How many right angles are there in a ${XeF}_{5}^{+}$ $"XeF"_5^"+"$ ion?

Question

15896 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-16T23:13:42

There are eight right angles in ${XeF}_{5}^{+}$ $"XeF"_5^+$ .

You must draw the Lewis structure of ${XeF}_{5}^{+}$ $"XeF"_5^+$ .

$Electrons available = 1Xe + 5F - 1 = 8 + 35 - 1 = 42$ $"Electrons available = 1Xe + 5F - 1 = 8 + 35 - 1 = 42"$

Five $Xe-F$ $"Xe-F"$ bonds account for 40 valence electrons, but you actually have 42 electrons available.

Your Lewis structure will have 5 $Xe-F$ $"Xe-F"$ bonds and a lone pair on the $Xe$ $"Xe"$ .

XeF5+ XeF5+
(Adapted from SlideShare)

This is an ${AX}_{5} E$ $"AX"_5"E"$ molecule.

The electron geometry is octahedral

dublinsciotochemistry.wikispaces.com

The molecular geometry is square pyramidal.

Square pyramidal
(Adapted from Ask the TA)

Now, we count the right angles.

Using the numbers of the atoms, we get the angles:

That makes eight right angles.

How many right angles are there in a XeF+5"XeF"_5^"+" ion?