This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Step 1: Understand the properties of Group II elements. Group II elements, also known as alkaline earth metals, include Beryllium (Be), Magnesium (Mg), Calcium (Ca), Strontium (Sr), Barium (Ba), and Radium (Ra). These elements are metals. Step 2: Analyze the electron configuration and reactivity of Group II elements. Group II elements have two valence electrons in their outermost shell. To achieve a stable electron configuration (like a noble gas), they readily lose these two electrons to form a +2 ion. For example, Mg Mg^2+ + 2e^-. Step 3: Evaluate the given options. • Option A: Catalysts only. While some metals can act as catalysts, Group II elements are not primarily known for their catalytic properties. Transition metals are more commonly used as catalysts. • Option B: Acids. Group II elements are metals. Metals typically form basic oxides and hydroxides, not acids. • Option C: Reducing agents. A reducing agent is a substance that readily loses electrons (gets oxidized) and causes another substance to be reduced. Since Group II elements readily lose two electrons, they are easily oxidized and thus act as strong reducing agents. • Option D: Non-reactive substances. Group II elements are quite reactive metals, especially as you move down the group. They react with water, acids, and oxygen, among other substances. Step 4: Conclude the correct option. Based on their tendency to readily lose electrons, Group II elements are good reducing agents. The final answer is Option C: Reducing agents. That's 2 down. 3 left today — send the next one.