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19 June, 14:19

2.16 Determine the oxidation states of the element emboldened in each of the following species: (a) S03, (b) NO+, (c) Cr 04, (d) VO, e) PCI

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  1. 19 June, 17:11
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    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
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