The average oxidation state of tungsten, W in Na₃PW₁₂O₄₀ is +6.67
The correct answer to the question is Option D. +6.67
The oxidation number of tungsten, W in Na₃PW₁₂O₄₀ can be obtained as follow:
Na₃PW₁₂O₄₀ = 0 (group state)
3Na + P + 12W + 40O = 0
3(1) + (–3) + 12W + 40(–2) = 0
3 – 3 + 12W – 80 = 0
12W – 80 = 0
Collect like terms
12W = 0 + 80
12W = 80
Divide both side by 12
W = 80 / 12
W = +6.67
Thus, the oxidation number of tungsten, W in Na₃PW₁₂O₄₀ is +6.67
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