The question is about finding the element with the highest first ionization enthalpy among the given options. Ionization enthalpy is the energy required to remove an electron from a gaseous atom or ion.
Explanation:
The given options are Silicon (Si), Aluminium (Al), Nitrogen (N), and Carbon (C).
Ionization enthalpy generally increases across a period in the periodic table because the nuclear charge increases, making it more difficult to remove an electron.
Ionization enthalpy decreases down a group as the atomic size increases, the outer electrons are further from the nucleus, and the increase in shielding reduces the effective nuclear charge felt by the valence electrons, making them easier to remove.
Let's analyze each element:
Silicon (Si): It is in the same group as Carbon but below it in the periodic table. Its ionization enthalpy is lower than C because it's further down the group.
Aluminium (Al): It is in Group 13, after Nitrogen and Carbon in the periodic table. Its first ionization enthalpy is less than that of N and C due to its position.
Nitrogen (N): It is in Group 15 and has one more proton than Carbon, making it more efficiently hold on to its electrons than Al or Si. Its ionization enthalpy is quite high due to its small atomic size and half-filled p orbital, making it stable.
Carbon (C): It lies before Nitrogen in the periodic table and has fewer protons, thus its ionization enthalpy is lower than that of Nitrogen.
From the above analysis, Nitrogen (N) has the highest first ionization enthalpy among the given elements due to its position in the periodic table, small atomic size, and stable half-filled configuration.
Conclusion: The correct answer is \(N\).
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Approach Solution -2
The order of first ionization enthalpy (\( I_{E_1} \)) for these elements is: \(\text{Al} < \text{Si} < \text{C} < \text{N}\)
Thus, nitrogen (\( \text{N} \)) has the highest first ionization enthalpy among the given options.
