The shape of SF4 and XeF2 respectively are:

To determine the shape of SF4 and XeF2, we need to consider their molecular geometries.

1. SF4:

• Start by drawing the Lewis structure of SF4, which shows that sulfur (S) is bonded to four fluorine (F) atoms.
• Count the total number of electron groups around the central atom (S). In this case, there are four electron groups (four bonds to F atoms).
• Determine the electron group arrangement. Since there are four electron groups, the arrangement is tetrahedral.
• Next, consider the lone pairs of electrons on the central atom. In SF4, there is one lone pair on S.
• Determine the molecular geometry by considering the lone pairs. The presence of one lone pair causes the molecular geometry to be trigonal bipyramidal.
• However, the lone pair occupies more space than the bonding pairs, resulting in a distorted shape. Therefore, the shape of SF4 is seesaw or distorted tetrahedral.

2. XeF2:

• Draw the Lewis structure of XeF2, which shows that xenon (Xe) is bonded to two fluorine (F) atoms.
• Count the total number of electron groups around the central atom (Xe). In this case, there are two electron groups (two bonds to F atoms).
• Determine the electron group arrangement. Since there are two electron groups, the arrangement is linear.
• Since there are no lone pairs on the central atom, the molecular geometry is the same as the electron group arrangement, which is linear.
• Therefore, the shape of XeF2 is linear.

In summary, the shape of SF4 is seesaw or distorted tetrahedral, while the shape of XeF2 is linear.