Gen Chem Question Valence Electrions

[bnleong]

From the following statements, which is NOT correct?
A. Reduction involves the gain of electrons.
B. Oxidation is associated with the loss of electrons.
C. The oxidized particle (atom) shows a decrease in valence number.
D. Oxidation-reduction reactions involve the loss and gain of electrons.
E. All of the above are CORRECT.

A particle loses electrons when it is oxidized. Thus oxidation is associated with electron loss. Meanwhile, electron gain is associated with reduction. Since oxidation and reduction processes occur simultaneously, loss and or gain of electrons is to be expected. An oxidized particle therefore loses its (valence) electrons. This results in an increase in valence number

Why is C the answer? Is this a mistake?

---

Members don't see this ad.

 

[UndergradGuy7]

I think they mean oxidation makes the "oxidation number" more positive or increases it.

Reduction makes it more negative or decreases it.

See here
http://pubs.acs.org/doi/abs/10.1021/ed083p791

Terms such as valence, oxidation number, coordination number, formal charge, and number of bonds are widely used throughout chemistry. Although each of these terms has a distinct meaning, their utility is lessened by imprecise usage. Valence, for example, is frequently taken to be synonymous with oxidation number. However, while valence and oxidation number are coincidentally equal for many compounds, the relationship breaks down in many cases; in such situations, the use of oxidation number instead of valence can result in misleading descriptions of molecules.

Anyway, it does sounds like a misleadingly worded question.

 

[ZFT]

this question caught my eyes a while ago, i wrote it down on a post-it and finally researched it just now.

i skimmed my textbook and basically bnleong's reasoning is correct (oxid = lose e-), but Guy7 had the correct vocabulary (lose e- = more positive valence number = increase).

i guess the question tests your familiarity with chemistry speak, which is fair...but tricky.