Question

Oxidation number of potassium in \( K_{2}O \), \( K_{2}O_{2} \) and \( KO_{2} \), respectively, is:

• A. \(+2\), \(+1\) and \(+\frac{1}{2}\)
• B. \(+1\), \(+1\) and \(+1\)
• C. \(+1\), \(+4\) and \(+2\)
• D. \(+1\), \(+2\) and \(+4\)
• E.

Hint:

Don't let the complex oxygen states (peroxides and superoxides) confuse you. In almost every chemical environment, Group 1 metals (Li, Na, K, Rb, Cs) are strictly \(+1\). It is the oxidation state of oxygen that varies to balance the charge.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The question asks for the oxidation state of potassium (K) in three different oxygen-containing compounds.
Potassium is an alkali metal, belonging to Group 1 of the periodic table.
A fundamental rule of assigning oxidation numbers is that alkali metals always have an oxidation state of +1 in all of their compounds.
Step 2: Key Formula or Approach:
1. The oxidation state of Group 1 metals (Li, Na, K, Rb, Cs) in their compounds is strictly $+1$.
2. The algebraic sum of oxidation states of all atoms in a neutral molecule must be zero.
We can use the second rule to verify the varying oxidation states of oxygen in these compounds, while K remains constant.
Step 3: Detailed Explanation:
According to the rules, since potassium is a Group 1 element, its oxidation state must be $+1$ in $K_2O$, $K_2O_2$, and $KO_2$.
Let's verify this by calculating the oxidation state of oxygen in each compound to see how they differ:
1. In Potassium oxide ($K_2O$): Let the oxidation state of O be $x$.
\[ 2(+1) + x = 0 \implies x = -2 \] Here, oxygen is a normal oxide ion ($O^{2-}$). Potassium is $+1$.
2. In Potassium peroxide ($K_2O_2$): Let the oxidation state of O be $y$.
\[ 2(+1) + 2y = 0 \implies 2y = -2 \implies y = -1 \] Here, oxygen is a peroxide ion ($O_2^{2-}$). Potassium is $+1$.
3. In Potassium superoxide ($KO_2$): Let the oxidation state of O be $z$.
\[ 1(+1) + 2z = 0 \implies 2z = -1 \implies z = -\frac{1}{2} \] Here, oxygen is a superoxide ion ($O_2^{-}$). Potassium is $+1$.
In every single case, regardless of the type of oxide formed, the oxidation state of potassium remains fixed at $+1$.
Step 4: Final Answer:
The oxidation numbers of potassium are $+1$, $+1$, and $+1$ respectively.