The oxidation number of ..V.. in `Rb_(4) Na[HV_(10)O_(28)]` is

Options

A. `+3`
B. `+5`
C. `+7`
D. `+6`

Answer

B

Explanation

`Rb_4Na[HV_(10)O_(28)]implies[HV_(10)O_(28)]^(-5)`
`({:(4Rb^(+)=+4),(Na^(+)=+1),(" "bar" "),(" "+5),(" "ul" "):})`
O.s of `V =1 (1)+10(x) +28 (-2)=-5`
`10 x - 55 =-5 implies 10x = 50 implies x = 5`