To determine the order of elements in terms of increasing metallic character, we need to understand the trends in the periodic table regarding metallic character. Here’s a step-by-step solution:
### Step 1: Understand Metallic Character
Metallic character refers to the tendency of an element to lose electrons and form positive ions (cations). Elements with high metallic character are typically good conductors of electricity and heat, and they tend to be malleable and ductile.
### Step 2: Identify the Trend
- **Down a Group**: Metallic character increases as you move down a group in the periodic table. This is because the atomic size increases, making it easier for the element to lose electrons.
- **Across a Period**: Metallic character decreases as you move from left to right across a period. This is due to the increase in nuclear charge, which makes it harder for the elements to lose electrons.
### Step 3: Analyze the Given Sets
Let's analyze the sets of elements given in the question:
1. **Beryllium (Be), Magnesium (Mg), Calcium (Ca)**:
- All belong to Group 2.
- As you go down the group from Be to Ca, metallic character increases.
- Order: Be < Mg < Ca (correct order).
2. **Lithium (Li), Sodium (Na), Potassium (K)**:
- All belong to Group 1.
- As you go down the group from Li to K, metallic character increases.
- Order: Li < Na < K (incorrect order as given).
3. **Magnesium (Mg), Aluminum (Al), Silicon (Si)**:
- All belong to Period 3.
- As you move from Mg to Si across the period, metallic character decreases.
- Order: Si < Al < Mg (incorrect order as given).
4. **Carbon (C), Oxygen (O), Nitrogen (N)**:
- All belong to Period 2.
- As you move from C to O across the period, metallic character decreases.
- Order: O < N < C (incorrect order as given).
### Step 4: Conclusion
The only set of elements that is correctly arranged in order of increasing metallic character is the first set: **Beryllium (Be), Magnesium (Mg), Calcium (Ca)**.
### Final Answer
The correct answer is: **Beryllium (Be), Magnesium (Mg), Calcium (Ca)**.
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