Question

Predict the shape of XeF4 molecule according to VSEPR theory

Answer 1

In XeF4, the central atom is Xe.

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The general concept used to find the central atom is that, the atom which has high electronegativity is taken as the central atom. But this example disproved the above statement because here, fluorine is more electronegative than xenon.

So, there is a simple concept or trick for you.

The central atom is that atom which is in the lowest quantity. Here, there is only 1 Xe and 4 F, so according to the above concept, Xe is the central atom.

If at some times, there is a case where there are 2 atoms with the least quantity, then the atom which has more valence electrons is taken as the central atom. For example, in the compound XeOF4, both Xe and O has 1 atoms each. But, Xe has more valence electrons (8) than oxygen (6). So, according to the above concept, Xe is taken as the central atom.

As you are clear with the above concepts for finding or identifying the central atom, let's move on to predict the shape of XeF4.

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The total number of electron pairs around Xe in XeF4 is 6. This can be found out from the equation,

$electron \: pair \: = \frac{1}{2} (v + l + a - c)$

Here, v = valence electrons of the central atom

l = number of ligands ( number of monovalent atoms attached to the central atom )

Monovalent atoms - atoms which can only form single bonds.

eg:- F, Cl, etc

Divalent atoms - atoms which can form double bonds.

eg:- Oxygen

a = anionic charge ( if the compound is an anion )

c = cationic charge ( if the compound is a cation )

But, XeF4 is a neutral species. So, a and c = 0

v = valence electrons of Xe = 8

l = 4

So, electron pairs = 1/2 ( 8 + 4 + 0 + 0)

= 6

But there are only 4 electron pairs which takes part in bonding. So, the 2 electron pairs will be lone pairs ( electron pairs which do not take part in bonding).

So, the geometry of 6 electron pairs is octahedral or square bipyramidal.

But, out of the 6, 2 are lone pairs which will be placed at the axial position of the octahedral structure.

So, the shape of the molecule XeF4 will become square planar.

[THANK YOU]

Answer 2

The shape of the XeF4 molecule according to VSEPR theory is square planar.

• The shape and geometry of the compound are determined by the formula is given:
• $electron \: pair \: =\frac{v + m - c + a}{2}$
• Here, v is valence shell electrons, m is several monovalent atoms, c is the cationic charge and a is an anionic charge.
• So, In XeF4, Xe has 8 valence electrons and 4 monovalent atoms.
• Now,
• $electron \: pair \: = \frac{8 + 4}{2} = \frac{12}{2} = 6$
• It has four bond pairs.
• So, lone pairs = electron pairs - bond pairs
• Lone pairs = 6 - 4 = 2
• XeF4 has 6 electron pairs, 4 bond pairs and 2 lone pairs so, the shape of XeF4 is square planar.

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