Asked by Cynthia
Since the oxidation number of Chromium is +3, and the oxidation number of Silicon is -4. Why is the chemical formula for Chromium III Silicide Cr3Si2?
Can somebody please explain to me please???????THANKS A LOT!!!
Can somebody please explain to me please???????THANKS A LOT!!!
Answers
Answered by
DrBob222
The best answer I can give is that although the oxidation number for Cr is +3 in most of its compounds and the oxidation number for Si, in most of its compounds is (I think +4 in SiO2, SiCl4, etc), the oxidation number for Si is not +4 (or -4) in Cr3Si2. In Cr3Si2, if we call Cr +3, then each Si is -4.5.
Oxidation numbers are nothing more than a set of man-made rules and is really a book keeping procedure to keep up with loss and gain of electrons. Let me show you what I mean.
CN^- ==> CNO^-
Suppose we call C+4, then N is -5 on the left and on the right it is (+4C, -2 for O, so N is -3) So the N atom changed from -5 to -3 which is a loss of 2 electrons . But suppose I'm on a test and forgot the rules. Simple. I make up my own.
Let's call C+19. then
CN^- gives C +19 and N must be -20.
In CNO^-, C is +19, O is -2, so N must be -18. The nitrogen atom changed from -20 to -18 which is a loss of 2 electrons.
Oxidation numbers are nothing more than a set of man-made rules and is really a book keeping procedure to keep up with loss and gain of electrons. Let me show you what I mean.
CN^- ==> CNO^-
Suppose we call C+4, then N is -5 on the left and on the right it is (+4C, -2 for O, so N is -3) So the N atom changed from -5 to -3 which is a loss of 2 electrons . But suppose I'm on a test and forgot the rules. Simple. I make up my own.
Let's call C+19. then
CN^- gives C +19 and N must be -20.
In CNO^-, C is +19, O is -2, so N must be -18. The nitrogen atom changed from -20 to -18 which is a loss of 2 electrons.
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