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Hello students, hope you are doing great.
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In this question we are asked to find out the oxidation number of given species in the given compounds.
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In the first example we are asked to find out oxidation number of barium in barium nitrate.
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We know that barium nitrate is made up of a cation that is barium and it is made up of an anion that is nitrate which is having minus one charge on it.
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So let x be the oxidation state of barium minus we have two nitrate ions so the charge will be of minus two which is equals to zero as the overall charge of the compound is zero.
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Solving this x is equals to plus two which means barium is here present in plus two oxidation state.
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In the next part of the question we are asked to find out oxidation state of sulfur in na2so4.
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Now here let x be the oxidation state of sulfur.
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We know that sodium is present in plus one oxidation state and oxygen is present in minus two oxidation state and let x be the oxidation state of sulfur.
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So there are two sodium so two plus x there are four oxygen and four into twos are eight minus eight is equals to zero as the overall charge of the material of this compound is zero.
01:29
Solving for the value of x, x is equals to plus six which means sulfur is present in plus six oxidation state in the given compound.
01:42
Next part of the question says that we have to find out oxidation state of chromium in k2cr2o4...