Question

What is the oxidation state of Chromium in ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$?
(A) 12
(B) 1
(C) 6
(D) 3
(E) 2

Answer 1

Hint: Oxidation state shows the total number of electrons which have been removed from an element (a positive oxidation state) or added to an element (a negative oxidation state) to get to its present state.

Step-by-Step Solution:
Let us first try and define what the oxidation state of an element in a compound really is before getting to the specifics of this question.
The oxidation state, sometimes referred to as oxidation number, describes the degree of oxidation (loss of electrons) of an atom in a chemical compound. Conceptually, the oxidation state, which may be positive, negative or zero, is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic, with no covalent component.
For instance, in NaCl, Na has the oxidation state of +1 since it is a Group 1 Alkali metal. Cl would have an oxidation number of -1 to make the sum of the oxidation states 0. In \[MgC{{l}_{2}}\], Mg has the oxidation state of +2, since it is a Group 2 Alkaline Earth metal. Cl would have an oxidation state of -1, and since there are 2 Cl atoms, the overall charge of the species would again be 0. Another example is that the oxidation state of Oxygen is -2 in \[O{{H}^{-}}\], \[{{H}_{2}}O\]
Using this idea, we now know that the oxidation state of K is +1 and that O is -2 Now, let us use this data to compute the OS of Cr in the given compound.
$\begin{align}
  & 2\times (OS\text{ }of\text{ }K)+2\times (OS\text{ }of\text{ Cr})+7\times (OS\text{ }of\text{ O})=0 \\
 & \Rightarrow 2\times (+1)+2\times (OS\text{ }of\text{ Cr})+7\times (-2)=0 \\
 & \Rightarrow 2+2\times (OS\text{ }of\text{ Cr})-14=0 \\
 & \Rightarrow 2\times (OS\text{ }of\text{ Cr})=12 \\
 & \Rightarrow (OS\text{ }of\text{ Cr})=6 \\
\end{align}$
Therefore, the answer to this question is c) 6.

Note: The terms “oxidation state” and “oxidation number” are often used interchangeably. The transfer of electrons is described by the oxidation state of the molecule. One might mistake formal charge for oxidation state but they are different. Oxidation state is commonly used to determine the changes in redox reactions and is numerically similar to valence electrons, but different from formal charge. Formal charge determines the arrangement of atoms and the likelihood of the molecule existing.