By definition "Ionic or saline hydrides are composed of hydride bound to an electropositive metal, generally an alkali metal or alkaline earth metal."

  • What is the relative order of reactivities of all the ionic hydrides of alkali and alkaline earth metals?Is there any general trend?

  • Is the reactivity of alkali/alkaline-earth metal hydrides greater than $\ce{LiAlH4}$ ?

  • Is the reactivity of alkali/alkaline-earth metal hydrides greater than $\ce{NaBH4}$ ?

I know reducing nature of $\ce{LiAlH4}$>$\ce{NaBH4}$.

  • $\begingroup$ Can't NaH reduce carbonyls? @orthocresol and cant NaBH4 deprotonate compounds? $\endgroup$
    – user14857
    Sep 27, 2016 at 18:01
  • $\begingroup$ Can you explain why NaBH4 or LiAlH4 is slower than the s block hydrides ? $\endgroup$
    – user14857
    Sep 27, 2016 at 18:07
  • $\begingroup$ @orthocresol Hey,sorry for the late follow-up question.But can you clarify something?Why is breaking B-H bond tougher than breaking Na-H bond?Aren't ionic bonds stronger than covalent bonds?Or is it because NaH is ionic it already dissociates in water? $\endgroup$
    – user14857
    Sep 29, 2016 at 7:35

2 Answers 2


This is a very generalized answer but I'll go for it anyhow, who knows how it'll turn out to help (if it does add to what you already know, I suspect you may already know about what I'm going to type....)

The metrics you're considering are 'Reactivity' and (since you brought up LiAlH) 'Reducing character'. Clump these together, and it turns out what you're looking for is the ability for these hydrides to donate/release their hydrogen.

Now I've reasons to believe you already know the 'reactivity' trend for p-block hydrides (correct me if I'm wrong). Figuring out the reducing character of the s-block ionic hydrides involves similar logic.

In the p-block hydrides, you've probably learn't that as you move down the group, the E-H bond length increases (I use 'E' to denote p-block elements). If the bond length increases, the easier it is to cleave, therefore the hydrogen is released more readily. Here's an analogy if that bit seems a bit hard to grasp. Think of the E-H bond as a pencil. The longer it is, the easier it is to break; the shorter it gets the more difficult it is to break (try it......P.S-this credits for this analogy goes to my Chem. Teacher).

In a nutshell, for the p-block hydrides, as you move down the group the tendency to donate hydrogen increases.

Now back to the s-block. You can't apply the logic you used for p-block hydrides here (at least not directly ;) ).

At first glance, the main impediment you'd see is "s-block hydrides are ionic but p-block hydrides are covalent, so I can't figure out the trend the same way...". You'd be correct, but if you trying extrapolating (in reverse) the logic you use for p-block hydrides, see what happens.....

Since they're all ionic hydrides, and since you want the reactivity trend, what you'd want to do is compare the covalent nature of each s-block hydride.

As you'd already know, an ionic bond isn't 100% ionic, and the same can be said for covalent bonds. You call them ionic or covalent, if the ionic or covalent nature of a bond is relatively significant.

So by using Fajan's rules (assuming you know those as well) you can figure out that LiH is the most 'covalent' of the ionic s-block hydrides, and therefore more prone to donating hydrogen thereby acting as a reducing agent. This Wikipedia page just confirms that...

As you move down Groups 1 and 2, the reducing character of the s-block hydrides decreases.

Also LiAlH and NaBH are much, much stronger reducing agents than plain ole' s-block hydrides.

  • $\begingroup$ Can you give reasons for your last sentence? $\endgroup$
    – user14857
    Sep 27, 2016 at 17:57
  • 1
    $\begingroup$ $\ce{BeH2}$ is the most covalent hydride among s-block hydrides. $\endgroup$
    – Apurvium
    Sep 13, 2021 at 9:30

@Aaron Abraham has written a nice answer to the original question.However I would like to combine his arguments along with the points made by @orthocresol.

  • In case of covalent compounds larger and weaker bonds in hydrides facilitates donation of hydrogen for reduction reactions.

  • S-block elements too have some covalent nature.So the previous fact can be applied for them too using Fajan's rules.

  • In general the top most element of both the groups in s-block has maximum covalent nature.Covalent bonds are generally weaker than ionic bonds (with some exceptions).So the reducing power is maximum at top and minimum at bottom.

  • Alkaline earth metal hydrides tend to be more covalent than alkali metals hydrides due to higher charge density on alkaline earth metals.So Alkaline earth metal hydrides have more reducing power compared to its counterpart.

  • S-block hydrides are usually never used in reduction of carbonyls and similar compunds.However they are used for deprotonation.

  • Even $\ce{NaBH_4}$ can be used to deprotonate.However they are much weaker and slower than s-block hydrides as for them to react the $\ce{B-H}$ bond has to be broken which is difficult and slow.
  • $\begingroup$ Al-H bond is really weak though, LiAlH4 deprotonates stuff really fast. $\endgroup$ Sep 29, 2016 at 7:48
  • $\begingroup$ But slower than s-block hydrides?Right ? @orthocresol $\endgroup$
    – user14857
    Sep 29, 2016 at 7:49
  • $\begingroup$ No idea actually. $\endgroup$ Sep 29, 2016 at 7:50
  • 1
    $\begingroup$ @orthocresol Are any values available on the net to comapare the rates?Well even I have not much idea... $\endgroup$
    – user14857
    Sep 29, 2016 at 7:52
  • 1
    $\begingroup$ I don't know. I don't think that such a thing would exist, deprotonations/protonations occur at a very fast rate, and for the most part we are more concerned about the extent of de/protonation (a thermodynamic question) rather than the rate (a kinetic question). $\endgroup$ Sep 29, 2016 at 7:56

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