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Answer:
- P in PBr3 is +3.
- N in N2 is 0.
- As in H3AsO4 is +5.
Explanation:
Hello!
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:
[tex]P^xBr_3^-\\\\x-3=0\\\\x=+3[/tex]
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:
[tex]H_3^+As^xO_4^{-2}\\\\3+x-8=0\\\\x=8-3\\\\x=5[/tex]
Thus, the oxidation state is +5.
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