Oxidation state of 'V' in Rb4Na[HV10O28] is:
Correct answer: A. +5
- A. +5
- B. +6
- C. +7/5
- D. +4
Explanation
The oxidation state of vanadium in Rb4Na[HV10O28] can be determined by considering the overall charge of the compound. The compound is neutral, meaning the sum of the oxidation states of all elements must equal zero. Rubidium (Rb) and sodium (Na) typically have oxidation states of +1 each, and hydrogen (H) has +1. Oxygen (O) has an oxidation state of -2. Therefore, we have:4(+1) + 1(+1) + 1(+1) + 10(x) + 28(-2) = 0Solving for x (oxidation state of vanadium), we find that x = +5. Thus, the correct oxidation state of vanadium in this compound is +5.The other options are incorrect because they do not satisfy the charge balance of the compound.
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