📚 Learning Guide
Ionization Potential
hard

Which of the following statements best explains the trend in successive ionization energies for an element with the electron configuration of [Ne]3s²?

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Choose the Best Answer

A

The first ionization energy is lower than the second due to increased shielding from the first electron removed.

B

The second ionization energy is significantly higher than the first due to the removal of an electron from a fully filled subshell.

C

Successive ionization energies are generally constant for all elements regardless of electron configuration.

D

The first ionization energy is the highest due to the stability of the electron configuration after the first electron is removed.

Understanding the Answer

Let's break down why this is correct

Answer

The first ionization removes one 3s electron, giving a Mg⁺ ion with a [Ne]3s¹ configuration; the second removes the last 3s electron, yielding a Mg²⁺ ion with a closed‑shell [Ne] configuration. Because both 3s electrons are in the same outer shell, their ionization energies are relatively low and close in value. After the second electron is removed, any further ionization must take an electron from the inner 2p shell, which is tightly bound to the nucleus. Consequently, the third ionization energy jumps sharply, reflecting the higher energy needed to remove an inner‑shell electron. Thus, the trend shows two similar, low ionization energies followed by a large increase.

Detailed Explanation

Removing the first 3s electron is easier because the remaining electron is still outside a noble gas core. Other options are incorrect because Shielding does not increase when one outer electron is removed; the inner electrons still block the nucleus from the outer electron; Ionization energies vary from element to element; they are not constant.

Key Concepts

electron configuration
subshells
successive ionization energies
Topic

Ionization Potential

Difficulty

hard level question

Cognitive Level

understand

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