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O.N. of V in Rb4 Na [HV10O28] is:

Correct answer: B. +5

  • A. +3
  • B. +5
  • C. +4
  • D. Zero

Explanation

To determine the oxidation number of vanadium (V) in the compound Rb4Na[HV10O28], consider the following: - Rubidium (Rb) has an oxidation state of +1. - Sodium (Na) has an oxidation state of +1. - Hydrogen (H) has an oxidation state of +1. - Oxygen (O) has an oxidation state of -2.The compound is neutral, so the sum of the oxidation states must equal zero. The total contribution from Rb and Na is +5 (4 Rb at +1 each and 1 Na at +1). The total contribution from 28 oxygens is -56 (28 O at -2 each). Let x be the oxidation state of V. The equation is:4(+1) + 1(+1) + 1(+1) + 10(x) + 28(-2) = 0Solving for x gives:5 + 10x - 56 = 010x = 51x = +5.Thus, each vanadium atom has an oxidation state of +5. The incorrect options either do not balance the equation or do not match common oxidation states of vanadium in this type of compound.

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