Calculate the oxidation number of [1+1] i) N in NH4OH ii) Cr in K2Cr2O7

i) Oxidation number of N in $\mathrm{NH}{}_4\mathrm{OH}$

1. Identify the oxidation numbers of known elements:

   • H = +1 (in compounds, except in metal hydrides)
   • O = −2 (in compounds, except in peroxides)

2. Write the formula with oxidation numbers: $$\mathrm{N}{}^{\times }\mathrm{H}{}_4{}^{+}\mathrm{O}{}^{-2}\mathrm{H}{}^{+}$$ Here, $\mathrm{NH}{}_4{}^{+}$ is a cation with a total charge of +1, and $\mathrm{OH}{}^{-}$ is an anion with a total charge of −1.

3. Calculate the oxidation number of N in $\mathrm{NH}{}_4{}^{+}$: $$x+4(+1)=+1⟹x=+1-4=-3$$

Oxidation number of N in $\mathrm{NH}{}_4\mathrm{OH}$ = −3

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ii) Oxidation number of Cr in $\mathrm{K}{}_2\mathrm{Cr}{}_2\mathrm{O}{}_7$

1. Identify the oxidation numbers of known elements:

   • K = +1 (alkali metal)
   • O = −2 (in compounds, except in peroxides)

2. Write the formula with oxidation numbers: $$\mathrm{K}{}_2{}^{+}\mathrm{Cr}{}_2{}^y\mathrm{O}{}_7{}^{-2}$$

3. Set up the equation for charge neutrality: $$2(+1)+2y+7(-2)=0⟹2+2y-14=0$$

4. Solve for $y$: $$2y-12=0⟹2y=+12⟹y=+6$$

Oxidation number of Cr in $\mathrm{K}{}_2\mathrm{Cr}{}_2\mathrm{O}{}_7$ = +6