Identify the Lewis acid and Lewis base among the reactants in eac...

Question

Identify the Lewis acid and Lewis base among the reactants in each of the following reactions: 
(d) HIO1lq2 + NH2-1lq2 Δ NH31lq2 + IO-1lq2(lq denotes liquid ammonia as solvent)

Accepted Answer

Hello everyone. So in this video we're looking at this reaction right here and we're trying to identify the lewis acid and Lewis base among the reactant. So that's going to be our starting materials on the left side of our equilibrium era. Alright, so let's go ahead and refresh our memory of what a lewis acid and lewis bases. So by definition our lewis acid is going to be our electron paris sectors. As for our louis space, that's going to be our electron pair donors. So we can go ahead and draw out the lewis dot structure to clearly see it visually. However, we can also use our knowledge given the definitions and just background of chemistry to go ahead and find our lewis acid and Lewis space. So we can see here that we have our H B R and we have our p H two minus. So one thing that I recognize really quickly is our Hbr that's one of our known six strong acids. And of course that means this has to be our lewis acid. But let's say I don't have that list memorized and I want to go ahead and look at this one right here. It's the only one with the charge we have a neutral molecule and a negatively charged molecule. If we have this negative here, this means that this is going to be very electron dance. It's gonna go wanna go ahead and give up some of his electrons. This just means that it will donate or share electrons never donate electrons. That fits our very definition of a Lewis base. And just by default, then the other, the only other reactant is going to be our acid because we have to have a lewis acid and louis space. Right, So here is our Lewis base. Here is our Lewis acid. And to write this more formally, Our final answer is that our Lewis acid, it's going to be our h b R and the Lewis space. It's going to be our PH two negative. And this, right here is going to be my final answer for this problem.

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