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

Updated On: Apr 17, 2024
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  • - 5, + 3 and + 4
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The Correct Option is D

Solution and Explanation

Key Idea Oxidation state of H is + 1 and that of O is - 2. Let the oxidation state of P in the given compounds isx. In $ H_4 P_2 O_5 $ $ ( + 1) \times 4 + 2 \times x + ( - 2) \times 5 = 0 $ $4 + 2x - 10 = 0$ $2x = 8$ $\therefore$ $ x = + 4$ In $ H_4 P_2 O_7$ $ ( + 1 ) \times 4 + 2 \times x + ( - 2) \times 7 = 0 $ $4 + 2x - 14 = 0$ $2x = 10$ $\therefore$ $ x = + 5$ Thus, the oxidation states of $P$ in $ H_4 P_2 O_6, \, H_4 P_2 O_6 $ and $ H_4 P_2 O_7 $ are $+3, + 4$ and $+5$ respectively
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Concepts Used:

P-Block Elements

  • P block elements are those in which the last electron enters any of the three p-orbitals of their respective shells. Since a p-subshell has three degenerate p-orbitals each of which can accommodate two electrons, therefore in all there are six groups of p-block elements.
  • P block elements are shiny and usually a good conductor of electricity and heat as they have a tendency to lose an electron. You will find some amazing properties of elements in a P-block element like gallium. It’s a metal that can melt in the palm of your hand. Silicon is also one of the most important metalloids of the p-block group as it is an important component of glass.

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