Drago's Rule and Applications of Drago's Rule

Drago's Rule and Applications of Drago's Rule

Drago's Rule and Applications of Drago's Rule

Drago's Rule

If the central atom of a molecule belongs to the 15th or 16th group and is in the 3rd period or below, and is bonded to a atoms having electronegativity less than or equal to 2.5, then the lone pair present in the s-orbital doesn't participate in hybridisation and bonds are formed by pure p-orbitals. As pure p-orbitals are at 90° to each other, the angle between bond pairs is close to 90°. A lone pair of s-orbitals is found in a stereo chemically inactive s-orbital.

Conditions for Drago's Rule

Molecules satisfying the following conditions fall under the class of Drago's molecules.
1. The central atom should be large. That is, from the third period or beyond.
2. The peripheral atoms should be small and less electronegative (EN ≤ 2.5).
3. There should be a lone pair present on central atom. Lone pairs are usually present in s-orbitals and do not participate in hybridisation.

Examples of Dragos molecules

PH3, AsH3, SbH3, H2S, H2Se, H2Te etc. are the exampples of Drago's molecules

Applications of Drago's Rule

Drago's rule has the following applications-

Basicity

The Dragos Rule can be used to predict the basicity of molecules. NH3 is more basic than PH3 because the lone pair of nitrogen atoms are present in their hybrid orbitals, whereas those of phosphorus are in unhybridised orbitals. Nitrogen is sp3 hybridised. Hence, these electrons are involved in hybridisation. In PH3, the lone pair is stereo-chemically inactive and not involved in hybridisation. Dragos Rule and Applications of Dragos Rule | NH3 is more basic than PH3

Reducing Character

Drago's molecules act as reducing agents. The size of the atom increases down a group. Hence, the bond strength decrease because of the poor overlapping of atomic orbitals.
SbH3 > AsH3 > PH3 > NH3

Percentage s-Character

The Drago's molecule has no hybridisation, hence the percentage of s-characters is zero. Also, it is found that these molecules have a very small bond angle. The bond angle has an inverse relationship with the percentage s-character. Some other relations of percentage s-character are-
% s-character ∝ Bond angle
% s-character ∝ Bond strength

Which of the following statements regarding Drago's rule with respect to MH3 and MH2 (where M belongs to group 15 and group 16 elements respectively except N and O ) are correct?

A. sp3 Hybridisation is not possible in these compounds.
B. s-orbital containing lone pair is stereochemically inert.
C. In these compounds. bond angle is approximately 90∘.
D. All of the above    


What is the percentage p-character in lone pair of PH3 having bond angle 94° ?

A. 82
B. 18
C. 75
D. 25

Answer


Hints: We know that cosθ = s/(s − 1)
where s is the fraction of s-character and θ is the bond angle
cos94° = s/(s − 1)
−0.069 = s/(s − 1)
s = 0.06
% of s in lone pair = 100 − 3x(0.06) = 82%
% of p in lone pair = 100 − 82 = 18%


What is the percentage of s-character in the orbital occupied by lone pair of NH3? It is given that : cos107° = −0.292

A. 22.6
B. 32.2
C. 25
D. 75

Answer


Hints: cos107° = s/(s − 1)
−0.292 = s/(s − 1)
s = 0.226 (in bond pair)
s in lone pair = 1 − 3x(0.226) = 0.322
% of s-character= 32.2%


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