• # question_answer 12) 1, 2 Assign oxidation number to the underlined elements in each of the following species: (a) $\text{K}{{\underset{\text{-}}{\mathop{\text{I}}}\,}_{\text{3}}}$ (b) $Ca{{\underset{-}{\mathop{O}}\,}_{2}}$ (c)$Na{{H}_{2}}\underset{-}{\mathop{P}}\,{{O}_{4}}$ (d)${{H}_{4}}{{\underset{-}{\mathop{P}}\,}_{2}}{{O}_{7}}$ (e) $Na\underset{-}{\mathop{B}}\,{{H}_{4}}$ (f) ${{H}_{2}}{{\underset{-}{\mathop{S}}\,}_{2}}{{O}_{7}}$ (g)$KAl{{(\underset{-}{\mathop{S}}\,{{O}_{4}})}_{2}}l2{{H}_{2}}O$ (h)${{H}_{2}}{{\underset{-}{\mathop{S}}\,}_{4}}{{O}_{6}}$ (i) $\underset{-}{\mathop{C}}\,{{H}_{3}}\underset{-}{\mathop{C}}\,{{H}_{2}}OH$ (j) $\underset{-}{\mathop{C}}\,{{H}_{3}}\underset{-}{\mathop{C}}\,OOH$

(a) $K{{\underset{-}{\mathop{I}}\,}_{3}}$ $+1+3x=0$ $x=-\frac{1}{3}$ (b) $Ca{{\underset{-}{\mathop{O}}\,}_{2}}$ $+2+2x=0$ $x=-1$ (c) $Na{{H}_{2}}\underset{-}{\mathop{P}}\,{{O}_{4}}$ $+1+2+x-8=0$ $x=+5$ (d) ${{H}_{4}}{{\underset{-}{\mathop{P}}\,}_{2}}{{O}_{7}}$ $+4+2x-14=0$ $x=+5$ (e) $Na\underset{-}{\mathop{B}}\,{{H}_{4}}$ $+1+x+(-4)=0$ $x=+3$ (f) ${{H}_{2}}{{\underset{-}{\mathop{S}}\,}_{2}}{{O}_{7}}$ $HO-\underset{\underset{O}{\mathop{\downarrow }}\,}{\overset{\overset{O}{\mathop{\uparrow }}\,}{\mathop{S}}}\,-O-\underset{\underset{O}{\mathop{\downarrow }}\,}{\overset{\overset{O}{\mathop{\uparrow }}\,}{\mathop{S}}}\,-OH$ $+2+2x-14=0$ $x=+6$ (g) $KAl{{(\underset{-}{\mathop{S}}\,{{O}_{4}})}_{2}}.12{{H}_{2}}O$ $+1+3+2(x-8)+12\times 0=0$ $4+2x-16=0$ $x=+6$ (h) ${{H}_{2}}{{\underset{-}{\mathop{S}}\,}_{4}}{{O}_{6}}$ $HO-O-\underset{\underset{O}{\mathop{\downarrow }}\,}{\overset{\overset{O}{\mathop{\uparrow }}\,}{\mathop{S}}}\,-S-S-\underset{\underset{O}{\mathop{\downarrow }}\,}{\overset{\overset{O}{\mathop{\uparrow }}\,}{\mathop{S}}}\,-O-H$ $+2+4x-12=0$ $x=+2.5$ The two donor sulphur atoms have +5 state, while non-donor atoms have zero oxidation state. (i) $\underset{1}{\mathop{\underset{-}{\mathop{C}}\,}}\,{{H}_{3}}\underset{2}{\mathop{\underset{-}{\mathop{C}}\,}}\,{{H}_{2}}OH\,\,\,\,or\,{{\underset{-}{\mathop{C}}\,}_{2}}{{H}_{5}}OH$ Oxidation number of carbon $=\frac{2{{n}_{O}}-{{n}_{H}}}{{{n}_{C}}}=\frac{2\times 1-6}{2}=-2$ $\frac{\frac{\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }}{\text{Note:}\,\,\text{Carbon no}\text{. }\left( \text{1} \right)\text{ has }\left( \text{-3} \right)\text{ and carbon no}\text{. }\left( \text{2} \right)\text{ has }\left( \text{-1} \right)\text{ state}\text{.}}}{{}}$ (i) $\underset{1}{\mathop{\underset{-}{\mathop{C}}\,}}\,{{H}_{3}}\underset{2}{\mathop{\underset{-}{\mathop{C}}\,}}\,OOH\,\,\,or\,\,{{C}_{2}}{{H}_{4}}{{O}_{2}}$ Oxidation number of carbon $=\frac{2{{n}_{O}}-{{n}_{H}}}{{{n}_{C}}}=\frac{2\times 2-4}{2}=0$ $\frac{\frac{\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }}{\text{Note:}\,\,\text{Carbon no}\text{. }\left( \text{1} \right)\text{ has }\left( \text{-3} \right)\text{ and carbon no}\text{. }\left( \text{2} \right)\text{ has }\left( \text{+3} \right)\text{ state}\text{.}}}{{}}$