What does the K° signify in the above reaction. Is it just a typo, or is it supposed to symbolize something?
EDIT: This screenshot is from the solutions manual. In the original textbook, the author has used $K ^ 0$.
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$\begingroup$ How do you write the oxidation of a pure element (metal or non-metal)? $\endgroup$– Mathew MahindaratneCommented Jul 10, 2020 at 6:16
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$\begingroup$ Gauri agrawal, Please tell us the full reference from where you saw this? $\endgroup$– ACRCommented Jul 10, 2020 at 8:48
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$\begingroup$ @M.Farooq This was practice problem 11.14 in Solomon's Organic Chemistry $\endgroup$– gauri agrawalCommented Jul 11, 2020 at 4:24
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$\begingroup$ @gauriagrawal, Thanks. Which edition? $\endgroup$– ACRCommented Jul 11, 2020 at 4:26
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1$\begingroup$ @M.Farooq It is the 11th edition $\endgroup$– gauri agrawalCommented Jul 11, 2020 at 4:26
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2 Answers
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K° here means elemental potassium (zero oxidation state). The author wanted to avoid the confusion of K with the equilibrium constant. There are more quantities than we have symbols, so letters and symbols get re-used. It is called a pigeon hole effect.
Now, that the OP has posted the reference, Solomons, Organic Chemistry, Ed 11, Problem 11.14 Solution manual, it is clear that the circle is a typo for zero. In the 11 and the 12th edition, the symbol stands corrected to proper numerical "zero" in the main textbook (pg 510). It is not unusual to find typos in solution manuals and textbooks. We are all humans-after all.
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1$\begingroup$ Thank you for finding the source, and for another proof that this is not a good textbook:) I scrolled through the solution manual looking for the use of uppercase circle and I wasn't disappointed. They also have $\ce{Mg^\circ},$ as well as three different notations for Celsius scale: $\mathrm{5~^\circ C},$ $\mathrm{5^\circ~C}$ and $\mathrm{5~^\circ~C}.$ And I was clearly mistaken about $\mathrm\LaTeX,$ the main textbook was prepared in InDesign, which could explain why editors missed the inconsistency in symbols. $\endgroup$– andselisk ♦Commented Jul 11, 2020 at 5:39
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1$\begingroup$ I have invariably noticed that the solution manuals rarely have the proper formatting. The authors and editors don't want to spend money on it. $\endgroup$– ACRCommented Jul 11, 2020 at 5:42
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1$\begingroup$ Exactly, they either type the solution manuals themselves in MS Word as best as they can and then publish at the university's typography, or outsource their handwritten papers to the freelancers or a typography on the other end of the world. And this is a problem. $\endgroup$– andselisk ♦Commented Jul 11, 2020 at 5:46
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It is possible that author's intent was to denote that the reaction with potassium metal is carried out under ambient conditions. One could use a superscript circle to refer to the standard state of a pure substance, and even though only the pressure is standardized, the notation of standard state is often utilized for denoting a temperature of 25 °C.
This is non-conventional use of a standard state notation and should not be encouraged.
Why it's probably not a notation for a zerovalent metal?
First, a zerovalency is denoted with, well, zero, not a circle.
Second, it appears that the image for the reaction scheme was produced using a $\mathrm\LaTeX$ compiler, and usually those who are using such typesetting system know the difference between zero “0” and a circle “○”.
One does not occasionally type K^\circ $(\ce{K^\circ})$ in place of K^0 $(\ce{K^0}).$
Why it's probably not an equilibrium constant? Balance aside, it could be a standard equilibrium constant $K^\circ,$ but since the symbol is upright $(\ce{K^\circ})$ and not slanted/italicized, it eliminates this possibility. There is no need for extra symbols to distinguish between the symbols for an element and for a variable. Same goes, e.g. for phosphorus $\ce{P}$ and power $P,$ or nitrogen $\ce{N}$ and number of particles $N.$
answered Jul 10, 2020 at 6:49
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1$\begingroup$ Dear andselisk, nobody is as perfect in notation as you. People are sloppy and we should not over interpret that notation for *elemental potassium" to mean anything else. The author meant to denote that circle as zero, rather, he should have used a numerical zero. Although it is all unnecessary. He could written state symbol with potassium or he should written the full name. $\endgroup$– ACRCommented Jul 10, 2020 at 7:03
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1$\begingroup$ @M.Farooq The notations are the part of math which is the language of science. If one wouldn't speak it, or would speak incorrectly, they will not be able to communicate. Unfortunately, I don't have any information about how literate the author is, so I cannot operate with phrases like "author wanted" or "author meant". There must be "probably" in between unless you know more than me about the author. Nobody's perfect, but I try to assume the the best in people. Am I having difficulties because of this? Sure, but it's better to overestimate a person rather than to underestimate. $\endgroup$– andselisk ♦Commented Jul 10, 2020 at 7:11
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1$\begingroup$ @andselisk - I'm hesitant to join this debate, but I wonder how you determined that the image appears to be compiled from LaTeX? Many people still use chemistry-specific drawing programs like ChemDraw for reaction diagrams, and I don't know of a way to tell from the final image how it was created. In a drawing program, inserting a degree symbol is a quick way to make something that looks somewhat like a superscript 0. I'm not condoning that, but I agree with M Farooq that an intended meaning of 0 oxidation state seems most likely. That said, I appreciate you raising an alternate possibility. $\endgroup$– AndrewCommented Jul 10, 2020 at 10:59
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1$\begingroup$ @Andrew I don't think there is much to debate about. Here, our answers are just opinions (equally plausible, I think, despite M.Farooq's self-confident manner of expression) and only the author of the text can clarify what the original intention were. $\endgroup$– andselisk ♦Commented Jul 10, 2020 at 11:18
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2$\begingroup$ @andselisk, I got the screenshot from the solutions manual, in the original book it shows as $K ^ 0$. So now it makes sense. Thanks! $\endgroup$ Commented Jul 11, 2020 at 4:29