I have learnt that in a group , say group $15$, the acidic strength increases on going down the group as

$NH_3 < PH_3<AsH_3<SbH_3<BiH_3$

This is due to decrease in the thermal stability on going down the group. But we can also reason it as follows : on going down the group , the electronegativity of the central atom decreases , and the bond between the central atom and hydrogen becomes less polar , so the bond strength also decreases hence it becomes easier to lose a hydrogen atom thus increasing the acidity.

Now if we compare the hydrides of a period, that is

$PH_3 , HCl , H_2S$ , according to the above logic , the order should’ve been $PH_3>H_2S>HCl$ because the electronegativity increases across a period. However , the correct order seems to be the reverse. Could you please explain why ? Is there a flaw in my reasoning?


The reasoning you gave for the increasing acidity of hydrides of group $15$ down the group is not the best one. You claim that less polar bonds are easy to break which is absolutely wrong. Non-polar covalent bonds are very difficult to break (think of beaking $\ce{C-H , N-H , O-H }$ bond, which is easier to break ?)The logic can be modified as follows.

Actually the basicity of those hydrides decrease down the group, which in turn results in the increase of acidity of them. The reason for decrease in their basicity down the group is the increase in inability to donate their lone pair to accept the proton. In $\ce{NH3}$, the lone pairs recide in the $sp^3$ hybrid orbitals, whereas for other hydrides like $\ce{PH3 , AsH3 , BiH3}$ the lone pairs recide almost in pure $s$ oritals (i.e. the hydrogen atoms are bonded to almost pure $p$ orbitals). So, as the $s$ character of the orbital where the lone pair is residing increases, it becomes closer and tightly bound to the nucleus, and thus difficult to protonate. Thus the basicity of group $15$ hydrides decreases down the group, alternatively, the acidity increases.

Coming to the case of $\ce{PH3 , H2S}$ and $\ce{HCl}$, two factors govern the acidity of them. One is the ease of bond-dissociation, and other is the solvation of the medium. When you consider the gas-phase acidity, i.e. no solvation is present, the bond-energy will only be the determining factor and bond energy is given as, $\ce{P-H < S-H < Cl-H}$ (easily justifiable by decreasing size of central atom), which tells us in gas phase the acidity of $\ce{PH3}$ should be the highest. But normally when we talk about acidity, we talk about $pK_a$ in water. Whenever there is a medium, it stabilises the dissociated species via solvation (in water , this is hydration). As. the size of the conjugate base decreases across the period, the extent of hydration energy released will dominate over the bond-dissociation energy. Chloride is the smallest ion among them having higher electronic charge density, and $\Delta H_{hyd} \propto \frac{1}{r} $, It is highest for hydrolysis of $\ce{HCl}$, making it a stronger acid among three.

  • $\begingroup$ The paragraph on s character increasing down the group,is it based on Drago's empirical rule? $\endgroup$ – YUSUF HASAN Jan 5 at 5:28
  • $\begingroup$ Also, for basicity increasing down the group, I generally use the rationale that as more orbitals(d and f) become available, the atom will be able to diffuse it's electrons in those empty orbitals, and hence they will be unavailable for functioning as Lewis base lone pairs. But since you have presented a different argument,is my reasoning justified? $\endgroup$ – YUSUF HASAN Jan 5 at 5:35
  • $\begingroup$ @Yusuf The heavier elements because of their vacant orbitals form the coordination compounds. Nitrogen does not form coordination compounds. I’m not sure if the lone pairs of electrons will be promoted to the empty orbitals . $\endgroup$ – Aditi Jan 5 at 6:06
  • $\begingroup$ @Aditi Yep, this promotion of electrons will be absent in nitrogen,hence it's lone pairs will be more susceptible(available) to Lewis basicity. Also, I'm talking of the elements per se, not in their complexed forms. After complexation,it will exhibit entirely different properties. $\endgroup$ – YUSUF HASAN Jan 5 at 8:52

@aditi I would like to suggest another dimension to look at it...acidity means the ability to evolve H+...now it's well know that EVERYTHING in this universe wants stability...now if ph3 evolves H+ it leaves Ph2-...now observe that phosphorus being more electronegative draws the electron cloud with hydrogen towards itself...causing high charge density on it in addition to the already present negative charge...but not being VERY electronegative it's unable to hold it...this thing is known as inductive effect you shall study in organic...but on the other hand look at cl-...no additional charge due to inductive effect as no hydrogen is present...plus it's far more electronegative than phosphorus...so it will accommodate negative charge far better...now you might be thinking how electronegative atoms accomodate negative charge better than comparitively electropositive ones...well it's cause greater effective nuclear charge causes them to attract electrons towards themselves...They LOVE charge...but remember...this is the case in a solvent...IN CASE WE ARE DISCUSSING WHAT IS HAPPENING IN VAPOUR STATE...acidity of PH3 is very much due to bond length...AS H-O bond is very strong hence in protic solvents (in this case water) will remove h from compounds...in that case cl- is more stable and hence HCl is most acidic...but otherwise h-cl bond is weak due to smaller size of cl and hence cl doesn't kick away h while PH3 and H2S do comparitively more.


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