I'm thinking: $\ce{CO}$ is neutral so $\ce{O}$ is -2 and $\ce{C}$ +2

$\ce{CN}$ is negative one so $\ce{C}$ is +2 and $\ce{N}$ -3

$\ce{CH3}$ is positive one so $\ce{H}$ is +1 and $\ce{C}$ is -2

$\ce{W}$ is neutral so $\ce{W}$=0

is this right?

  • $\begingroup$ Why is the compound a compound? Why is the neutral W there? And what exactly does your class use as the definition of "oxidation state" anyway? What would the oxidation states of molecular oxygen and nitrogen be? (i.e. <O=O> and |N≡N|) $\endgroup$
    – MSalters
    Commented Sep 25, 2014 at 20:07

2 Answers 2



Compound is neutral, So the sum of oxidation state of all the element should be ZERO.

In what you have got, is sum of oxidation state of all the element is zero? This way you can check whether you are right or wrong.


I know its an old question but just wanted to make a technical correction . The final integral values deduced by you are quiet correct but the method of arriving at OS of W looks clumsy . As Freddy says , the compound is neutral , hence sum of all OS must be zero . So we get OS of W as zero not by

"W is neutral so W=0"

but rather using the fact that the compound is neutral and the other elements (which are in the form of ligands) form neutral ligands .

  • $\begingroup$ This is marginally an answer, but please be cautious that the answer space is for answers only, not for new questions and not for comments. Once you attain 50 rep, you can comment everywhere on the site. I mention this because you've had a few instances where you've misused the answer area. $\endgroup$
    – jonsca
    Commented Feb 14, 2015 at 10:31
  • $\begingroup$ @jonsca the OP's question was whether the answers were correct . I have mentioned in my answer that they are correct but just the methodology needed to be refined . I am sorry if I did something wrong . $\endgroup$
    – Del Pate
    Commented Feb 14, 2015 at 12:25
  • $\begingroup$ Perhaps my choice of the word "marginally" was a poor one. I just meant that you'd had a couple of instances where there had been some confusion, and I wanted to just give you a reminder about it. I apologize if it came out in a condescending manner. $\endgroup$
    – jonsca
    Commented Feb 14, 2015 at 23:27

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