Give oxidation numbers for the underlined element in each of the following molec

Question

Give oxidation numbers for the underlined element in each of the following molecules or ions. (a) C_4H_10O_2: ____________ (b) Fe_3O_4: _________ (c) VO^+: _____ (d) TiF_4: ___________ (e) AsH_4 (f) Ce(SO_4)_2: Use the following data to place the fictitious metals Q, R, T, x, A, and E into an activity series. R + A^+ rightarrow N.R. ________ Q + T^+ rightarrow T + Q^+ __________ E + T^+ rightarrow N.R. _________ R + E^+ rightarrow N.R. __________ E + X^+ rightarrow E^+ + X ______ E + Q^+ rightarrow N.R. _________ X^+ + A rightarrow A + X^+ ________ E + A^+ rightarrow E^+ + A

Explanation / Answer

a). C4H10O2

We know that oxidation number of oxygen is -2 and hydrogen is +1 respectively. Moreover, the overall molecule is neutral. Let us take the oxidation number of C as X

Then we will get a simple equation as

4X + 10 (+1)+ 2 (-2) = 0 (obtained by the products oxidation number and the number of each atoms)

Thus

4X= -6 or

X = -6/4 = -1.5

b) Fe3O4

As discussed earlier, the oxidation number of oxygen is -2 and let the oxidation number of Fe be X

Then

3X + 4 (-2) = 0 (Right hand side is zero since the overall molecule is neutral)

3X = 8 or

X = 8/3 = +2.66

c). VO+,

As discussed earlier, the oxidation number of oxygen is -2 and let the oxidation number of Fe be X

Then

X+ -2 = 1 (Right hand side is one since the overall molecule is charged by +1)

Or

X = +3

d). TiF4

F is a halogen and its oxidation number is -1

X + 4 (-1) = 0

X = 4

e) AsH4+

In the case of metal hydrides, the oxidation number of H is -1

Thus,

X + 4 (-1) = +1

X = +5

f) Ce(SO4)2

In the case of SO42-, S has an oxidation number of + 6 and O as -2

Thus we will get an equation for unknown Cs

X + 2(1 (+6)+ 4 (-2)) = 0

X = +4

As per the policy of Chegg first question is answered.

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