
Which of the following is incorrect about \[{H_4}{P_2}{O_5}\]?
(A) it contains \[P\] in \[ + 5\]oxidation state
(B) it is a dibasic acid
(C) it is strongly reducing in nature
(D) it contains one \[P - O - P\] bond
Answer
585k+ views
Hint: When compounds have two hydroxyl groups attached to the central atom, then they are known as dibasic and the reducing property of compounds depends upon the number of hydrogens which are attached directly to the central atom.
Complete step by step answer:
The compound \[{H_4}{P_2}{O_5}\] known as pyrophosphorous acid is represented as shown below.
The oxidation state of \[P\] in this acid is calculated by the formal charge of the hydrogen atom and the oxygen atoms which are \[ + 1\] and \[ - 2\]respectively. Suppose \[y\] is the oxidation state of\[P\]then,
\[\begin{array}{c}
4( + 1) + 2y + \left( {5 \times \left( { - 2} \right)} \right) = 0\\
2y = + 6\\
y = + 3
\end{array}\]
Therefore, the oxidation state of\[P\] is \[ + 3\] and thus the option (A) is incorrect.
Let’s see the other options,
As we can see in the picture of \[{H_4}{P_2}{O_5}\] ,there are two hydroxyl groups( \[OH\]) which are attached with two phosphorus atoms so \[{H_4}{P_2}{O_5}\] is a dibasic acid .The \[OH\] group donates the lone pair of electron to \[P\] atom .therefore, the option (B) is correct.
Reducing property-when the compound has tendency to release \[{H^ + }\] ion which is directly attached to the central atom as in \[{H_4}{P_2}{O_5}\] there are two \[P - H\] bond from which \[P\] can release \[{H^ + }\] ion. Therefore, the option (C) is also correct.
The option (D) is also correct because as we can see in the above diagram, \[{H_4}{P_2}{O_5}\] contains one\[P - O - P\] bond.
The correct options are B, C & D.
Note:
Reducing property of compounds is due to the electron deficiency of the central atom and it attracts electrons of hydrogen and makes it \[{H^ + }\] ion.
Complete step by step answer:
The compound \[{H_4}{P_2}{O_5}\] known as pyrophosphorous acid is represented as shown below.
The oxidation state of \[P\] in this acid is calculated by the formal charge of the hydrogen atom and the oxygen atoms which are \[ + 1\] and \[ - 2\]respectively. Suppose \[y\] is the oxidation state of\[P\]then,
\[\begin{array}{c}
4( + 1) + 2y + \left( {5 \times \left( { - 2} \right)} \right) = 0\\
2y = + 6\\
y = + 3
\end{array}\]
Therefore, the oxidation state of\[P\] is \[ + 3\] and thus the option (A) is incorrect.
Let’s see the other options,
As we can see in the picture of \[{H_4}{P_2}{O_5}\] ,there are two hydroxyl groups( \[OH\]) which are attached with two phosphorus atoms so \[{H_4}{P_2}{O_5}\] is a dibasic acid .The \[OH\] group donates the lone pair of electron to \[P\] atom .therefore, the option (B) is correct.
Reducing property-when the compound has tendency to release \[{H^ + }\] ion which is directly attached to the central atom as in \[{H_4}{P_2}{O_5}\] there are two \[P - H\] bond from which \[P\] can release \[{H^ + }\] ion. Therefore, the option (C) is also correct.
The option (D) is also correct because as we can see in the above diagram, \[{H_4}{P_2}{O_5}\] contains one\[P - O - P\] bond.
The correct options are B, C & D.
Note:
Reducing property of compounds is due to the electron deficiency of the central atom and it attracts electrons of hydrogen and makes it \[{H^ + }\] ion.
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