Questions: Draw the Lewis dot structure for NI₃. Determine the electron geometry of NI₃. trigonal planar tetrahedral linear

Transcript text: Draw the Lewis dot structure for $\mathrm{NI}_{3}$. Determine the electron geometry of $\mathrm{NI}_{3}$. trigonal planar tetrahedral linear

Solution

Solution Steps

Step 1: Determine the Total Number of Valence Electrons

To draw the Lewis dot structure for $\mathrm{NI}_3$, we first need to determine the total number of valence electrons. Nitrogen (N) has 5 valence electrons, and each iodine (I) atom has 7 valence electrons. Therefore, the total number of valence electrons is:

\[ 5 + 3 \times 7 = 26 \]

Step 2: Draw the Basic Structure

Place the nitrogen atom in the center and arrange the three iodine atoms around it. Connect each iodine atom to the nitrogen atom with a single bond. This uses up 6 electrons (3 bonds $\times$ 2 electrons per bond).

Step 3: Distribute Remaining Electrons

Distribute the remaining 20 electrons (26 total - 6 used in bonds) to satisfy the octet rule for each iodine atom. Each iodine atom will have 6 additional electrons (3 lone pairs), using up all 18 electrons. The nitrogen atom will have one lone pair.

Step 4: Determine the Electron Geometry

The electron geometry considers both the bonding pairs and lone pairs around the central atom. Nitrogen in $\mathrm{NI}_3$ has three bonding pairs and one lone pair, which corresponds to a tetrahedral electron geometry.