Step 1: Understanding the Concept
First ionization enthalpy is the minimum energy required to remove the outermost electron from a neutral gaseous atom. The question asks to identify which alkali metal (Group 1 element) has the highest value for this property. This requires understanding the periodic trends for ionization enthalpy.
Step 2: Key Formula or Approach
The primary trend for first ionization enthalpy within a group of the periodic table is that it decreases as you move down the group. This trend is explained by two main factors: atomic size and shielding.
Step 3: Detailed Explanation
1. Identify the Alkali Metals' Position:
The alkali metals are in Group 1. Their order from top to bottom in the periodic table is: Lithium (Li), Sodium (Na), Potassium (K), Rubidium (Rb), Caesium (Cs).
2. Analyze the Trend Down the Group:
- Atomic Radius: As we move down the group, each element has one more electron shell than the one above it. This leads to a significant increase in atomic radius.
- Shielding Effect: The inner shell electrons shield the outermost valence electron from the full attractive force of the nucleus. This shielding effect increases down the group as more inner shells are added.
- Effective Nuclear Charge: The combination of increasing atomic radius and increased shielding means that the outermost electron is held less tightly by the nucleus in larger atoms.
As a result of this weaker attraction, less energy is needed to remove the outermost electron. Thus, the first ionization enthalpy decreases as we descend the group.
3. Conclusion:
The element at the top of the group will have the highest ionization enthalpy. Among the given options, Lithium (Li) is the first element in the alkali metal group.
The trend for first ionization enthalpy is: \(Li>Na>K>Rb>Cs\).
Therefore, Lithium (Li) has the highest first ionization enthalpy.
Step 4: Final Answer
The alkali metal with the highest first enthalpy of ionization is Li.