A) +5, +6 and +6
B) +3, +6 and +5
C) +5, +3 and +6
D) -3, +6 and +6
Correct Answer: A
Solution :
| Key Idea (i) Sum of oxidation states of all atoms = charge of ion. |
| (ii) Oxidation number of oxygen = - 2. |
| Let the oxidation state of P in \[{{C}_{2}},{{C}_{3}},{{C}_{5}}\]is x. |
| \[{{C}_{6}}\] |
| \[{{F}_{2}}\]\[B{{r}_{2}}\] |
| \[{{I}_{2}}\] |
| \[C{{l}_{2}}\] |
| Let the oxidation state of S in \[\frac{M}{32}\] is y. |
| \[1.25\times {{10}^{-5}}\] |
| \[1.25\times {{10}^{-6}}\]\[6.25\times {{10}^{-4}}\] |
| \[1.25\times {{10}^{-4}}\] |
| \[-C{{F}_{2}}-C{{F}_{2}}-{{)}_{n}}\] |
| Let the oxidation state of Cr in \[{{\left( -C{{H}_{2}}-\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\mathop{C}}\,=CH-C{{H}_{2}}-C{{H}_{2}}- \right)}_{n}}\] is z. |
| \[-{{[NH{{(C{{H}_{2}})}_{6}}NHCO{{(C{{H}_{2}})}_{4}}-CO-]}_{n}}\] |
| \[PO_{4}^{3-},\]\[SO_{4}^{2-}\] |
| \[C{{r}_{2}}O_{7}^{2-}\] |
| \[CH\equiv C-CH=C{{H}_{2}}\] |
| Hence, oxidadon state of P, S and Cr are +5, + 6 and + 6 respectively. |
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