Why is the methoxide ($\ce{CH_{3}O^{-}}$) anion negatively charged?

It has 13 valence electrons and has 13 free electrons. In spite of that the methoxide anion is negatively charged.

  • 2
    $\begingroup$ I couldn't figure out the count of 13 free electrons. What you did actually mean by free electron. $\endgroup$
    – blackSmith
    Aug 3 '13 at 19:20
  • 2
    $\begingroup$ Probably, you are having problem with formal charge calculation. $\endgroup$
    – ashu
    Aug 3 '13 at 20:50

The negative charge is due to the unpaired electron in the oxygen atom, remained after all bonds(molecular orbitals) are filled. Oxygen has 2 unpaired electron in its valence shell and is singly bonded to carbon in this molecule. Hence being un-bonded, the other electron exhibits the charge.

Check this link, hover the cursor on the $\ce{O}$-atom, you will get a graphical representation of the same.


Simply put: there are more electrons than protons.


  • Hydrogen x 3 = 1 x 3 = 3
  • Oxygen x 1 = 8 x 1 = 8
  • Carbon x 1 = 6 x 1 = 6

3 + 8 + 6 = 17 protons.


Looking at the electron shells and covalent bonds in the Lewis diagram provided by blackSmith, you have the following:

  • 4 single covalent bonds = 4 x 2 = 8
  • 3 lone pairs on oxygen = 3 x 2 = 6
  • Unseen $1s^{2}$ pair on both carbon and oxygen = 2 x 2 = 4

8 + 6 + 4 = 18 electrons.

As there is 1 more electron than proton, it has a charge of -1.

As to where that negative charge tends to be, one must create and look at the Lewis diagram where you see that the oxygen atom has an extra valence electron.


From what you have given, this question can not be answered in terms of electrons.

It is meaningless to do any electron calculation, because you have to know it has one negative charge before you can do the electron counting. How can you get a correct Lewis structure without knowing that it has one extra electron? The negative charge is part of the given condition, not your result you get from calculation.

Since I saw you are talking about 13 valence electrons, I guess you want to count valence electrons. If you add valence electron from all atoms, you will get 13 valence electron: 4 (C) + 3 * 1 (H) + 6 (O) = 13. Then because you have one negative charge, you have to add one more electron and then you get 13 + 1 = 14 valence electrons. Now everything agree with each other.

If you want to find out which atom the charge is on, you have to do the formal charge calculation. Find it here


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