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 = 0in 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

explanation:

detials on each model with digrams

elements make up everything, and they combine to make compounds.

the answer is c