Question

According to periodic trend, which of the following most likely has the highest ionization energy? Na Al Si Ar

Answer 1

Answer: The correct answer is Argon.

Explanation:

Ionization energy is defined as the energy required to remove valence electron from an isolated gaseous atom. It is expressed as $E_i$

$X(g)\rightarrow X^{n+}(g)+ne^-$

Energy required for a stable molecule (having fully filled and half filled electronic configuration)

The electronic configuration for the elements given are:

1. Na: $1s^22s^22p^63s^1$

As, this element does not have stable electronic configuration. So, this will not have highest ionization energy.

2. Al: $1s^22s^22p^63s^23p^1$

As, this element does not have stable electronic configuration. So, this will not have highest ionization energy.

3. Si: $1s^22s^22p^63s^23p^4$

As, this element does not have stable electronic configuration. So, this will not have highest ionization energy.

4. Ar: $1s^22s^22p^63s^23p^6$

As, this element have stable electronic configuration where all the orbitals are fully filled. thus, this element will require more energy to remove the valence electron. So, this have highest ionization energy.

Hence, the correct answer is Argon.

Answer 2

electrons in the outer shell have the lowest ionization energy. (not even gonna consider the quantum model which does funny things with d-sublevel). Highest ionization energy in descending order is: Ar = 1520.6 Si = 786.5 Al = 577.44 Na 495.8