59.CHEMICAL BONDING (6)- Covalent Bonding(5)- VESPR model(3).

 

Lone pairs – AXIAL OR EQUATORIAL POSITION?

In the structures mentioned in earlier post, we have two positions, namely – Axial and equatorial.

Equatorial positions lie on the equatorial plane – the plane with maximum number of atoms, which lies in the equatorial ( like equator , in the middle) position.

Axial positions are perpendicular to the equatorial plane – one above one below the plane.

592
Trigonal bi pyramidal
593
Octahedral

 

The following figure shows a trigonal bi pyramidal structure , where 4 atoms are in the equatorial plane ( 3 equatorial positions and one central atom) and 2 atoms in axial positions.

 


590

Now the question is , which positions will the lone pairs occupy in a molecule ? There are 3 possibilities as shown below –

594

 

(1) –  Both lone pairs occupy axial positions – In this possibility, the BPs occupy the equatorial positions and LPs occupy axial position. The angle between two LPs  is 180º , so they are far apart from each other and the LP- LP repulsion is insignificant (repulsions above 90º are insignificant). However, in this position there are 6 LP – BP repulsions .All three BP electrons experience repulsion from both lone pairs as they are only 90º (perpendicular) apart from the bonding pair of electrons. Thus, this possibility is not favored.

(2) – One lone pair occupies an axial position and the other one occupies equatorial position.In this case, two lone pairs are 90º apart and so the high intensity LP-LP repulsion is maximum.As molecules tend to avoid maximum repulsions, this case is not favored too.

(3) – Two LPs occupy the equatorial positions and thus they are 120º apart (They form vertices of an equilateral triangle .An equilateral triangle has 120º angle). Also, there are only 4 LP-BP repulsions. So, this case is favored and thus, LONE PAIRS PREFERABLY OCCUPY EQUATORIAL POSITIONS.

EXAMPLES 

A] HCN.

Structure:  595
Central atom:

Carbon, C (6) [He] 2s2 2p2

Valence electrons on central atom:

4

H  contributes 1 electron , N contributes 3 electrons 3+1 = 4. However, we consider triple bond as single bond so we subtract 4 electrons(2 bonds have 4 electron pairs).
∴8-4 = 4

TOTAL:

 4

Divide by 2 to give electron pairs

4/2 = 2

VSEPR geometry(AX2E0)

Linear

Bond angle ( found out experimentally)

180º

596

B] PO43-

Structure:  597
Central atom: Phosphorous, P(15) 1s2 2s2 2p6 3s2 3p3
Valence electrons on central atom:

5

4 O  contributes 2 electrons each , 3 electrons of (-3) charge 5+8+3 = 16. However, we consider a double bond as a single bond so we subtract 8 electrons(4 double bonds have 8 electron pairs).
∴16- 8 = 8

TOTAL:

8

Divide by 2 to give electron pairs

8/2 = 4

VSEPR geometry(AX4E0)

Tetrahedral

Bond angle ( found out experimentally)

109.5º

C] XeOF4

Structure  598.jpg
Central atom: Xenon , Xe (54) [Kr] 4d10 5s2 5p6
Valence electrons on central atom:

8

4F  contribute 4 electrons , O contributes 2 electrons, and 2 lone pair electrons on Xe atom 8+4+2-2 = 12. ( But a double bond is considered as single bond, so subtract 2 electrons.Only LP on central atom is considered)

TOTAL:

12

Divide by 2 to give electron pairs

12/2 = 6

VSEPR geometry(AX2E0)

Square pyramidal ( not octahedral due to the presence of LP on Xe)

Bond angle ( found out experimentally)

 ≈ 90º

Although VSEPR model is great at predicting geometries for most molecules, it fails for many other molecules like the transition metal compounds. So, in our next post we start discussing another very important theory , which is developed on a very different approach. Till then ,

Be a perpetual student of life and keep learning…

Good Day !

References and Further Reading –

  1. https://chem.libretexts.org/Core/Inorganic_Chemistry/Molecular_Geometry/Limitations_of_VSEPR
  2. http://nptel.ac.in/courses/104103069/14

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