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I was just pondering about the reaction mechanism of Cannizzaro reaction.

$\ce{H2C^{+}-O- + OH- -> H2C(OH)O-}$

$\ce{HCHO + H2C(OH)O- -> H3CO- + HCOOH}$

$\ce{H3CO- + HCOOH -> HCOO- + H3COH}$

I wondered why in 2nd step instead of receiving $\ce{H-}$ from $\ce{H2C(OH)O-}$; $\ce{H2C^{+}-O-}$ wouldn't react with NaOH to form $\ce{H2C(OH)ONa}$.

I don't think this compound will be able to exist. But don't know why! Everything seems right about it.

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    $\begingroup$ It's not clear what compounds you are talking about. The species you seem to be implying $\ce{H2C+O-}$ does not exist - the central carbon only has 6 electrons! Please clarifying what steps in the reaction you mean. $\endgroup$ – Jordan Epstein Jul 28 '16 at 18:20
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    $\begingroup$ Geez, it's imply mesomeric structure written so it should be more clear. One molecule accepts hydroxide and becomes H- donor, other functions as acceptor. $\endgroup$ – Mithoron Jul 28 '16 at 21:41
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Of course, you shouldn’t start from $\ce{H2C+-O-}$ unless you note that it is a mesomeric form of $\ce{H2C=O}$ — the more relevant mesmeric structure.

The mechanism as you wrote it and as it is typically written is given without any counterion, i.e. using just $\ce{OH-}$ and not $\ce{NaOH}$. You could instead write the mechanism using a counterion for $\ce{OH-}$ — $\ce{Na+}$, which would result in the first step being:

$$\ce{H2C=O + NaOH -> H2C(ONa)(OH)}\tag{1}$$

This is essentially the compound you are suggesting. Hence it exists, but it is a minor constituent of the overall solution. It’s not part of the second step, it is the first step. For the second step to happen, you need another molecule of $\ce{H2C=O}$, which has not yet reacted to the geminal diolate $\ce{H2C(ONa)(OH)}$.

See this scheme for the transition state of the Cannizzaro reaction and where $\ce{H2C(ONa)(OH)}$ comes into play:

Mechanism of addition and transitino state

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While the mechanism that you refer to is not correct (as Jordan mentioned in his comment), the question is why this compound - $\ce{H2C(ONa)(OH)}$ - doesn't exist. As you say yourself, you see nothing wrong with this compound. Neither do I! However, I see that this compound can undergo simple reactions in either acid or base to become simple formaldehyde. (If you don't see them yourself, I can explain further, but it may be better to just try figuring out yourself.)

There is a small note on this matter that I think is worth mentioning. While in most cases, compounds with carbonyls are much more stable than the corresponding geminal diols, formaldehyde is an exception to this rule (one of the driving forces for the equilibrium towards the carbonyls is the steric interactions in the geminal diol which are greatly reduced in $\ce{H2C(OH)2}$). Therefore, the compound that you drew up is even more likely to occur than if you had an $\ce{R}$ group instead of one or both of the hydrogens.

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    $\begingroup$ Mechanism is ok, simply OP has problem with mesomeric structures. Gem-diolate is present as reactive species - in small amount which answers another question. $\endgroup$ – Mithoron Jul 28 '16 at 21:46

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