Question

The oxidation number of Cr in \[\left[ Cr{{\left( N{{H}_{3}} \right)}_{6}} \right]C{{l}_{3}}\]is
(A) 8
(B) 6
(C) 4
(D) 3

Answer 1

Hint: The given compound is a coordinate complex in which the first part of the compound is non-ionizable (cation) and another part is an anion. In the non-ionizable part, central metal Cr is surrounded by six neutral (no charge) ammine ligands. Thus, the oxidation state of Cr is the same as the charge on the whole non-ionizable (cation) part, and the charge on chlorine is -1.

Complete Step by Step Solution:
The first part (non-ionizable part) of a compound or salt is always positive and the second one is negative. Write down both separately such as \[{{\left[ Cr{{\left( N{{H}_{3}} \right)}_{6}} \right]}^{+}}\]and \[C{{l}^{-}}\].

Now cross multiply with both charges,the number of an element means the number of chlorine atoms (3) will multiply with a positive charge of the first part of the compound and the number of the first part (1) will multiply with a negative sign of chlorine atom such as \[{{\left[ Co{{\left( N{{H}_{3}} \right)}_{6}} \right]}^{+3}}\]and \[C{{l}^{-}}\]. As ammonia legend is neutral thus oxidation state of central metal (Co) will be the same as the charge on the first part of the compound which is equal to +3 such as

Let the oxidation state of Cr is x such as
x + 6(0) = 3 (by cross multiply)
x = 3
Thus, the correct option is D.

Note: Also another way to find the oxidation state of the compound is to just add all the oxidation state (multiply by the number of elements) element and equate to zero because all compounds composed of positive (cation) and negative (anion) charge, making compound completely neutral such as:
Let the oxidation state of Cr is x, the charge on ammonia is neutral, and the charge on chlorine is -1 thus
x + 6(0) + 3(-1) = 0
x – 3 = 0
x = 3