The correct order of hybridisation of the central atom in the following species.
`NH_(3),XeO_(2)F_(2),SeF_(4),NO_(2)^(+)`

Options

A. `sp^(3),sp^(3),sp^(3)d,sp`
B. `sp^(3),sp^(3)d,sp^(3)d,sp`
C. `sp^(3),sp^(3)d^(2),sp^(3)d,sp^(2)`
D. `sp^(2),sp^(3)d,sp^(3)d^(2),sp`

Answer

To determine the correct order of hybridization of the central atom in the given species (NH3, XeO2F2, SeF4, NO2+), we will follow these steps:

### Step 1: Determine the hybridization of NH3
1. **Identify the central atom**: Nitrogen (N).
2. **Count the valence electrons**: Nitrogen has 5 valence electrons.
3. **Draw the Lewis structure**: NH3 has three hydrogen atoms bonded to nitrogen and one lone pair on nitrogen.
4. **Calculate the steric number**:
- Number of sigma bonds = 3 (N-H bonds)
- Number of lone pairs = 1
- Steric number = 3 (sigma bonds) + 1 (lone pair) = 4.
5. **Determine hybridization**:
- A steric number of 4 corresponds to sp³ hybridization.

### Step 2: Determine the hybridization of XeO2F2
1. **Identify the central atom**: Xenon (Xe).
2. **Count the valence electrons**: Xenon has 8 valence electrons.
3. **Draw the Lewis structure**: XeO2F2 has two oxygen and two fluorine atoms bonded to xenon, with one lone pair.
4. **Calculate the steric number**:
- Number of sigma bonds = 4 (2 O and 2 F)
- Number of lone pairs = 1
- Steric number = 4 (sigma bonds) + 1 (lone pair) = 5.
5. **Determine hybridization**:
- A steric number of 5 corresponds to sp³d hybridization.

### Step 3: Determine the hybridization of SeF4
1. **Identify the central atom**: Selenium (Se).
2. **Count the valence electrons**: Selenium has 6 valence electrons.
3. **Draw the Lewis structure**: SeF4 has four fluorine atoms bonded to selenium, with one lone pair.
4. **Calculate the steric number**:
- Number of sigma bonds = 4 (4 F)
- Number of lone pairs = 1
- Steric number = 4 (sigma bonds) + 1 (lone pair) = 5.
5. **Determine hybridization**:
- A steric number of 5 corresponds to sp³d hybridization.

### Step 4: Determine the hybridization of NO2+
1. **Identify the central atom**: Nitrogen (N).
2. **Count the valence electrons**: Nitrogen has 5 valence electrons, but with a +1 charge, it effectively has 4 electrons.
3. **Draw the Lewis structure**: NO2+ has one nitrogen atom bonded to two oxygen atoms (one double bond and one single bond).
4. **Calculate the steric number**:
- Number of sigma bonds = 1 (N-O single bond) + 1 (N=O double bond) = 2
- Number of lone pairs = 0
- Steric number = 2 (sigma bonds) + 0 (lone pairs) = 2.
5. **Determine hybridization**:
- A steric number of 2 corresponds to sp hybridization.

### Final Order of Hybridization
Now we can summarize the hybridizations:
- NH3: sp³
- XeO2F2: sp³d
- SeF4: sp³d
- NO2+: sp

Thus, the correct order of hybridization is:
**sp³, sp³d, sp³d, sp**