Question

Oxidation number of Co in $ {{\left[ Co{{(N{{H}_{3}})}_{3}}{{({{H}_{2}}O)}_{2}}Cl \right]}^{+}} $ is
(A) 1
(B) 2
(C) 3
(D) 4

Answer 1

Hint: Assume the charge on Co to be ’x’, the charge on ammonia and water is zero i.e. they are neutral ligands; charge on chlorine is $ -1 $ . Equate these charges equal to the charge on the compound which is $ +1 $ and find the charge on Co. This will give us the oxidation number of Co in the given compound.

Complete Step by Step Solution
Oxidation number is also known as oxidation state which gives the total number of electrons that an atom either gains or losses in order to form a chemical bond with another atom.
Ammonia and water are neutral ligands so charge on them is zero. For Co let’s assume charge to be ‘x’ and for chlorine we have charge $ -1 $ . So, we get,
 $ x+3\times 0+2\times 0+(-1)=+1 $
 $ \begin{align}
  & x+(-1)=+1 \\
 & x=1+1 \\
 & x=2 \\
\end{align} $
So, charge on Co is 2 which gives its oxidation state as 2.
So, option-(B) is correct. Here, we have 3 ammonia atoms so $ 3\times 0 $ is done and 2 water atoms so $ 2\times 0 $ is done.

Note
Atoms participating in redox reaction have oxidation number which tells about the ability of an atom to acquire, donate, or share electrons. The oxidation state can be positive, negative or zero. Redox reactions are the one in which oxidation and reduction occurs simultaneously. Metals undergo oxidation by losing electrons and non-metal undergo reduction by gaining electrons. In its pure elemental form, an atom has an oxidation number of zero. The oxidation number of an ion is the same as its charge. The sum of the oxidation numbers in a neutral compound is equal to zero.