when hydrogen is bonded to a metalloid what is its oxidation number?

[MedGrl@2022]

Consider the following reaction:

CaH2 + 2BH3 -----> 2BH4- +Ca2+

What type of reaction is this?

A. Lewis acid/Lewis base
B. Bronsted-Lowry acid/base
C. Oxidation/reduction
D. Combustion

I chose C but according to EK the answer is A because the oxidation numbers do not change. I understand that H in Ca has an oxidation number of -1. According to EK, H has a -1 oxidation state in BH3 and BH4-. How am I supposed to know the oxidation state of H in BH3 or BH4-? B is a metalloid not a metal or non-metal. Thus, how are we supposed to know the oxidation state of either atom in the compounds BH3 or BH4-?

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[polyatomic]

You should automatically know this is an acid/base reaction of some type, because H atoms are being transferred. And you know it's not a Bronsted-Lowry acid/base because you have a hydride transfer rather than a proton transfer. Proton transfers would be Bronsted-Lowry. Otherwise, you can figure out the oxidation states, which is more tedious.


How-to find oxidation states:
example: BH3

B is more electronegative than H, so for each H that B holds, we give B -1
Because we have 3 H's: (-1) * 3 = -3

Here is another approach where you use the overall charge and "known" oxidation states:
Use the following general rules:
H = +1 (sometimes -1
F = -1
O = -2
Alkali metals = +1
Alkaline metals = +2

overall charge = (charge of B) + (charge of H)
0 = (charge of B) + (3 * 1)
0 = (charge of B) + 3
0 - 3 = charge of B = -3


example: BH4-

This problem is a bit odd...because the hydrogen that has been added to BH3 was a hydride (a hydrogen with an electron pair) rather than a proton. So...you have to approach it a little differently.

Do the following: overall charge = (charge of B) + (charge of H) + (charge of hydride)
-1 = (charge of B) + (3 * 1) + (-1)
-1 = (charge of B) + 2
-1 - 2 = charge of B = -3

 

[sdm33]

To be honest, it is not that common to see Lewis acid/Lewis base exchanging Protons, normally they use electrons, but I guess the point of this question to recognize H with a negative charge is no longer a proton.

 

[ScientifiqueSucré]

You should automatically know this is an acid/base reaction of some type, because H atoms are being transferred. And you know it's not a Bronsted-Lowry acid/base because you have a hydride transfer rather than a proton transfer. Proton transfers would be Bronsted-Lowry. Otherwise, you can figure out the oxidation states, which is more tedious.


How-to find oxidation states:
example: BH3

B is more electronegative than H, so for each H that B holds, we give B -1
Because we have 3 H's: (-1) * 3 = -3

Here is another approach where you use the overall charge and "known" oxidation states:
Use the following general rules:
H = +1 (sometimes -1
F = -1
O = -2
Alkali metals = +1
Alkaline metals = +2

overall charge = (charge of B) + (charge of H)
0 = (charge of B) + (3 * 1)
0 = (charge of B) + 3
0 - 3 = charge of B = -3


example: BH4-

This problem is a bit odd...because the hydrogen that has been added to BH3 was a hydride (a hydrogen with an electron pair) rather than a proton. So...you have to approach it a little differently.

Do the following: overall charge = (charge of B) + (charge of H) + (charge of hydride)
-1 = (charge of B) + (3 * 1) + (-1)
-1 = (charge of B) + 2
-1 - 2 = charge of B = -3

Just want to state (since this thread pops up whenever anyone Google searches for the oxidation state of BH3) that Boron is not more electronegative to Hydrogen, at least not according to ptable.com.
http://www.ptable.com/#Property/Electronegativity

Hydrogen has a -1 oxidation state in BH3. Therefore B is +3, each H is -1. Boron is less electronegative than H, but does this have anything to do with the relative electronegativities or rather some other chemical/physical property?
In this EK problem where you have CaH2 + 2BH3 --> 2BH4- + Ca2+, you can calculate the oxidation states as:
Ca is a metal, so H automatically has a -1 oxidation state when bonded to it (as per the rule that H=+1 except when bonded to metals), so
Ca = +2 ox. state
H = -1 ox. state

For BH3, Boron is a metalloid...somewhere between a nonmetal and a metal. But it is less electronegative than H. So where does the oxidation state stand?
According to this oxidation state calculator, http://www.periodni.com/oxidation_numbers_calculator.php, H has a -1 oxidation state when bonded to many metalloids, regardless of their electronegativities. Take for example Selenium, which is more electronegative than H at 2.55 to H's 2.20. When plugging into the calculator (admittedly I don't know how this calculator works) as SeH4, it assigns H a -1 oxidation state. (According to Wikipedia, Selenium is 'sometimes considered a metalloid'). The same goes for TeH4, with Te being a metalloid but less electronegative than H.

So it seems to me that if you have hydrogen bonded to a metalloid or an atom with metalloid character, H will take on a -1 oxidation state, regardless of electronegativities.

Would love to get some input from an actual chemist.

 

[aldol16]

So it seems to me that if you have hydrogen bonded to a metalloid or an atom with metalloid character, H will take on a -1 oxidation state, regardless of electronegativities.

Would love to get some input from an actual chemist.

In boron-hydrogen bonds, boron and hydrogen have almost similar electronegativities, with hydrogen somewhat more electronegative. Remember that oxidation states are only one end of the spectrum for understanding bond polarization and is only useful when viewed in that context. At the other end is formal charge. Metalloid-hydrogen bonds are somewhere in between ionic and covalent, so it's hard to characterize it completely.

 

[BerkReviewTeach]

Would love to get some input from an actual chemist.

ScientifiqueSucré, you did a very good job with this question.

As a general rule I shy away from posting replies to questions from other sources, but I want to start by saying that EK's question here is excellent and apologize if I'm stepping on anyone's toes here.

As pointed out, in CaH2, Ca is +2 and H is -1. As for the oxidation states in BH3, there is no need for the over-thinking of this thread. Just go with the nomenclature. NaBH4 is called "sodium borohydride", which means it is a "hydride". A hydride is H-, so in both BH3 and BH4-, H has an oxidation state of -1, making B +3 in both compounds.

So let's consider the choices keeping it simple as possible.

Choice A: Lewis acid/base reactions involve a compound or atom with a lone pair attacking another atom to form a new covalent bond. In this reaction, H:- has attacked BH3 to form H—BH3-, which is described as a Lewis acid/base reaction where H:- is the Lewis base and BH3 is the Lewis acid.

Choice B: Bronsted-Lowry acid/base reactions involve an H+, not an H-, so it can be eliminated.

Choice C: The oxidation state of Ca stayed at +2, the oxidation state of B stayed at +3, and the oxidation state of H stayed at -1. Nothing changed in terms of oxidation state, so there was no redox involved. Choice C is out.

Choice D: There is no gain of oxygen involved in the reaction, so this is a throwaway wrong answer.

The simple eloquence of this question is wonderful. It was a simple question that inspired a great deal of over-thinking, which is right up AAMC's alley.

Hope this helped.