Concise Inorganic Chemistry for JEE(Main and Advanced) by JD Lee(Third Edition) writes

If we compare the rate of hydrolysis for $\ce{SiX4}$(where X= F,Cl,Br,I), it is in the same order as that of their Lewis acid strength order,i.e. $\ce{SiF4}$ > $\ce{SiCl4}$ > $\ce{SiBr4}$ > $\ce{SiI4}$ .This is because the effect of π back bonding is less dominant over the negative inductive effect.

My question is, why is the π back bonding neglected here even though 3dπ-2pπ back bonding is strong enough to be evoked in many other cases? Also, Wikipedia lists the bond strengths of $\ce{Si-X}$ as $\ce{Si-F}$ > $\ce{Si-Cl}$ > $\ce{Si-Br}$ > $\ce{Si-I}$ , and I feel that the rate of hydrolysis should be inversely proportional to this order as it will become increasingly difficult to break the silicon-halide bond when bond strength increases,and consequently difficult to hydrolyse the molecule


1 Answer 1


We can justify the rate of hydrolysis of Silicon tetrahalides if we just think on the statement carefully. The statement claims that the rate of hydrolysis is in the order $\ce{SiF4 > SiCl4 > SiBr4 > SiI4}$. So, we are mainly concerned about the rate of the reaction, not on the thermodynamic stability or any energetic criterion for this reaction( I have dealt with it in the next paragraph). If we recall the characteristics for rate of a reaction, it depends on the slowest step of the reaction. In the hydrolysis reaction of silicon tetrahalides, the first step, i.e the attack of a $\ce{H2O}$ molecule to form the trigonal bipyramidal intermidiate is the rate determining step. Also note that, no cleavage of $\ce{Si-X}$ bond happens in the rate determining step, but it actually happens in the next step, which is a fast step, so it doesn't contribute to the rate of the reaction. So, basically strength of $\ce{Si-X }$ bond doesn't affect the rate of hydrolysis reaction (the situation is exactly same as Nuleophilic aromatic substitution reactions ($S_NAr_2$) also). But if the $\ce{Si}$ atom becomes more and more electrophilic, the attack by a water molecule becomes more easier which subsequently results in the increase of the rate of the reaction, So the electrophilicity of $\ce{Si}$ in the tetrahalides increase in the order $\ce{SiF4 > SiCl4 > SiBr4 > SiI4}$, and so the rate of the hydrolysis also.

Here is the mechanism of the first out of the four subsequent substitutions in the hydrolysis reaction of Silicon tetrahalides.

enter image description here

Secondly, if we really want to judge the rate of the reaction by numerical calculations, observe that the ultimate product of hydrolysis reaction of all the tetrahalides are same. Also, all the reactions are exothermic in nature. So, $\Delta _r H^0$ (assuming the hydrolysis reactions are done in standard conditions; it's not a strict assumption, it may not be at standard conditions also, but the changes in the heat of reaction will be almost similar for all the tetrahalides) will be more if $\Delta_f H^0$ is more for the reactants (note that, $\Delta _r H^0 = \Delta_f H^0_{products} -\Delta_f H^0_{reactants} $ and we are bothered about reactants. If $\Delta_f H^0_{reacants}$ are more $\Delta_r H^0$ will be more negative and hence energetically favourable). The values for $\Delta_f H^0$ have the order, $\ce{\Delta_fH^0_{SiF4} < \Delta_fH^0_{SiCl4} < \Delta_fH^0_{SiBr4} < \Delta_fH^0_{SiI4}}$. So, the hydrolysis reaction of $\ce{SiF4}$ will be the least exothermic and that of $\ce{SiI4}$ will be the most exothermic, which is expected also from the bond energy considerations. So, if you have less release of heat energy, i.e thermodynamically less stable product, it will have lesser activation energy and higher rate constant, and thus the rate of the reaction will be highest for $\ce{SiF4}$ and least for $\ce{SiI4}$ as we will get the product along with relatively more thermodynamic stability. So, by thermodynamic consideration also, rate of hydrolysis should be, $\ce{SiF4 > SiCl4 > SiBr4 > SiI4}$.

Lastly, the $\pi$-bonding is not important in case of large size difference between $\ce{Si}$ and $\ce{F}$ ($2p$ and $3d$ have course greater size difference though you may not admit). The fluorine atom makes the silicon more electrophilic by its electron pulling ability rather than back-bonding.

  • 1
    $\begingroup$ Then why do we consider back bonding in case of the boron trihalides?( The order given for their rate of hydrolysis is fluoride<chloride<bromide<iodide, and although they have not mentioned the reason for this below, in my mind I have always thought that this is due to backbonding,or am I wrong here?) $\endgroup$ Feb 12, 2019 at 3:40
  • $\begingroup$ Also, I just checked in my book, and the author is performing an SN2 reaction for the hydrolysis of silicon tetrahalides,which goes through a single five membered transition state resulting in Walden inversion. So can you please clarify which mechanism is correct? $\endgroup$ Feb 12, 2019 at 4:07
  • $\begingroup$ @YUSUFHASAN We have to consider the back-bonding in case of Boron because it lowers the electrophilicity of the Boron atom, and if electrophilicity is lowered, obviously water molecule will be less prone to attack at the Boron centre which does affect the rate of the reaction. In case of Silicon, only the strength of the bond matters at the end of the reaction. Thus the two cases are diffferent. $\endgroup$
    – Soumik Das
    Feb 12, 2019 at 6:47
  • $\begingroup$ So you mean to say that the RDS of hydrolysis of boron trihalides does not concern the addition of water molecule,while it is so for silicon tetrahalides? $\endgroup$ Feb 12, 2019 at 6:48
  • 1
    $\begingroup$ Yes, Exactly..Nothing more than that.. $\endgroup$
    – Soumik Das
    Feb 12, 2019 at 7:10

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.