5
$\begingroup$

Which of the following is the most stable Lewis acid–base adduct?
(a) $\ce{H2O\bond{->}BCl3}$
(b) $\ce{H2S\bond{->}BCl3}$
(c) $\ce{H3N\bond{->}BCl3}$
(d) $\ce{H3P\bond{->}BCl3}$

This is a question from KVPY 2016 examination. In my first attempt, I focused on the positive charge that is going to appear on the donating atom. So the less electronegative the donor, the better it is at donating its lone pair to the vacant orbital on boron. However, the answer key states that the answer is option (c).

I am unable to rationalize why (c) is the most stable adduct.

| improve this question | | | | |
$\endgroup$
  • $\begingroup$ Since this is a Lewis acid-base interaction, the simplest way to go about it would have been to see the strongest Lewis base for this acid. As we move down a group, basicity decreases and as we move across a period,basicity again decreases. So ammonia is the strongest base here, and hence, answer would be (c) $\endgroup$ – Yusuf Hasan Dec 8 '18 at 2:50
2
$\begingroup$

Perhaps, what is expected of the question is for the student to use Pearson's Hard Soft Acid Base (HSAB) theory, which essentially states that hard acids have a preference to bind to hard bases and soft acids have a preference to bind to soft ligands. $\ce {BCl3}$ can be seen as a hard acid and would thus preferably interact with hard bases. Wikipedia classifies boranes as hard acids. Water and ammonia are much harder bases than phosphine and hydrogen sulfide since $\ce {O}$ and $\ce {N}$ are markedly more electronegative and possess higher electron density. Thus, we can likely expect either water or ammonia to be the correct option. Next, we need to compare between the two hard bases. Since they are both relatively hard, we need to compare now a different parameter, that is, their ability to donate the electron pair. In this aspect, nitrogen is clearly superior to oxygen due to the former's lower electronegativity. Thus, $\ce {NH3}$ is the correct option.

| improve this answer | | | | |
$\endgroup$
2
$\begingroup$

You have to consider also the inherent strength of the covalent bond. A strong covalent bond can entice an electronegative atom to accept some positive charge or give up some negative charge.

Roughly speaking, hydrogen and period 2 nonmetals form much stronger bonds among one another than with other elements, and this tends to override the electronegativity difference. For instance, fluorine is much more electronegative than chlorine, but fluorine bonds so much more strongly with hydrogen that hydrofluoric acid is only a weak acid in water whereas hydrochloric acid is strong. Note that the dissociation of hydrogen chloride in water creates hydrogen-oxygen covalent bonds.

Here, boron binds especially strongly with the period 2 nonmetals nitrogen and oxygen. So (a) and (c) have this advantage, and after that (c) wins between these two by electronegativity difference.

| improve this answer | | | | |
$\endgroup$
0
$\begingroup$

See b and d are out due to weak 3p-2p overlap as compared to 2p-2p...also that as oxygen is more electronegative than nitrogen, so even if it does share lone pair, it will develop a positive charge which won't make it happy and it will pull its lone pair back to a greater extent than nitrogen...hence the deficiency of electrons in BCl3 will be most compensated on adding ammonia, and this goes with the facts too

| improve this answer | | | | |
$\endgroup$

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.