Question

In which of the following molecular species axial bond length is shorter than equatorial bond length?
A. $P{F_2}C{l_3}$
B. $P{F_5}$
C. $Xe{O_3}{F_2}$
D. $I{O_2}{F_2}^ -$

Answer 1

The axial and equatorial are the two positions available in trigonal bipyramidal geometry. The less electronegative atoms in a molecule favor the equatorial positions.

Complete step by step answer:
Several compounds are known to possess trigonal bipyramidal geometry. In trigonal bipyramidal geometry consists of three positions which are in the same plane called equatorial positions and two positions one above and one below the plane called the axial position.
Of these compounds the atoms attached to the central atoms can attain either of the two positions available in trigonal bipyramidal geometry. The preference of the position depends on the electronegativity of the bonded atoms to the central atom.
The more electronegative atoms favor the axial position and the less electronegative atom favors the equatorial position. So let us consider the given molecules and determine the structure and respective bond lengths one by one.
A. $P{F_2}C{l_3}$. The phosphorous central atom is attached to two different atoms of which one is fluorine and the other is chlorine atom. Two fluorine atoms and three chlorine atoms are positioned in the five available positions of trigonal bipyramidal geometry. Out of fluorine and chlorine, fluorine is more electronegative and will be strongly attracted to the central atom than the chlorine atom. The fluorine is positioned in the axial positions and hence the axial bond length is shorter than equatorial bond length in $P{F_2}C{l_3}$.
B. $P{F_5}$. In this molecule all the five fluorine atoms are attached to phosphorus elements. So the electronegativity difference concept will not apply. The axially bonded two fluorine atoms will experience greater repulsion from the equatorially bonded fluorine atoms. So in this molecule axial bond length is longer than equatorial bond length.
C. $Xe{O_3}{F_2}$. In this molecule xenon is the central atom which is attached to three oxygen atoms by double bonds and two fluorine atoms by single bonds. The more electronegative atom fluorine favors the axial position and the double bonded oxygen atoms are positioned at the equatorial positions. The repulsion of the equatorial bonds makes the axial bonds longer in this molecule.
D.$I{O_2}{F_2}^ -$ . The central atom is iodine which is attached to two oxygen atoms (one by double and other by single bond) and two fluorine atoms. The oxygen atom occupies the equatorial position and the fluorine atom occupies the axial position. A lone pair of iodine is present at the equatorial position. Hence the axial bond feels more repulsion than the equatorial bonds and is longer.
And hence the correct answer is option A.

Note: The preference for the equatorial position is in order lone pair > double bonds > lower electronegative atom. The higher electronegative atom always is positioned at axial position. In other words atoms with bigger atomic size are positioned in the equatorial plane.