Question

what is the molecular geometry of icl₃?
trigonal bipyramidal
linear
t - shaped
trigonal planar
square pyramidal
square planar
trigonal pyramidal
bent
tetrahedral
octahedral
seesaw

Explanation:

• First, find the number of valence electrons. Iodine (\(I\)) has \(7\) valence electrons and each chlorine (\(Cl\)) has \(7\) valence electrons. So for \(ICl_3\), the total number of valence electrons is \(7+(3\times7)=28\).
• Then, determine the electron - pair geometry. Using the formula for the number of electron - pairs \(n=\frac{\text{valence electrons}}{2}\) (for simple covalent molecules), \(n = 14\) (but a better way is to use the VSEPR formula \(AX_nE_m\), where \(A\) is the central atom, \(X\) is the surrounding atom, \(E\) is the lone pair. For \(ICl_3\), \(I\) is the central atom. The number of bonding pairs \(n = 3\) (since there are three \(I - Cl\) bonds). The number of lone pairs on \(I\): \(I\) has \(7\) valence electrons, \(3\) are used in bonding, so the number of lone pairs \(m=\frac{7 - 3}{2}=2\).
• According to VSEPR theory, for \(AX_3E_2\) (where \(A = I\), \(X=Cl\)), the electron - pair geometry is trigonal bipyramidal (because the sum of bonding and lone pairs is \(3 + 2=5\)). But when considering the molecular geometry (ignoring lone pairs in the sense of the shape description, but considering their repulsion effects on the bonding pairs), the two lone pairs occupy the equatorial positions (to minimize repulsion, since lone - pair - lone - pair repulsion\(>\)lone - pair - bond - pair repulsion\(>\)bond - pair - bond - pair repulsion). So the molecular geometry of \(ICl_3\) is T - shaped.

Answer:

T - shaped