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We know that Mn shows variable oxidation states ranging from +2 to +7 but why is +1 oxidation state of Manganese(Mn) not stable? The +1 oxidation state of Mn would have a configuration of 4s1 3d5. What is that I am missing or wrong with?

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    $\begingroup$ You're underestimating things here chemistry.stackexchange.com/questions/118739/… for example. $\endgroup$ – Mithoron Dec 22 '19 at 17:58
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    $\begingroup$ @Abhishek There is no need to affront other users, especially if they actually pointed you in the right direction. Note that the question wasn't closed single-handedly; rather, it was a peer-review. In your previous (currently deleted) question, there also was a heated discussion in the comments: you refused to fix factological issues and typos; you ignored clues by other users and demanded an answer. This is not a constructive behavior and is not in line with the guidelines for a respectable discussion. $\endgroup$ – andselisk Dec 23 '19 at 16:53
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$\ce{Mn+}$ has configuration $\mathrm{(4s)^0(3d)^6},$ while on the other hand $\ce{Mn^2+}$ $\mathrm{(3d)^5(4s)^0}$ and also, $\ce{Mn}$ in ground state is $\mathrm{(4s)^2(3d)^5}.$

Therefore you can see in the oxidation state +2 orbitals are half-filled, and in +1 they are not half-filled. As half-filled is more stable, therefore +2 and 0 are more stable than +1.

Therefore Mn prefers not to stay in +1 due to higher stability of adjacent states.

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  • $\begingroup$ Why is the configuration of Mn+ 4s0 3d6 and not 4s1 3d5? $\endgroup$ – Abhishek Dec 22 '19 at 14:55
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    $\begingroup$ (-1) Writing that Mn(II) is "more stable" than Mn(I) is a very dangerous thing. Note that Fe(II)/(III) are isoelectronic to Mn(I)/(II) respectively. Why, then, is Fe(II) common, and is it "more stable" than Fe(III)? Why can you invoke the half-filled subshell thing to "explain" why Mn(I) is rare, but not why Fe(II) is common? A proper explanation needs to consider many energetic factors, e.g. ionisation energy, lattice energy, ... something similar to chemistry.stackexchange.com/questions/45153/… $\endgroup$ – orthocresol Dec 24 '19 at 18:41
  • $\begingroup$ I think you miss understood my question,I was asking when Mn+ can be in 4s1 3d5( a half filled stable state) , then why does it exist in 4s0 3d6? $\endgroup$ – Abhishek Dec 26 '19 at 3:40
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Compounds with Mn in formal oxidation state +1 and -1 are well known. They simply are not stable on air and in water.

In general, air-stable compounds of more active 3d-metals are ionic compounds. Their stability is a result of fine balance of energy of formation of relevant ions and energy of stabilization of said ions due to ion-ion and/or ion-dipole interactions. Naturally, $\ce{Mn^{2+}}$ interacts stronger than $\ce{Mn^{1+}}$, but isn't all that harder to make. However, by giving $\ce{Mn}$ a good partner that isn't hard dipole, but is still capable of interacting with the metal without oxidizing it, low oxidation states can be stabilized. Common partners of such type are $\ce{CO}$ and organic phosphines. If you are interested, go for introductory coordination chemistry book/chapter.

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