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