Question

Oxidation states of $P$ in $ H_4 \, P_2 \, O_5, H_4, P_2, O_6 , H_4 P_2 O_7 $ respectively are

• A. #ERROR!
• B. - 5, + 3 and + 4
• C. #ERROR!
• D. #ERROR!

Answer 1

The Correct Option is D

 To determine the oxidation states of phosphorus in the given compounds, we need to apply the rules for finding oxidation numbers. Let's calculate systematically for each compound:

1. $H_4P_2O_5$
   • Let the oxidation state of phosphorus (P) be \(x\).
   • The oxidation state of hydrogen (H) is +1 and that of oxygen (O) is -2.
   • There are 4 hydrogens, 2 phosphorus, and 5 oxygens. Thus, the compound is neutral, so we equate the sum of oxidation states to zero:
   • \(4(+1) + 2(x) + 5(-2) = 0\)
   • \(4 + 2x - 10 = 0\)
   • \(2x - 6 = 0\)
   • \(2x = 6\)
   • \(x = +3\)
2. $H_4P_2O_6$
   • Let the oxidation state of phosphorus (P) be \(x\).
   • The sum of the oxidation states will also be zero.
   • \(4(+1) + 2(x) + 6(-2) = 0\)
   • \(4 + 2x - 12 = 0\)
   • \(2x - 8 = 0\)
   • \(2x = 8\)
   • \(x = +4\)
3. $H_4P_2O_7$
   • Again, let the oxidation state of phosphorus (P) be \(x\).
   • \(4(+1) + 2(x) + 7(-2) = 0\)
   • \(4 + 2x - 14 = 0\)
   • \(2x - 10 = 0\)
   • \(2x = 10\)
   • \(x = +5\)

Therefore, the oxidation states of phosphorus in $H_4P_2O_5$, $H_4P_2O_6$, and $H_4P_2O_7$ are +3, +4, and +5 respectively. However, the original question specifies the option is an error, suggesting this may be a trick question with the necessity to recognize the lack of a proper option indicative of conceptual understanding. To summarize:

Compound	Oxidation State of P
\(H_4P_2O_5\)	+3
\(H_4P_2O_6\)	+4
\(H_4P_2O_7\)	+5