THE DRAGO RULE.
Drago rule states that , if –
- the central atom has at least one lone pair of electron on it
- the central atom belongs to group 13,14 ,15 or 16 and is from 3rd to 7th period.
- if electronegativity of central element is 2.5 or less
- no. of sigma bonds+ lone pair=4
then , THERE IS NO NEED TO CONSIDER HYBRIDIZATION OF THAT ELEMENT.
Consider phosphine . Phosphorous is a group 15 element and is in the third row of the periodic table as seen in the above figure.It has electronegativity value, χ = 2.19 and in phosphine molecule it has a lone pair of electron too. So, according to Drago’s rule, hybridization does not take place. There is only overlapping of atomic orbitals to form bonds.
If we closely look at the structure of Phosphine, we would know the above statement to be true.
The bond angle in phosphine is 94º . If we apply the formula from post 69 ,
∴fs= (-0.069) /( – 0.069-1)
∴fs= -0.069 / -1.069
%s character = fs×100 = 0.064 ×100 = 6.4 %
This means that % s character in the P- H bonds is just 6.4% ≈ 6%.
Thus, there is only 6% s- character in P-H bonds. So, s-character in three P-H bonds will be 6 × 3 = 18 % . So, we can conclude that , the lone pair of electrons are in orbitals which have 100 – 18 = 82% s character ! Now, no hybridized orbital ever has this high % of s- character(The maximum %s character is 50% in sp orbital) .This means that the lone pair of electrons are in pure s- orbital of the atom and NOT IN ANY HYBRIDIZED ORBITAL.
The same logic holds true for all the other elements too.Now that we have established that the lone pair is NOT in any hybrid orbital but in a s- orbital , we can explain all the observations we listed in the earlier post.
As the lone pair is in a s- orbital, it is closer to the nucleus and so very tightly bound to it.Thus, this lone pair of electron is NOT available for sharing or donating. So, it takes concentrated acid to react with phosphine. Also, as these lone pairs cannot be shared , these compounds do not form coordination compounds. Ammonia on the other hand can lend its lone pair to form coordinate covalent bond and so can form co ordination compounds.
Ammonia dissolves in water as it forms hydrogen bonds with water molecules. This is possible as the lone pair of electrons are available on nitrogen.
So all the questions which we considerd in the last post can be answered by Drago’s rule. The answer to all those questions is that, as the elements of group 13-16 and 3rd- 7th period donot have a lone pair in hybridized orbital, the lone pair is unavailabe .In elements of 2nd period though, hybridization takes place and so the lone pairs are in hybrid orbitals , which have less %s character and thus, these lone pairs are available for sharing or bonding.
In our next post we shall try to solve some problems related to hybridization theory. Till then,
Be a perpetual student of life and keep learning…
Happy Diwali !
References and Further Reading –
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