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In Concise Inorganic Chemistry by JD Lee (4th edition; adapted by Sudarshan Guha), page 76 under the topic "Back Bonding" it is given:

For $\ce{SCl2}$ the bond angle is 102° since the lone pair of $\ce{S}$ or $\ce{Cl}$ atoms need not be delocalised due to the availability of vacant d orbitals of their own.

Kindly note, the delocalisation the author is talking about is the delocalisation of electrons in back bonding.

I am unable to understand why back bonding doesn't take place even though the conditions are satisfied. Is it because of the availability of empty orbitals in both the atoms involved, the π-electrons are confused to which atom they must bond to, to form back bonding? Is it right to say that back bonding doesn't take place when the electron-donating atom itself has empty orbitals?

Kindly explain the above-mentioned statement and tell me whether my reasoning is correct or not.

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    $\begingroup$ The author is using an outdated concept of d orbital participation. It requires all sorts of somersaults like the one you just found. Forget it. Neither sulphur nor chlorine nor any main group element use their d orbitals in any noteworthy way, shape or form and that is why there can be no back bonding in $\ce{SCl2}$. $\endgroup$ – Jan Aug 24 at 17:44
  • $\begingroup$ Maybe this could help. The extent of back-bonding follows the order, $\ce{2p\pi-2p\pi > 3p\pi-3p\pi > 4p\pi-4p\pi}$ due to large distance of higher orbitals from nucleus. $\endgroup$ – rv7 Aug 25 at 3:10

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