Question

What is the value of n in the following equation?
$Cr{\left( {OH} \right)_4}^ - + O{H^ - } \to Cr{O_4}^{2 - } + {H_2}O + n{e^ - }$
A.$3$
B.$6$
C.$5$
D.$2$

Answer 1

Chromium is a chemical element involved in the formation of complex compounds. Chromium is a central metal atom and hydroxide molecules are ligands. The chromium complex undergoes oxidation to form chromium oxide. The electrons will be lost from the complex compound.

Complete answer:
Complex compounds are also known as coordination compounds. These are the special type of compounds as complex compounds do not lose their identity even when dissolved in water. These compounds do not give the test for those atoms or ions or groups present in that compound.
Chromium forms a complex compound with hydroxide ions. The chromium atom is a central metal atom and hydroxide ion is a ligand. The complex is $Cr{\left( {OH} \right)_4}^ -$ . This complex compound reacts with hydroxide ion and forms chromium oxide along with the liberation of water.
This chemical reaction is an oxidation process. Oxidation can be defined as loss of electrons. Thus, electrons will be lost from the chromium complex.
The reaction will be as follows:
$Cr{\left( {OH} \right)_4}^ - + O{H^ - } \to Cr{O_4}^{2 - } + {H_2}O + 6{e^ - }$
The oxidation number of chromium in $Cr{\left( {OH} \right)_4}^ -$ is $+ 3$
The oxidation number of chromium in $Cr{O_4}^{2 - }$ is $+ 6$
Thus, the number of electrons lost is $3$
But the balanced equation is different from the above equation and has the change of six electrons. Thus, the value of n in the given equation is $6$

Option B is the correct one.

Note:
The complex compound involved in the redox process. Redox process is the process where both reduction and oxidation take place. The oxidation takes place by this chromium complex. To balance the number of electrons the reaction multiplied by two. Thus, the electrons involved in the redox process is $6$.