The oxidation number of ''V'' in the complex Rb4Na[HV10O28] is

Rb4Na[HV10O28] 4(+1)+1+[1+10x+28(-2)] = 0 5+[10x - 55] = 0 10x = 50 x = +5 Alternate method [HV10O8]-5 +1+10(x)+28(-2)=-5 x = +5

The oxidation number of ''V'' in the complex Rb4Na[HV10O28] is

Rb4Na[HV10O28] 4(+1)+1+[1+10x+28(-2)] = 0 5+[10x - 55] = 0 10x = 50 x = +5 Alternate method [HV10O8]-5 +1+10(x)+28(-2)=-5 x = +5

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The oxidation number of ''V'' in the complex Rb₄Na[HV₁₀O₂₈] is

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answer is B.

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Detailed Solution

Rb₄Na[HV₁₀O₂₈]
4(+1)+1+[1+10x+28(-2)] = 0
5+[10x - 55] = 0
10x = 50
x = +5
Alternate method
$\large R{b_4}Na\left[ {H{V_{10}}{O_{28}}} \right]\;\; \rightleftharpoons \;\;4R{b^ + }\; + \;N{a^ + }\; + \;{\left[ {H{V_{10}}{O_{28}}} \right]^{ - 5}}$
[HV₁₀O₈]^-5
+1+10(x)+28(-2)=-5
x = +5