Question

draw the lewis structure of icl₄⁻ and then determine its electron domain and molecular geometries.
answer:
click to draw a new structure
a trigonal planar / pentagonal
b octahedral / square planar

Explanation:

Step1: Determine the central atom and valence electrons

The central atom is \(I\). \(I\) has \(7\) valence electrons, and each \(Cl\) has \(7\) valence electrons. The ion \(ICl_{4}^{-}\) has a charge of \(- 1\). So the total number of valence electrons is \(7+(4\times7)+1 = 36\)

Step2: Draw the Lewis structure

\(I\) forms single bonds with \(4\) \(Cl\) atoms. Each single bond uses \(2\) electrons. So \(4\times2 = 8\) electrons are used in bonding. The remaining \(36 - 8=28\) electrons are distributed as lone pairs. Each \(Cl\) atom gets \(6\) lone - pair electrons (\(4\times6 = 24\) electrons), and \(I\) gets \(2\) lone - pair electrons.

Step3: Determine electron - domain geometry

The formula for electron - domain geometry is based on \(AX_{n}E_{m}\), where \(A\) is the central atom, \(X\) is the surrounding atom, and \(E\) is the lone pair. For \(ICl_{4}^{-}\), \(n = 4\) (number of \(Cl\) atoms) and \(m = 2\) (number of lone pairs on \(I\)). The total number of electron domains is \(n + m=4 + 2=6\). The electron - domain geometry for \(6\) electron domains is octahedral.

Step4: Determine molecular geometry

When there are \(4\) bonding pairs and \(2\) lone pairs (\(AX_{4}E_{2}\)), the molecular geometry is square planar. The lone pairs are opposite each other to minimize repulsion.

Answer:

B. octahedral / square planar