Short Answer

The structure on the left is "preferred" because the structure on the right cannot exist. You cannot put 4 electrons in a p-orbital.

Detailed Explanation

In $\ce{ClF3}$ the central chlorine atom is roughly $\ce{sp^2}$ hybridized. This means that we will have 3 $\ce{sp^2}$ orbitals emanating from the chlorine; they will form an equatorial plane and will either contain a lone pair of electrons or bond to a ligand. Further in the case of $\ce{sp^2}$ hybridization, we also have 1 unhybridized p-orbital perpendicular to the equatorial plane.

The drawing on the right has two lone pairs placed in one p-orbital. A p-orbital cannot contain 4 electrons, therefore the drawing on the right does not represent a valid chemical structure.

The question really becomes, does $\ce{ClF3}$ prefer to exist with

• 2 axial fluorines and 1 equatorial fluorine

or

• 1 axial fluorine and 2 equatorial fluorines?

We can use Bent's rule to answer this question. Bent's rule tells us that more p character will be placed in those orbitals directed towards more electronegative substituents (or conversely, more s character will placed in those orbitals directed towards more electropositive substituents). This is because orbitals with more p-character are higher in energy than orbitals with more s-character (because a p-orbital is higher in energy than an s-orbital). An electronegative substituent will draw electron density away from the central atom. Consequently there is less electron density to stabilize so the orbital will rehybridize such that it contains more p-character, saving the s-character to stabilize another orbital that contains more electron density.

Since fluorine is very electronegative and an electron pair is very electropositive, applying Bent's rule we see that the two electronegative fluorines will prefer axial positions where the orbital is rich in p-character (in fact, it is roughly 100% p) and the two lone pairs will prefer to occupy two of the s-rich $\ce{sp^2}$ equatorial orbitals. Therefore, your structure on the left with 2 axial fluorines and 1 equatorial fluorine correctly depicts the preferred arrangement of fluorines and lone pairs in $\ce{ClF3}$.

A final note on the bonding in $\ce{ClF3}$, the two axial fluorines bond with the chlorine p-orbital to form what is referred to as a hypercoordinated or hypervalent or a 3-center, 4-electron bond (3-center because there are 3 atoms $\ce{F-Cl-F}$ involved in the bond and the bond contains 4 electrons, 2 from chlorine and 1 from each of the two fluorines). Hypercoordinated bonding is common in situations where the central atom appears to have more than an octet of electrons. See these earlier answers for more detail on hypercoordinate bonding.

• Why does F replace an axial bond in $\ce{PCl5}$?
• How can Iodine bond with 5 Fluorines in Iodine Pentafluoride?
• Hypervalency and the octet rule

Short Answer

The structure on the left is "preferred" because the structure on the right cannot exist. You cannot put 4 electrons in a p-orbital.

Detailed Explanation

In $\ce{ClF3}$ the central chlorine atom is roughly $\ce{sp^2}$ hybridized. This means that we will have 3 $\ce{sp^2}$ orbitals emanating from the chlorine; they will form an equatorial plane and will either contain a lone pair of electrons or bond to a ligand. Further in the case of $\ce{sp^2}$ hybridization, we also have 1 unhybridized p-orbital perpendicular to the equatorial plane.

The drawing on the right has two lone pairs placed in one p-orbital. A p-orbital cannot contain 4 electrons, therefore the drawing on the right does not represent a valid chemical structure.

The question really becomes, does $\ce{ClF3}$ prefer to exist with

• 2 axial fluorines and 1 equatorial fluorine

or

• 1 axial fluorine and 2 equatorial fluorines?

We can use Bent's rule to answer this question. Bent's rule tells us that more p character will be placed in those orbitals directed towards more electronegative substituents (or conversely, more s character will placed in those orbitals directed towards more electropositive substituents). This is because orbitals with more p-character are higher in energy than orbitals with more s-character (because a p-orbital is higher in energy than an s-orbital). An electronegative substituent will draw electron density away from the central atom. Consequently there is less electron density to stabilize so the orbital will rehybridize such that it contains more p-character, saving the s-character to stabilize another orbital that contains more electron density.

Since fluorine is very electronegative and an electron pair is very electropositive, applying Bent's rule we see that the two electronegative fluorines will prefer axial positions where the orbital is rich in p-character (in fact, it is roughly 100% p) and the two lone pairs will prefer to occupy two of the s-rich $\ce{sp^2}$ equatorial orbitals. Therefore, your structure on the left with 2 axial fluorines and 1 equatorial fluorine correctly depicts the preferred arrangement of fluorines and lone pairs in $\ce{ClF3}$.

A final note on the bonding in $\ce{ClF3}$, the two axial fluorines bond with the chlorine p-orbital to form what is referred to as a hypercoordinated or hypervalent or a 3-center, 4-electron bond (3-center because there are 3 atoms $\ce{F-Cl-F}$ involved in the bond and the bond contains 4 electrons, 2 from chlorine and 1 from each of the two fluorines). Hypercoordinated bonding is common in situations where the central atom appears to have more than an octet of electrons. See these earlier answers for more detail on hypercoordinate bonding.

• Why does F replace an axial bond in $\ce{PCl5}$?
• How can Iodine bond with 5 Fluorines in Iodine Pentafluoride?
• Hypervalency and the octet rule

In $\ce{ClF3}$ the central chlorine atom is roughly $\ce{sp^2}$ hybridized. This means that we will have 3 $\ce{sp^2}$ orbitals emanating from the chlorine; they will form an equatorial plane and will either contain a lone pair of electrons or bond to a ligand. Further in the case of $\ce{sp^2}$ hybridization, we also have 1 unhybridized p-orbital perpendicular to the equatorial plane.

In your drawings, E represents a lone pair of electrons. The drawing on the right has two lone pairs placed in one p-orbital. A p-orbital cannot contain 4 electrons, therefore the drawing on the right does not represent a valid chemical structure.

The question really becomes, does $\ce{ClF3}$ exist with

• 2 axial fluorines and 1 equatorial fluorine

or

• 1 axial and 2 equatorial fluorines?

We can use Bent's rule to answer this question. Bent's rule tells us that more p character will be placed in those orbitals directed towards more electronegative substituents (or conversely, more s character will placed in those orbitals directed towards more electropositive substituents). This is because orbitals with more p-character are higher in energy than orbitals with more s-character (because a p-orbital is higher in energy than an s-orbital). An electronegative substituent will draw electron density away from the central atom. Consequently there is less electron density to stabilize so the orbital will rehybridize such that it contains more p-character, saving the s-character to stabilize another orbital that contains more electron density.

Since fluorine is very electronegative and an electron pair is very electropositive, applying Bent's rule we see that the two electronegative fluorines will prefer axial positions where the orbital is rich in p-character (in fact, it is roughly 100% p) and the two lone pairs will prefer to occupy two of the s-rich $\ce{sp^2}$ equatorial orbitals. Therefore, your structure on the left shows the preferred arrangement of ligands in $\ce{ClF3}$.

One final note, the two axial fluorines bonding with the chlorine p-orbital form what is referred to as a hypercoordinated (or hypervalent) bond or a 3-center, 4-electron bond (3-center because there are 3 atoms $\ce{F-Cl-F}$ involved in the bond and the bond contains 4 electrons, 2 from chlorine and 1 from each of the two fluorines). Hypercoordinated bonding is common in situations where the central atom appears to have more than an octet of electrons. See these earlier answers for more detail on hypercoordinate bonding.

• Why does F replace an axial bond in $\ce{PCl5}$?
• How can Iodine bond with 5 Fluorines in Iodine Pentafluoride?
• Hypervalency and the octet rule

Short Answer

The structure on the left is "preferred" because the structure on the right cannot exist. You cannot put 4 electrons in a p-orbital.

Detailed Explanation

In $\ce{ClF3}$ the central chlorine atom is roughly $\ce{sp^2}$ hybridized. This means that we will have 3 $\ce{sp^2}$ orbitals emanating from the chlorine; they will form an equatorial plane and will either contain a lone pair of electrons or bond to a ligand. Further in the case of $\ce{sp^2}$ hybridization, we also have 1 unhybridized p-orbital perpendicular to the equatorial plane.

The drawing on the right has two lone pairs placed in one p-orbital. A p-orbital cannot contain 4 electrons, therefore the drawing on the right does not represent a valid chemical structure.

The question really becomes, does $\ce{ClF3}$ prefer to exist with

• 2 axial fluorines and 1 equatorial fluorine

or

• 1 axial fluorine and 2 equatorial fluorines?

We can use Bent's rule to answer this question. Bent's rule tells us that more p character will be placed in those orbitals directed towards more electronegative substituents (or conversely, more s character will placed in those orbitals directed towards more electropositive substituents). This is because orbitals with more p-character are higher in energy than orbitals with more s-character (because a p-orbital is higher in energy than an s-orbital). An electronegative substituent will draw electron density away from the central atom. Consequently there is less electron density to stabilize so the orbital will rehybridize such that it contains more p-character, saving the s-character to stabilize another orbital that contains more electron density.

Since fluorine is very electronegative and an electron pair is very electropositive, applying Bent's rule we see that the two electronegative fluorines will prefer axial positions where the orbital is rich in p-character (in fact, it is roughly 100% p) and the two lone pairs will prefer to occupy two of the s-rich $\ce{sp^2}$ equatorial orbitals. Therefore, your structure on the left with 2 axial fluorines and 1 equatorial fluorine correctly depicts the preferred arrangement of fluorines and lone pairs in $\ce{ClF3}$.

A final note on the bonding in $\ce{ClF3}$, the two axial fluorines bond with the chlorine p-orbital to form what is referred to as a hypercoordinated or hypervalent or a 3-center, 4-electron bond (3-center because there are 3 atoms $\ce{F-Cl-F}$ involved in the bond and the bond contains 4 electrons, 2 from chlorine and 1 from each of the two fluorines). Hypercoordinated bonding is common in situations where the central atom appears to have more than an octet of electrons. See these earlier answers for more detail on hypercoordinate bonding.

• Why does F replace an axial bond in $\ce{PCl5}$?
• How can Iodine bond with 5 Fluorines in Iodine Pentafluoride?
• Hypervalency and the octet rule