6
$\begingroup$

In Concise Inorganic Chemistry by J.D.Lee (Adapted by Sudarsan Guha, Fourth Edition), on page 75, under the topic "Effect of Electronegativity - When the surrounding atom is same with different central atom having lone pair" it is given:

The effect of electronegativity as postulated in VSEPR theory explains the order of the angle for the above molecules (Hydrides of groups 15 and 16) but cannot rationalize very small angles (~$90^\circ$) in $\ce{PH3,AsH3,SbH3}$ and $\ce{H2S,H2Se,H2Te}$ with respect to $\ce{NH3}$ and $\ce{H2O}$ respectively.

To explain this, Drago suggested an empirical rule$^{[1]}$ which is compatible with the energitics of the hybridisation. It states that if the central atom is in the third row or below in the periodic table, the lone pair will occupy a stereochemically inactive s orbital$^{[2]}$, and the bonding will be through p orbitals, and bond angles will be nearly $90^\circ$ if the electronegativity of the surrounding atom is less than or equal to $2.5$.

The above rule is based upon the relation between hybridisation and bond angle for two or more equivalent s-p hybrid orbitals, where the fraction of s character (S) or fraction of p character (P) is given by the relationship:

$$\cos \theta = \frac{S}{S-1}=\frac{P-1}{P}$$

for $\theta \in (90^\circ,180^\circ)$

The following are links to previous questions asked on Chemistry Stack Exchange regarding concepts in this quoted text.

[1] : Drago's Rule : What is Drago's rule? Does it really exist?

[2] : Stereochemically (in)active s orbital : What is a stereochemically active or inactive s orbital?

This formula,

$$\cos \theta = \frac{S}{S-1}=\frac{P-1}{P}$$

relates the bond angle $\theta$ and the fraction of s/p characters. I tried to plug in some standard values.

For example, On substituting the expression with S=0.25 or P=0.75, I get $\frac{S}{S-1}=-0.33...$ and $\frac{P-1}{P}=-0.33...$. I know they both give the same as in sp$^n$ hybridisation where only s and p orbitals are involved, the sum of fractions of the s character and p character i.e., S+P=1, and I was able to interconvert the expressions $\frac{S}{S-1}$ and $\frac{P-1}{P}$ with simple substitution - S=1-P or P=1-S.

On finding the cosine inverse of -0.33...,

$\cos ^{-1} (-0.33...)=109.471^{\circ}$

which is exactly equal to the tetrahedral angle ($109.471^{\circ}$). From this, we can obtain the bond angle in sp3 hybridisation. And similarly, we can do it for sp, and sp2 hybridisation which I've skipped here.

It is also possible to the reverse with this formula, i.e., given the bond angle we can compute the fraction of s or p character. For example, for $\ce{AsH3}$, the H-As-H angle is $91.8 ^{\circ}$, and from the calculation, it can be shown that each As-H bond consists of almost 97% p character and 3% s character.

My doubt is how did they arrive at the given formula? Is there any derivation* for this or is that simply an experimental result? Will this work for all values of S and P (under the constraint S+P=1)? What is the logic behind this formula?

I am unable to find any useful information about this on the internet.


*Simple or easy to understand derivation which can be understood by a high school student.

$\endgroup$
  • 1
    $\begingroup$ Possible duplicate of Sp5 hybridization in cyclopropane? $\endgroup$ – Mithoron Oct 18 at 23:51
  • 2
    $\begingroup$ @Mithoron, The question you referred was useful but did not clarify my doubt completely. I am asking for the relation between the bond angles and the amount of s/p character, whereas that answer/question is about s/p character itself and doesn't derive the formula, just uses it. $\endgroup$ – Intellex Oct 19 at 3:44
  • 1
    $\begingroup$ The full derivation should be available on Wikipedia: Bent's rule and is known as Coulson's theorem. $\endgroup$ – Martin - マーチン Oct 29 at 10:02
  • 1
    $\begingroup$ I currently have little time, but I will try to get to it. I remember that ron's answers using the theorem always confused me, too. The theorem itself can be applied to all compounds, but it obviously only makes sense for those that can be described with spⁿ hybridisation. Also, since it is derived from Bent's rule, it is also an observation, and might lead to wrong conclusions if used outside its scope. $\endgroup$ – Martin - マーチン Oct 30 at 10:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.