Answer:
in parenthesis the oxidation state:
a) S(+6)O(-2)3
b) (N(+3)O(-2))+
c) (Cr(+6)O(-2)4)2-
d) V(+2)O(-2)
e) P(+1)Cl(-1)
Explanation:
a) SO3:
∴ O: has oxidation state of ( - 2 ): ⇒ ( -2 ) * 3 = - 6
⇒ for the specie to be neutral, S must have the oxidation state as the value of + 6
⇒ S (+6)O(-2)3 = +6 - 6 = 0......in parenthesis the corresponding oxidation state
b) NO+:
∴ O (-2 )
⇒ for the specie to be ( +1 ), N must have the oxidation state as the value of + 3
⇒ (N(+3)O(-2))+ = +3 - 2 = +1
c) CrO4: this specie is generally in solution as CrO42-
∴ O (-2) ⇒ - 2 * 4 = - 8
⇒ Cr (+6)
⇒ (Cr(+6)O(-2)4)2- = +6 - 8 = - 2
d) VO:
∴ O (-2)
⇒ V (+2)
⇒ V(+2) O(-2) = +2 - 2 = 0
e) PCl:
∴ Cl (-1)
⇒ P (+1)
⇒ P(+1) Cl(-1) = +1 - 1 = 0
