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 \( N(SiH_3)_3 \), we will follow these steps:

### Step 1: Identify the central atom and its valence electrons
- The central atom in \( N(SiH_3)_3 \) is nitrogen (N).
- Nitrogen has an atomic number of 7, which gives it the electronic configuration of \( 1s^2, 2s^2, 2p^3 \). Thus, nitrogen has 5 valence electrons (2 from \( 2s \) and 3 from \( 2p \)).

**Hint:** Look at the periodic table to find the atomic number and valence electrons of nitrogen.

### Step 2: Identify the surrounding atoms and their valence electrons
- The surrounding atoms are three silicon (Si) atoms.
- Silicon has an atomic number of 14, with the electronic configuration \( 1s^2, 2s^2, 2p^6, 3s^2, 3p^2 \). This means silicon has 4 valence electrons (2 from \( 3s \) and 2 from \( 3p \)).

**Hint:** Remember that the number of valence electrons corresponds to the group number in the periodic table.

### Step 3: Determine the number of bonds and lone pairs
- Each silicon atom forms a single bond with nitrogen, resulting in three \( Si-H \) bonds and one lone pair on nitrogen.
- Therefore, the total number of sigma bonds is 3 (from \( SiH_3 \)) and 1 lone pair on nitrogen.

**Hint:** Count the number of bonds formed by the central atom and the lone pairs to find the steric number.

### Step 4: Calculate the steric number
- The steric number is calculated as the number of sigma bonds plus the number of lone pairs:
\[ \text{Steric Number} = \text{Number of Sigma Bonds} + \text{Number of Lone Pairs} = 3 + 1 = 4 \]

**Hint:** Use the formula for steric number to determine the hybridization.

### Step 5: Determine the hybridization
- A steric number of 4 corresponds to \( sp^3 \) hybridization.

**Hint:** Recall that \( sp^3 \) hybridization results in a tetrahedral arrangement of electron pairs.

### Step 6: Determine the molecular geometry
- With \( sp^3 \) hybridization and one lone pair, the molecular geometry is trigonal pyramidal.

**Hint:** Visualize the arrangement of bonds and lone pairs to determine the shape of the molecule.

### Step 7: Identify any back-bonding
- In \( N(SiH_3)_3 \), back-bonding occurs between the lone pair on nitrogen and the empty d-orbitals of silicon, leading to \( p\pi - d\pi \) back-bonding.

**Hint:** Consider how lone pairs can interact with empty orbitals of adjacent atoms.

### Conclusion
The structure of \( N(SiH_3)_3 \) is trigonal pyramidal with \( sp^3 \) hybridization.

**Final Answer:** The structure and hybridization of \( N(SiH_3)_3 \) is trigonal pyramidal and \( sp^3 \) respectively.