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Drago's Rule | Chemical Bonding | Inorganic Chemistry for Jee Advance, AIIMS, NEET and BITSAT Video

Drago’s rule is an empirical rule and is used to explain the bond angles of hydrides of group 15 and 16 of or below 3rd period. So essentially it is only applicable for PH3, AsH3,SbH3 and H2S, H2Se, H2Te.

According to drago,s rule when the following conditions are satisfied, then the energy difference between the participating atomic orbitals will be very high and thus no mixing of orbitals or hybridization takes place.

1. Atleast one lone pair must be present on the central atom.
2. The central atom must be of or below 3rd period.
3. The electronegativity of the surrounding atom must be less than equal to 2.5.
No hybridisation means bonding will take place through pure atomic p orbitals in PH3 and thus bond angle will be approximately 90.
Explaination:-
For any atomic system with more than one electron, quantum chemistry predicts the energies of the s and p subshells of a shell to differ — while for hydrogen-like systems (one-electron systems) all subshells of a given shell have the same energy. The s subshell of a certain shell in multi-electron systems has a lower energy than the p subshell which in turn has a (far) lower energy than the d subshell and so on and so forth. Since s subshell is determined to be more stable by this method, it is not surprising that the first two electrons of a given shell are added to said s subshell; the next six electrons are added to p subshells whereafter a new shell is used.
This immediately explains that it is most stable for any atom to treat the s electrons of its valence shell as ‘extended core electrons’; i.e. we can predict any configuration that does not touch the s electrons to be energetically favourable with respect to any configuration that does a priori. (You can consider that part of the quoted statement to be correct.) Therefore, the a priori predicted most stable configurations of certain coordination numbers are:
1. mono-, di- and tri-coordination: use 1, 2 or 3 p orbitals to form bonds; keep one lone pair in an s type orbital. (Predicted bond angle: 90∘
2. tetracoordination (no additional lone pairs): since the former is not possible, hybridise s and p orbitals to form sp3sp3 hybrid orbitals. (Predicted bond angle: 109.5∘109.5∘.) Actually, in MO terms you should consider the four ligands to supply four group orbitals whose symmetry always fits one of the three central atom’s orbitals, resulting in one lower bond energy and one higher bond energy corresponding to triple-degenerate bonds. The sp3 description is mathematically equivalent, though.
3. penta- and hexacoordination; tetracoordination with additioal lone pairs and related: attempt to form as many normal bonds with p orbitals as possible; keep one lone pair in an s orbital if possible. Use remaining p orbitals to construct four-electron-3-centre bonds to the remaining atoms. (Predicted bond angles: diverse. 90∘ going from 2e2c bonds to 4e3c bonds; 180∘ between a pair of coordinating atoms contributing to the same 4e3c bond.)

The theories I have outlined so far do not adequately predict the bond angles of water, ammonia and related compounds. Indeed, these compounds are remarkably complicated although they seem simple. The gist, however, is that a bond angle of 90∘, while being most favourable on a pure orbital basis, introduces steric strain between the outer atoms which approach each other too closely if the central atom is relatively small. To combat this destabilisation, the bond angle is extended by mixing s-contribution into the bonding orbitals — which in turn means that the lone pair will be less s-like. The ideal bond angle for dicoordinated systems based on sterics is 180∘, showing that the electronic contribution (90∘) is more important than the steric one (180∘). Larger central atoms than oxygen and nitrogen, such as phosphorus, antimony, selenium, etc., allow for smaller bond angles since the bond lengths are larger increasing the spacing between the ligands.
The reason for the special behaviour of nitrogen and oxygen is not their high electronegativity but their small size. Chlorine, which has an electronegativity similar to nitrogen, behaves as predicted for large atoms.

We hope you will enjoy this lecture.
Thanks
Team IITian Explains.

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Please watch: "Tricks for dpi - ppi Bonding | Explained by IITian | Jee Mains, Advance, BITSAT, NEET & AIIMS"
https://www.youtube.com/watch?v=ca1GZPqvcw8
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