The reaction involving potassium dichromate, potassium chloride, and concentrated sulfuric acid typically leads to the formation of chromyl chloride (\(CrO_2Cl_2\)). In this process, let's consider the reaction:
\(K_2Cr_2O_7 + 4KCl + 6H_2SO_4 \rightarrow 2CrO_2Cl_2 + 6H_2O + 3K_2SO_4\)
In potassium dichromate (\(K_2Cr_2O_7\)), the oxidation state of chromium is \(+6\). During the formation of chromyl chloride (\(CrO_2Cl_2\)), each chromium atom maintains its oxidation state. We verify this by calculating the oxidation state in \(CrO_2Cl_2\):
Let the oxidation state of \(Cr\) be \(x\). Oxygen has an oxidation state of \(-2\), and chlorine typically exhibits \(-1\):
\[x + 2(-2) + 2(-1) = 0\]
\[x - 4 - 2 = 0\]
\[x = +6\]
This computation confirms chromium retains an oxidation state of \(+6\) in the product \(CrO_2Cl_2\), matching the given range (6,6). Therefore, the oxidation state of chromium in the product is correctly \(+6\).
The reaction is as follows:
\[ K_2\text{Cr}_2\text{O}_7 + 4\text{KCl} + 6\text{H}_2\text{SO}_4 \rightarrow 2\text{CrO}_2\text{Cl}_2 + 6\text{KHSO}_4 + 3\text{H}_2\text{O} \]This reaction is known as the chromyl chloride test. In this reaction, the oxidation state of chromium in \(\text{CrO}_2\text{Cl}_2\) is \(+6\).
Let \( \alpha = \dfrac{-1 + i\sqrt{3}}{2} \) and \( \beta = \dfrac{-1 - i\sqrt{3}}{2} \), where \( i = \sqrt{-1} \). If
\[ (7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}, \] then the value of \( m \) is ___________.