A) + 3, + 5, + 4
B) + 5, + 3, + 4
C) + 5, + 4, + 3
D) + 3, + 4, + 5
Correct Answer: D
Solution :
Key Idea Oxidation state of H is + 1 and that of O is - 2. Let the oxidation state of P in the given compounds is x. In \[{{\text{H}}_{\text{4}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}\text{,}\] \[(+1)\times 4+2\times x=(-2)\times 5=0\] \[4+2x-10=0\] \[2x=6\] \[x=+3\] \[\therefore \] In\[{{H}_{4}}{{P}_{2}}{{O}_{6}},\] \[(+1)\times 4+2\times x+(-2)\times 6=0\] \[4+2x-12=0\] \[2x=8\] \[x=+4\] \[\therefore \] 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 \[{{H}_{4}}{{P}_{2}}{{O}_{5}},\] \[{{H}_{4}}{{P}_{2}}{{O}_{6}}\]and \[{{H}_{4}}{{P}_{2}}{{O}_{7}}\] are + 3, + 4 and + 5 respectively.
You need to login to perform this action.
You will be redirected in
3 sec
