Question

The oxidation number of chromium in sodium tetraflurooxochromate complex is:
(A) II
(B) IV
(C) VI
(D) III

Answer 1

Hint: Given compound,\[N{{a}_{2}}\left[ Cr{{F}_{4}}O \right]\]is a coordinate complex. To calculate the oxidation state of chromium, Cr just assumes its oxidation state, let’s say x, and adds to it the oxidation state of ligands and sodium (cation). Then equate the addition of oxidation state of all elements of the coordinate complex to zero and find the value of x.

Complete Step by Step Solution:
As any compound or any salt is formed with a cation and an anion, where cation is in first place and anion in second place. Thus, in a given coordinate complex, \[N{{a}_{2}}\]is a cation and\[Cr{{F}_{4}}O\] is an anion which is a non-ionizable part. Chromium, Cr which is a central metal surrounded by five ligands (four fluorine atoms and one oxygen atom) forming a non-ionizable anionic part. And sodium which is in the first place is a cation forming a complex neutral.

To get the oxidation state of chromium, just assume the oxidation state of chromium, say x, and add to it the oxidation state of its ligands (fluorine has -1 charge, oxygen has -2 charge and sodium has +1 charge). Also, multiply the oxidation state by the number of atoms present.

There are four fluorine (\[{{F}_{4}}\]) present thus, multiply its charge by four such as 4(-1), oxygen (O) charge will multiply with one as one oxygen ligand is present such as 1(-2) and two sodium (\[N{{a}_{2}}\]) are present so its charge will multiply by two such as 2(+1). Now add all the oxidation states of all the elements in the coordinate complex so that the whole compound will be electrically neutral such as
2(+1) + x + 4(-1) + 1(-2) = 0
+2 + x – 4 – 2 = 0
2 + x – 6 = 0
x – 4 = 0
x = +4
Thus, the correct option is B.

Note: To calculate the oxidation state of chromium, you can also use the trick of cross multiplication. To do this, just separate both anionic and cationic parts such as\[N{{a}^{+}}\]and \[{{\left[ Cr{{F}_{4}}O \right]}^{-}}\]. Then cross multiple subscripts (number of cation and number of anion) such as\[N{{a}^{+}}\]and \[{{\left[ Cr{{F}_{4}}O \right]}^{-2}}\]. Now to calculate the oxidation state of chromium(say x) just add the oxidation state of ligand to it (elements of non-ionizable part) and equate it to the modified oxidation state of the same non-ionizable part such as
x + 4(-1) + 1(-2) = -2
x – 4 – 2 = -2
x – 6 = -2
x = -2 + 6
x = +4