anonymous
  • anonymous
oxidation state of Se in MgSeO3 b. N in N2H4 (hydrazine) c. P in Na2H2P2O7 d. CinHCN
Chemistry
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schrodinger
  • schrodinger
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anonymous
  • anonymous
x=4
anonymous
  • anonymous
in se
anonymous
  • anonymous
N=-2

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.Sam.
  • .Sam.
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
anonymous
  • anonymous
p=10
anonymous
  • anonymous
@Jerry43094

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