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Find the oxidation states of Cr in $[\text{Cr(H}_2\text{O)}_6]\text{Cl}_3$, $[\text{Cr(C}_6\text{H}_6)_2]$, and $\text{K}_2[\text{Cr(CN)}_2(\text{O})_2(\text{O}_2)(\text{NH}_3)]$ respectively.
$[\text{Cr(H}_2\text{O)}_6]\text{Cl}_3$: H$_2$O is neutral, Cl$^-$ is $-1$, complex has charge 3+. So Cr = +3.
$[\text{Cr(C}_6\text{H}_6)_2]$: Benzene is a neutral ligand (0 charge). Complex is neutral. So Cr = 0.
$\text{K}_2[\text{Cr(CN)}_2(\text{O})_2(\text{O}_2)(\text{NH}_3)]$: K$^+$ gives $+2$ outside; complex ion has $-2$ charge. CN$^-$ ($-1$ each, $2\times-1=-2$); O$^{2-}$ ($-2$ each, $2\times-2=-4$); O$_2^{2-}$ (peroxo, $-2$); NH$_3$ (0). Let Cr = $x$: $x + (-2) + (-4) + (-2) + 0 = -2 \Rightarrow x - 8 = -2 \Rightarrow x = +6$.
Oxidation states: +3, 0, +6.
Let $x$ and $m$ be positive numbers, where $m$ is a multiple of 3. Compare:
Quantity A: $\dfrac{x^m}{x^3}$
Quantity B: $x^{m/3}$
Simplify Quantity A: $\dfrac{x^m}{x^3} = x^{m-3}$.
Quantity B: $x^{m/3}$.
We need to compare $x^{m-3}$ with $x^{m/3}$. The relationship depends on the base $x$ and the exponent comparison.
Case 1: $x = 1$. Both quantities $= 1$. Equal.
Case 2: $x = 2,\ m = 6$. Qty A $= 2^{6-3} = 2^3 = 8$. Qty B $= 2^{6/3} = 2^2 = 4$. A > B.
Case 3: $x = 2,\ m = 3$. Qty A $= 2^{3-3} = 2^0 = 1$. Qty B $= 2^{3/3} = 2^1 = 2$. B > A.
Because different values give different results, the relationship cannot be determined.
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