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.

  • 1
    $\begingroup$ Possible duplicate of Sp5 hybridization in cyclopropane? $\endgroup$
    – Mithoron
    Commented Oct 18, 2019 at 23:51
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    $\begingroup$ @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. $\endgroup$
    – Vishnu
    Commented Oct 19, 2019 at 3:44
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    $\begingroup$ The full derivation should be available on Wikipedia: Bent's rule and is known as Coulson's theorem. $\endgroup$ Commented Oct 29, 2019 at 10:02
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    $\begingroup$ 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. $\endgroup$ Commented Oct 30, 2019 at 10:41

1 Answer 1


Just like most other theories Bent's rule also has an formal theory which can be used to derive mathematical equations following from the rule.

Bents rule is based on quantum mechanics and thus derivation of its mathematical form inherently uses quantum mechanical nomenclature and thus might not fit your "understood by high school student" criteria.

But still I am referencing the derivation of the mathematical form of Bent's rule from Wikipedia[1] just in case you need it. Feel free to ignore the quoted text if it's too complex to comprehend.

To construct hybrid $s$ and $p$ orbitals, let the first hybrid orbital be given by $s + \sqrt{λ_i}p_i$, where $p_i$ is directed towards a bonding group and $λ_i$ determines the amount of p character this hybrid orbital has. This is a weighted sum of the wavefunctions. Now choose a second hybrid orbital $s + \sqrt{λ_j}p_j$, where $p_j$ is directed in some way and $λ_j$ is the amount of $p$ character in this second orbital. The value of $λ_j$ and direction of $p_j$ must be determined so that the resulting orbital can be normalized and so that it is orthogonal to the first hybrid orbital. The hybrid can certainly be normalized, as it is the sum of two normalized wavefunctions. Orthogonality must be established so that the two hybrid orbitals can be involved in separate covalent bonds. The inner product of orthogonal orbitals must be zero and computing the inner product of the constructed hybrids gives the following calculation. $$ \begin{align} \langle s + \sqrt{λ_i}p_i | s + \sqrt{λ_j}p_j \rangle &= \langle s |s \rangle + \sqrt{λ_i}\langle s | p_i\rangle + \sqrt{λ_j}\langle s | p_j\rangle + \sqrt{λ_iλ_j}\langle p_i | p_j\rangle\\ &= 1+0+0+\sqrt{λ_iλ_j}\cos{\omega_{ij}}=1+\sqrt{λ_iλ_j}\cos{\omega_{ij}} \end{align} $$ The s orbital is normalized and so the inner product $⟨ s | s ⟩ = 1$. Also, the $s$ orbital is orthogonal to the $p_i$ and $p_j$ orbitals, which leads to two terms in the above equaling zero. Finally, the last term is the inner product of two normalized functions that are at an angle of $ω_{ij}$ to each other, which gives $\cos{ω_{ij}}$ by definition. However, the orthogonality of bonding orbitals demands that $1 + \sqrt{λ_iλ_j}\cos{ω_{ij}} = 0$, so we get Coulson's theorem as a result:[15] $$\cos{\omega_{ij}}=-\frac{1}{\sqrt{λ_iλ_j}}$$

Now at this point you need to know only one thing:

For any $\mathrm{sp^n}$ hybridized species with symmetrically hybridized orbitals, $n=\lambda_i=\lambda_j$.


$$\cos\omega=-\frac{1}{n} \tag{2} \label{eqn:2}$$

Now as $n$ is the ratio of $p$ character to the $s$ character in the hybridized orbital.


Which when substituted in $(\ref{eqn:2})$ gives your desired results.


(1) Wikipedia contributors. Bent’s rule https://en.wikipedia.org/w/index.php?title=Bent%27s_rule&oldid=992423483 (accessed Jun 18, 2021).


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