My textbooks states the Drago's rule in inorganic chemistry as follows:

The more electronegative atom prefers the orbital having more p character, and lone-pairs or less electronegative elements prefer such orbitals, as have more s character.

Now, according to hybridization, $\mathrm{sp^3d = sp^2 + pd}$, $\mathrm{sp^3d^2 = sp + p^2d^2}$, $\mathrm{sp^3d^3 = sp + p^2d^3}$

So, in $\mathrm{sp^3d}$ more electronegative elements like Fluorine go to axial position, but such elements go to equatorial position in $\mathrm{sp^3d^2}$ or $\mathrm{sp^3d^3}$.

Also, as a result of Drago's effect when atoms of group $14, 15$ and $16$ (belonging to third period and above) are attached to atoms having Electronegativity less than $2.5$ then it has valence electrons in pure s-orbitals. And apparently no hybridization effect is observed in such cases. That is why in $\ce{NH3}$ the bond angles are $107.8^\circ$ but for $\ce{PH3},\ce{AsH3},\ce{SbH3},\ce{BiH3}$ the bond angles are very close to $90^\circ$.

I couldn't make much sense out of the above rule. My questions are as follows:

  • What does the following part mean?

    according to hybridization, $\mathrm{sp^3d = sp^2 + pd}$, $\mathrm{sp^3d^2 = sp + p^2d^2}$, $\mathrm{sp^3d^3 = sp + p^2d^3}$

    Does it mean that the hybridized orbitals split in this way? If so, why should they split?

  • Why would the more electronegative atom prefer the orbital with more p character?

  • Even if Drago's rule holds, why is the statement in the paragraph beginning with "Also, as a result of Drago's effect when atoms …" true? I could not understand why the same won't occur for atoms with electronegativity $>2.5$.

I could not find much information regarding Drago's rule on the internet. Not even on Wikipedia. One of my acquaintances (who is a chemist) said that no such rule exists. But I didn't get much time to discuss the matter with him.

So does such a rule really exist? And, in that case, how can we answer the above two questions given as bullets using the existing theories in chemistry?


Going through what you posted, I think ‘Drago’s rule’ (which I never encountered, either at school or at university) gives good predictions but uses a largely terrible set of arguments.

For any atomic system with more than one electron, quantum chemistry predicts the energies of the $\mathrm{s}$ and $\mathrm{p}$ 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 $\mathrm{s}$ subshell of a certain shell in multi-electron systems has a lower energy than the $\mathrm{p}$ subshell, which in turn has a (far) lower energy than the $\mathrm{d}$ subshell and so on and so forth. Since $\mathrm{s}$ 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 $\mathrm{s}$ subshell; the next six electrons are added to $\mathrm{s}$ subshells whereafter a new shell is used.

This immediately explains that it is most stable for any atom to treat the $\mathrm{s}$ electrons of its valence shell as ‘extended core electrons’; i.e. we can predict any configuration that does not touch the $\mathrm{s}$ 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:

  • mono-, di- and tri-coordination: use 1, 2 or 3 $\mathrm{p}$ orbitals to form bonds; keep one lone pair in an $\mathrm{s}$ type orbital. (Predicted bond angle: $90^\circ$.)

  • tetracoordination (no additional lone pairs): since the former is not possible, hybridise $\mathrm{s}$ and $\mathrm{s}$ orbitals to form $\mathrm{sp^3}$ hybrid orbitals. (Predicted bond angle: $109.5^\circ$.) 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 $\mathrm{sp^3}$ description is mathematically equivalent, though.

  • 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 $\mathrm{s}$ orbital if possible. Use remaining $\mathrm{p}$ orbitals to construct four-electron-3-centre bonds to the remaining atoms. (Predicted bond angles: diverse. $90^\circ$ going from 2e2c bonds to 4e3c bonds; $180^\circ$ between a pair of coordinating atoms contributing to the same 4e3c bond.)

Using the aforementioned 4e3c bonds nicely explains why electronegative atoms prefer to occupy these positions. You should always remember that 4e3c bonds can be described by the following resonance structures (drawn for $\ce{ClF3}$): $$\ce{F-Cl^+\bond{...}F- <-> F^-\bond{...}Cl^+-F}$$ In this resonance, each outer atom taking part in a 4e3c bond has a formal average charge of $-0.5$; it is generally better to assign charges to electronegative atoms.

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 $90^\circ$, 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 $\mathrm{s}$-contribution into the bonding orbitals — which in turn means that the lone pair will be less $\mathrm{s}$-like. The ideal bond angle for dicoordinated systems based on sterics is $180^\circ$, showing that the electronic contribution ($90^\circ$) is more important than the steric one ($180^\circ$). 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.

It has been commented that you should not invoke $\mathrm{d}$ orbital participation for main group elements. Remember what I said in the second paragraph: a shell’s $\mathrm{d}$ subshell has a higher energy than the next shell’s s subshell. Thus, before any same-shell $\mathrm{d}$ subshell is added to hybridisation, the next shell’s $\mathrm{s}$ subshell should be. Since hybridisations involving two different shells’ $\mathrm{s}$ subshells are not observed, we should reject the possibility of using $\mathrm{d}$ orbitals for hybridisation. Therefore, we can also flat-out reject the central two paragraphs of your quotation.

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    $\begingroup$ "...you should not invoke d orbital participation for main group elements". Does that mean all hybridizations like $\mathrm{sp^3d}$ ,$\mathrm {sp^3d^2}$ and $\mathrm{sp^3d^3}$ are invalid for main group elements ? And which elements do you consider as "main group" elements ? $\endgroup$ – user37316 Dec 17 '16 at 4:36
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    $\begingroup$ And, in that case, how do you explain the hybridization of compounds like $\ce{PCl5}$ ? If it is not $\ce{sp3d}$ then what is it ? How can you explain its hybridization using only $s$ and $p$ orbitals ? $\endgroup$ – user37316 Dec 17 '16 at 4:40
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    $\begingroup$ @dora chemistry.stackexchange.com/q/52179/4945 hybridisation simply isn't a good enough concept for these cases. $\endgroup$ – Martin - マーチン Dec 17 '16 at 5:38
  • $\begingroup$ ' Since hybridisations involving two different shells’ s subshells are not observed, ...' No one ever observed any hybridisation. But try mixing 1s and 2s (etc), they have the same symmetry... $\endgroup$ – Martin - マーチン Dec 17 '16 at 5:41
  • $\begingroup$ @Doraemonドラえもん $\ce{PCl5}$ is $\mathrm{sp^2}$ hybridised. The two axial chlorine atoms form a 4-electron-3-centre bond with phosphorus. $\endgroup$ – Jan Dec 17 '16 at 17:32

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