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(PH3). The central atom in phosphine is phosphorous.
i) belongs to group 15 and is in the third row of the periodic table (see the table above).
ii) has an electronegativity value, χ = 2.19 and
iii) in phosphine molecule, it has a lone pair of electrons too.
So, according to Drago’s rule, hybridization does not take place. The atomic orbitals just overlap 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 a 6% s- character in P-H bonds. So, the 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 a pure s- orbital 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 electrons 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 bonds and so can form coordination compounds.
e.g.- 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.
All the questions which we considered in the last post, can be answered by Drago’s rule. The simple explanation to those questions is, the elements of group 13-16 and 3rd- 7th period do not have a lone pair in any hybridized orbital. This lone pair is thus unavailable.
As opposed to this, hybridization takes place in the elements of the 2nd period. Thus, the lone pairs occupy the hybrid orbitals, which have less %s character. Naturally, they are less tightly bound to the nucleus and 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…
References and Further Reading –
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