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Ch.22 - The Main Group Elements

Chapter 22, Problem 10

Consider the following oxoacids: HClO, HClO2, HClO3, and HClO4. In which oxoacid does chlorine have an oxidation

state of +5? Which oxoacid is the strongest?

(LO 22.17)

(a) HClO has a Cl oxidation state of +5, and HClO4 is the strongest acid.

(b) HClO2 has a Cl oxidation state of +5, and HClO is the strongest acid.

(c) HClO3 has a Cl oxidation state of +5, and HClO4 is the strongest acid.

(d) HClO4 has a Cl oxidation state of +5, and HClO is the strongest acid.

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All right. Hi, everyone. So this question says given the oxo Aids of Bromine, Hbrohbro two, HBRO three and HBRO four identify the Oxo aid where Bromine exhibits a plus five or positive five oxidation number. Additionally determine which oxo aid among the given possesses the highest acidic strength. So here we've got four different answer. Choices. Each proposing a different one whose bromate has an oxidation number of positive five as well as a different one for who is the strongest acid of the four or which is the strongest acid. So recall that in this case, the oxidation number of bromine is unknown, but we can solve for it by using the oxidation numbers of the other atoms in these molecules. Because recall that when you add up the oxidation numbers of each atom multiplied by the number of each atom in the molecule itself, that total should equal the charge of your molecule. Now, in this case, it just so happens that all of these oxo acids are neutral, which means that I'm going to make all of my equations equal to zero, right. Starting off with HBO, starting off with HBRO by remembering the oxidation states of hydrogen and oxygen we can use them to solve for bromine. In this case, right, recall it, hydrogen can have an oxidation state of either positive one or negative one depending on what it is connected to. When hydrogen is combined with non metals, it has an oxidation state of positive one. However, when it's combined with either metals or boron, it's going to have an oxidation state of negative one. So hydrogen is going to be a positive one. Whereas oxygen, if you recall is generally going to have an oxidation state of negative two. Now exceptions do apply if the compound is a peroxide or a superoxide, but those exceptions do not apply in this case. So our oxidation state for oxygen is negative too. So now to set up the equation right X is going to be the oxidation state of bromine and I can add that to the oxidation state of hydrogen which is positive one multiplied by the number of hydrogens in the given compound. And then I add that by the oxidation state of oxygen which is negative two multiplied by the number of oxygen atoms. And so this entire expression is going to be equal to zero because that is the charge of my compound overall. So in this case, I can make some substitutions because it just so happened that I have one hydrogen atom as well as one oxygen atom. So here when I solve for X, it just so happens that X is going to be equal to positive one, which means that the oxidation state of bromine and HBO is positive one. But in this case, we're looking for an oxidation state of positive five. So here's where I proceed to the next one, which is HBRO two. So here I'm going to use the same equation as last time. But now it just so happens that I have two oxygens instead of one. However, I still have one hydrogen. So the second term is going to stay the same and it's still equal to zero. So when I solve for this expression, I get that X is equal to positive three, which is still not quite what I'm looking for. So I can proceed on to the next Oxo aid, which is H bro 03. And I'm also going to scroll down briefly, right? So once again, the second term is not going to change because I still have only one hydrogen in this compound. However, my third term where oxygen is going to change because now I have three oxygens. And after I solve for this, I do actually get that X meaning the oxidation state of bromine is equal to positive five. So HBRO three is the correct Oxo aid because it has bromine with an oxidation state of positive five. But regardless, I still want to show the calculation for our fourth and final Oxo aid, that's HBRO four. So here is my expression. And now I have four oxygens. Therefore, when I solve for eggs, I get positive seven. So again, right? HBRO three is the Oxo aid that we're looking for. But I just wanted to show all four calculations regardless, however we're not quite done. And because we still have to discuss which of the Oxo Aids listed is the strongest one. So recall that there is actually a general trend that we can follow for Oxo Aids that contain halogens, right? Because generally speaking, as the number of oxygen atoms increases, so does the strength of the given acid, right? Because recall that the term oxo in oxo aid refers to the presence of doubly bonded oxygen atoms. So when it comes to halogen oxo aids, right, as the number of oxo groups increases. So does the strength of the acid? Why? Because oxygen being electron is going to pull electron density away from the bond between oxygen and hydrogen, which makes that proton easier to be removed, therefore more acidic. So of the four Oxo aids discussed in this question, HBRO four is the strongest because it has the highest number of oxo groups. So our answer if I scroll up once more is going to be option c in the multiple choice because bromine in H bro three exhibits an oxidation number of positive five and HB four is the strongest acid because it has the highest number of oxo groups. So with that being said, if you stuck around to the end. Thank you so very much for watching and I hope you found this helpful.