 4 valence electrons left, which form 2 lone pairs.
4. **Hybridization**: The steric number (number of bond pairs + number of lone pairs) is 4 (2 bond pairs + 2 lone pairs), leading to sp³ hybridization.
5. **Shape**: With 2 lone pairs, the shape is angular (or bent).

### Step 2: Analyze BCl3 (Boron Trichloride)
1. **Valence Electrons**: Boron has 3 valence electrons, and each chlorine has 7 valence electrons. Therefore, total valence electrons = 3 (B) + 7 (Cl) + 7 (Cl) + 7 (Cl) = 24.
2. **Bonding**: Boron forms 3 single bonds with 3 chlorine atoms, using all 3 of its valence electrons.
3. **Lone Pairs**: Boron has no lone pairs left.
4. **Hybridization**: The steric number is 3 (3 bond pairs + 0 lone pairs), leading to sp² hybridization.
5. **Shape**: The shape is trigonal planar.

### Step 3: Analyze NH4⁺ (Ammonium Ion)
1. **Valence Electrons**: Nitrogen has 5 valence electrons, and each hydrogen has 1 valence electron. The positive charge means nitrogen effectively has 4 valence electrons. Therefore, total valence electrons = 4 (N) + 1 (H) + 1 (H) + 1 (H) + 1 (H) = 8.
2. **Bonding**: Nitrogen forms 4 single bonds with 4 hydrogen atoms, using all 4 of its valence electrons.
3. **Lone Pairs**: There are no lone pairs left.
4. **Hybridization**: The steric number is 4 (4 bond pairs + 0 lone pairs), leading to sp³ hybridization.
5. **Shape**: The shape is tetrahedral.

### Step 4: Analyze CH4 (Methane)
1. **Valence Electrons**: Carbon has 4 valence electrons, and each hydrogen has 1 valence electron. Therefore, total valence electrons = 4 (C) + 1 (H) + 1 (H) + 1 (H) + 1 (H) = 8.
2. **Bonding**: Carbon forms 4 single bonds with 4 hydrogen atoms, using all 4 of its valence electrons.
3. **Lone Pairs**: There are no lone pairs left.
4. **Hybridization**: The steric number is 4 (4 bond pairs + 0 lone pairs), leading to sp³ hybridization.
5. **Shape**: The shape is tetrahedral.

### Conclusion
- H2O: sp³ hybridization, angular shape (correct)
- BCl3: sp² hybridization, trigonal planar shape (correct)
- NH4⁺: sp³ hybridization, tetrahedral shape (correct)
- CH4: sp³ hybridization, tetrahedral shape (correct)

The correct set is:
- H2O: sp³, Angular
- BCl3: sp², Trigonal planar
- NH4⁺: sp³, Tetrahedral
- CH4: sp³, Tetrahedral

### Final Answer
The correct set is:
- H2O (sp³, Angular)
- BCl3 (sp², Trigonal planar)
- NH4⁺ (sp³, Tetrahedral)
- CH4 (sp³, Tetrahedral)