The structure and hybridization of `N(SiH_3)_3` is :

Options

A. bent , sp
B. trigonal planar , `sp^2`
C. octahedral , `sp^3 d^2`
D. tetrahedral , `sp^3`.

Answer

To determine the structure and hybridization of the molecule \( N(SiH_3)_3 \), we can follow these steps:

### Step 1: Identify the central atom and its valence electrons
The central atom in this molecule is nitrogen (N). Nitrogen belongs to group 15 of the periodic table and has 5 valence electrons.

### Step 2: Determine the bonding and lone pairs
In \( N(SiH_3)_3 \), nitrogen forms three sigma bonds with three silicon (Si) atoms from the three \( SiH_3 \) groups. Since nitrogen has 5 valence electrons and it uses 3 of them to form bonds, it will have 2 electrons left, which will form a lone pair.

### Step 3: Calculate the steric number
The steric number is calculated as the number of sigma bonds plus the number of lone pairs. Here, nitrogen has:
- 3 sigma bonds (to Si)
- 1 lone pair

Thus, the steric number is \( 3 + 1 = 4 \).

### Step 4: Determine the hybridization
For a steric number of 4, the hybridization is \( sp^3 \). However, since there is a lone pair, the molecular geometry will be affected.

### Step 5: Identify the molecular geometry
With one lone pair and three bonding pairs, the molecular geometry will be trigonal pyramidal. This is because the lone pair occupies more space and pushes the bonding pairs down.

### Step 6: Consider back bonding
In \( N(SiH_3)_3 \), there is a possibility of back bonding. Nitrogen can donate its lone pair to the empty d-orbitals of silicon, which can lead to a decrease in hybridization.

### Step 7: Adjust the hybridization due to back bonding
Due to back bonding, the effective hybridization of nitrogen can be considered as \( sp^2 \) instead of \( sp^3 \). This means that the nitrogen will have a trigonal planar arrangement around it.

### Final Answer
Thus, the structure of \( N(SiH_3)_3 \) is trigonal pyramidal with \( sp^2 \) hybridization.

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