Which chemical reactions are possible?

[jdoyle]

The question in the attachment is from an old entrance exam for medical school. The correct answer is B.

I realize that if I assign a 'point value' to the levels of reactivity for each of the letters i.e. the more reactive the higher the number

T = 1
Q = 2
M = 3
J = 4

and that if the corresponding point value for the added element is less than that of the unknown element (letter) in the original molecule, then the reaction will not take place. For example


This reaction would not take place
MCl₂ + T → TCl₂ + M

because when substituting the point values for the two unknowns, the value of T<M.
3Cl₂ + 1 → 1Cl₂ + 3

On the other hand, the following reaction would take place because J>M
MSO₄ + J → JSO₄ + M

3SO₄ + 4 → 4SO₄ + 3

My problem is that I don't understand why. Is it because the addition of an element with a lower reactivity will not be able to break the bonds forming the original molecule? I would very much appreciate it if someone could give me an idiot-proof answeras I want to explain this to my daughter. Thanks in advance for your help.

John

 

[Mike H]

First thing of note - there's no attachment. At least I'm not seeing one.

Based on what you wrote, though, it seems as if it inquires about the chemistry of various salts. Beyond that, I wouldn't want to speculate without the actual question in front of me.

 

[jdoyle]

Hi Mike,

Thanks for your quick reply. I have uploaded the attachment again. I don't understand why it wasn't attached the first time as it was there when I checked the original post.

Thanks ahgain.

John

 

[Acut]

@jdoyle:

Essentially, there is a series of reactivity, which can be used to determine whether or not one element can displace the other in a compound (if you take a look at the reactions on the attachment, there's always one element displacing the other - in the first one, T displaces M for instance). The question gives you the series of reactivities. In a nutshell, a more reactive element will displace a less reactive one.

Search "displacement reactions" on Google and see what you get.

 

[jdoyle]

Hi ACut,

Thanks. That was exactly what I was looking for.

Best wishes

John