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

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