how to calculate oxidation state of pt in NH3
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There is no such thing as the "oxidation state of a compound". Each element has an oxidation state. We sort of stretch the issue to say that an ion has an oxidation state. The sulfate ion (SO4^2-) has an "effective oxidation state" of -2. In sulfate, sulfur has an oxidation state of +6 and oxygen, as it always has in a compound, is -2
The sum of all of the oxidation states in a compound is zero. That's a given.
You can skip the water in the first one. The salt appears to be hydrated, but we don't use a "color" (:) to indicate hydration. V2(SO4)3 is the anhydrous salt. We know that the sulfate ion has an "oxidation state" of -2, and since the sum of the oxidation states is zero, vanadium must have an oxidation state of +3.
V2(SO4)3 ==> 2(+3) + 3(-2) = 0
In K4V(CN)6, the cyanide ion is -1, and K is always +1, so vanadium must have an oxidation state of +2.
K4V(CN)6 ==> 4(+1) + 2 + 6(-1) = 0
In [Pt(NH3)5Cl]Br3 the three Br- gives us -3, so that everything inside [ ] must be +3. NH3 is neutral and Cl is -1, so Pt is +4. Again, the sum of all the oxidation numbers is zero.
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Oxidation state of N is −III.
Oxidation number is the charge left on the central atom when each of the bonding pairs are broken, with the charge going to the most electronegative atom. If I break the N−H bonds I am left with N3− and 3×H+.
The sum of all of the oxidation states in a compound is zero. That's a given.
You can skip the water in the first one. The salt appears to be hydrated, but we don't use a "color" (:) to indicate hydration. V2(SO4)3 is the anhydrous salt. We know that the sulfate ion has an "oxidation state" of -2, and since the sum of the oxidation states is zero, vanadium must have an oxidation state of +3.
V2(SO4)3 ==> 2(+3) + 3(-2) = 0
In K4V(CN)6, the cyanide ion is -1, and K is always +1, so vanadium must have an oxidation state of +2.
K4V(CN)6 ==> 4(+1) + 2 + 6(-1) = 0
In [Pt(NH3)5Cl]Br3 the three Br- gives us -3, so that everything inside [ ] must be +3. NH3 is neutral and Cl is -1, so Pt is +4. Again, the sum of all the oxidation numbers is zero.
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Oxidation state of N is −III.
Oxidation number is the charge left on the central atom when each of the bonding pairs are broken, with the charge going to the most electronegative atom. If I break the N−H bonds I am left with N3− and 3×H+.
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