To determine the correct order of increasing oxidation numbers of the transition metal ions in the compounds \( K_2Cr_2O_7 \), \( V_2O_5 \), and \( KMnO_4 \), we first need to identify the oxidation states of the transition metals in each compound. ### Step 1: Determine the oxidation states 1. **For \( K_2Cr_2O_7 \) (Potassium dichromate)**: - Potassium (K) has an oxidation state of +1. - Oxygen (O) typically has an oxidation state of -2. - The formula can be set up as follows: \[ 2(+1) + 2(\text{Cr}) + 7(-2) = 0 \] Simplifying this gives: \[ 2 + 2\text{Cr} - 14 = 0 \implies 2\text{Cr} = 12 \implies \text{Cr} = +6 \] - Therefore, the oxidation state of chromium (Cr) in \( K_2Cr_2O_7 \) is +6. 2. **For \( V_2O_5 \) (Vanadium pentoxide)**: - Again, oxygen has an oxidation state of -2. - The formula can be set up as follows: \[ 2(\text{V}) + 5(-2) = 0 \] Simplifying this gives: \[ 2\text{V} - 10 = 0 \implies 2\text{V} = 10 \implies \text{V} = +5 \] - Therefore, the oxidation state of vanadium (V) in \( V_2O_5 \) is +5. 3. **For \( KMnO_4 \) (Potassium permanganate)**: - Potassium (K) has an oxidation state of +1, and oxygen has an oxidation state of -2. - The formula can be set up as follows: \[ +1 + \text{Mn} + 4(-2) = 0 \] Simplifying this gives: \[ 1 + \text{Mn} - 8 = 0 \implies \text{Mn} - 7 = 0 \implies \text{Mn} = +7 \] - Therefore, the oxidation state of manganese (Mn) in \( KMnO_4 \) is +7. ### Step 2: Compare the oxidation states Now we have the oxidation states for each transition metal: - Chromium in \( K_2Cr_2O_7 \): +6 - Vanadium in \( V_2O_5 \): +5 - Manganese in \( KMnO_4 \): +7 ### Step 3: Order the oxidation states To find the order of increasing oxidation numbers: - The lowest oxidation state is for vanadium (+5). - Next is chromium (+6). - The highest oxidation state is for manganese (+7). Thus, the correct order of increasing oxidation number is: \[ V_2O_5 < K_2Cr_2O_7 < KMnO_4 \] ### Conclusion: Correct Option The correct option is **A. \( V_2O_5 < K_2Cr_2O_7 < KMnO_4 \)**. ### Explanation of Other Options - **Option B: \( K_2Cr_2O_7 < KMnO_4 < V_2O_5 \)**: This is incorrect because it places \( K_2Cr_2O_7 \) with a higher oxidation state than \( KMnO_4 \), which is not true. - **Option C: \( KMnO_4 < K_2Cr_2O_7 < V_2O_5 \)**: This is incorrect as it suggests that \( KMnO_4 \) has a lower oxidation state than both \( K_2Cr_2O_7 \) and \( V_2O_5 \), which is also not true. - **Option D: \( KMnO_4 < V_2O_5 < K_2Cr_2O_7 \)**: This is incorrect because it places \( KMnO_4 \) with a lower oxidation state than both \( V_2O_5 \) and \( K_2Cr_2O_7 \), which contradicts our findings. ### Revision Summary - The oxidation states are: Cr in \( K_2Cr_2O_7 \) is +6, V in \( V_2O_5 \) is +5, and Mn in \( KMnO_4 \) is +7. - The correct order of increasing oxidation numbers is \( V_2O_5 < K_2Cr_2O_7 < KMnO_4 \). - Always remember to account for the oxidation states of all elements in a compound when determining the oxidation state of transition metals. - Double-check calculations to avoid common pitfalls in oxidation state determination.