1.

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

Answer»

`+3`
`+5`
`+7`
`+6`

Solution :`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`


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