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Question

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Answer: A

💡 Solution & Explanation

4 2 2 6 2 XeF O F XeF O       2 5 6 XeF PF XeF PF     6 2 4 XeF H O XeOF 2HF    6 2 2 2 XeF 2H O XeO F 4HF   

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