Question

The oxidation number of oxygen in \(O_2\), \(O_2F_2\) and \(RbO_2\) are respectively}

• A. 0, +1, +2
• B. 0, +2, -1/2
• C. 0, +1, -1/2
• D. 0, 0, +1/2
• E. 0, -1, -1/2

Hint:

Remember oxygen exceptions:

• elemental oxygen: \(0\)
• superoxide: \(-1/2\)
• peroxide: \(-1\)
• with fluorine, oxygen can be positive

Answer 1

The Correct Option is C

For \(O_2\), oxygen is in elemental form: \[ \text{Oxidation number} = 0 \] For \(O_2F_2\), fluorine is always \(-1\). Let oxidation number of each oxygen be \(x\): \[ 2x + 2(-1)=0 \] \[ 2x-2=0 \Rightarrow x=+1 \] For \(RbO_2\), this is superoxide. The superoxide ion is: \[ O_2^- \] So oxidation number of each oxygen is: \[ -\frac{1}{2} \]
Hence, the correct answer is: \[ \boxed{(C)\ 0,\ +1,\ -\frac12} \]