Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

The oxidation number of cobalt in $K[Co{{(CO)}_{4}}]$ is _______.
(a) +1
(b) +3
(c) -1
(d) -3

seo-qna
SearchIcon
Answer
VerifiedVerified
456k+ views
Hint: Oxidation state or oxidation number of a species is the total number of electrons that an atom can either gain or lose in order to form a chemical bond with another atom. By knowing the oxidation state of the other atoms present in a compound, we can calculate the oxidation state of the central atom of the compounds. Here, the oxidation state of CO is zero and the oxidation state of potassium (K) is +1.

 Complete step by step answer:
Oxidation number can also be described as the degree of oxidation of an atom in a chemical compound. It can be negative, positive and zero and are integers.
For calculating the oxidation state of $K[Co{{(CO)}_{4}}]$ we should know the oxidation state of potassium, and carbon monoxide.
For potassium, the oxidation state is always +1 and for CO oxidation state is 0.
We can also confirm that the given compound is chargeless or neutral because there is no charge mentioned in its molecular formula. Thus, the total oxidation state of the compound is 0.
Let us consider the oxidation state of cobalt in the compound as X. So, the total oxidation state is given as,
(Oxidation state of potassium) + X + 4(Oxidation state of CO) = Total oxidation state of the complex.
Oxidation state of K = +1.
Oxidation state of CO = 0.
Total oxidation state of the compound = 0.
Substituting these values we get,
1 + X + (4x0) = 0,
X = -1.

Therefore, the oxidation state of Cobalt X is equal to -1.

So, the correct answer is “Option (C)”.

Note: While calculating the oxidation state of any compound, we should always remember to multiply the no. of atoms with the oxidation state of the same atom present in the compound. For example, in the above complex, there are 4 CO ligands and it is multiplied with its oxidation state 0 while calculating the oxidation state of cobalt. If there were two potassium present in the complex, we have to multiply 2 with +1, so the total oxidation state becomes +2.