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To find the correct oxidations state of P in PO4 3 the Phosphate ion and each element in the ion we use a few rules and some simple math more

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In this post we will determine the oxidation numbers in the phosphate ion PO 43 To determine the oxidation number of an atom s in a molecule or an ion start with the known oxidation numbers and

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In PO4 3 there are 5 atoms sharing 4 bonds for which we need to assign the electrons in order to find the oxidation numbers Once you have drawn the lewis diagram for PO 4 3 you can look at

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We know that the charge on an oxygen atom is 2 PO3 4 x 4 2 3 x 3 8 5 x 5 Thus the oxidation number of P in PO3 4 is 5 Oxidation number in simple

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Therefore the oxidation number of phosphorus in P O 4 3 P 4 O 10 and P 2 O 7 4 is 5 So the correct answer is Option A Note Phosphorous is a group 15 element It has 5 electrons in the valence

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To determine the oxidation number of phosphorus P in the phosphate ion PO4 3 we need to use the rules for assigning oxidation numbers The sum of the oxidation numbers of all

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The oxidation number of phosphorus in the phosphate ion P O43 is 5 This is determined by calculating the contributions from the oxygen atoms and setting up an equation to find

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Since there are four oxygen atoms in the phosphate ion PO43 the total oxidation number contribution from oxygen is 8 This leaves phosphorus P with an oxidation number of 5 to

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The correct option is d 5 6 and 6Explanation Let oxidation number of P in PO43 be x x 4 2 3 x 5Let oxidation number of S in SO42 be y y 4 2 2 y 6Let oxidation number of