Answer:
- P in PBr3 is +3.
- N in N2 is 0.
- As in H3AsO4 is +5.
Step-by-step explanation:
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In this case, since the determination of the oxidation states is performed by using the well-known charge balances, we can proceed as shown below:
- P in PBr3: Here, bromide ions have an oxidation state of -1, so we follow:
![P^xBr_3^-\\\\x-3=0\\\\x=+3](https://img.qammunity.org/2022/formulas/chemistry/college/ejnksukipx8fumc8bejo247bqher6smgei.png)
Thus, the oxidation state is +3.
- N in N2: Here, since nitrogen is bonding with nitrogen and it is neutral, we infer its oxidation state is 0.
- As in H3AsO4: Here, oxygen is -2 and hydrogen +1, so we follow:
![H_3^+As^xO_4^(-2)\\\\3+x-8=0\\\\x=8-3\\\\x=5](https://img.qammunity.org/2022/formulas/chemistry/college/2s3vqaeb2kxih2ataaz01mlljkckta9u67.png)
Thus, the oxidation state is +5.
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