During hydrolysis of a halide of carbon via $\mathrm{S_N2}$ mechanism, the incoming neutral water molecule attacks the partially positive carbon centre from the side opposite to the halide side and forms a transition state where the hybridisation of centre is $\mathrm{sp^2}$. The nucleophile and the leaving group (halide) thus are attached to opposite sides of the same p orbital (now elongated). Ultimately the nucleophile kicks off the halide from the back.

But when we are carrying hydrolysis on $\ce{PCl3}$ (phosphorus trichloride) why does the water attack an empty d-orbital of the central phosphorus atom and not do the same mechanism as it does in the previous case mentioned. Why can't the attack take place like above leading to formation of an elongated p orbital?


1 Answer 1


I don’t know where you heard about the water molecule interacting with d orbitals before displacement takes place, but the idea is wrong. Phosphorus’ d orbitals are energetically too far removed to take part in bonding. Remember that according to the aufbau principle, 3d orbitals are higher in energy than 4s! They exist and you can excite electrons into them but that’s it (and you need high energies for that excitation).

Rather, the oxygen of water interacts with the $\sigma^*_{\ce{P-Cl}}$ orbital. Since phosphorus uses an orbital with a very high p-content, it has a notable lobe on the far side allowing for interaction. An intermediate tetracoordinated phosphorus species is formed which displaces the better leaving group (chloride). Water can lose a proton at any point in this mechanism.

The mechanism is basically identical to that of a carbon-centred $\mathrm{S_N2}$ reaction except that the intermediate breaks down much faster in carbon-based reactions.

  • $\begingroup$ Thank you ! The d orbital concept was given to me by my school teacher, thus I couldn't corelate it to SN2 . But now it's clear. $\endgroup$ Dec 23, 2016 at 4:22

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