Question

oxidation state of
Se in MgSeO3
b. N in N2H4 (hydrazine)
c. P in Na2H2P2O7
d. CinHCN

Answer 1

x=4

Answer 2

in se

Answer 3

N=-2

Answer 4

MgSeO3

You know that Mg is in group II, All group II elements have an oxidation state of +2.
Also, Oxygen always have an oxidation state of -2. (Execpt for \(H_2O_2\)).

So, since MgSeO3 is in ground state, all oxidation states in MgSeO3 must add up to zero.

2+Se+3(-2)=0

Se=+4

You can apply the same rule for others,

b. Hydrogen in N2H4 can be +1 or -1, because All Group I elements have an oxidation state of +1 (Hydrogen is an exception here). Since Hydrogen is on the right hand side of the compound, usually it will be negative, so Hydrogen is -1, try to figure that out.

c. P in Na2H2P2O7, Sodium is in Group I, Hydrogen is more to the left, we use our convention, +1 since its more to the left, Oxygen as I've said its -2. So, apply the same technique as the first one,

2(1)+2(1)+2P+7(-2)=0
P=5

Answer 5

p=10

Answer 6

@Jerry43094