Problem 95 - A Complete Resource Book in Chemistry for JEE Main

Question

The correct order of second ionization potential of carbon, nitrogen, oxygen and fluorine is (a) \(\mathrm{O}>\mathrm{N}>\mathrm{F}>\mathrm{C}\) (b) \(\mathrm{O}>\mathrm{F}>\mathrm{N}>\mathrm{C}\) (c) \(\mathrm{F}>\mathrm{O}>\mathrm{N}>\mathrm{C}\) (d) \(C>N>0>F\)

Step-by-Step Solution

Answer

The correct order is (c) \(\mathrm{F}>\mathrm{O}>\mathrm{N}>\mathrm{C}\).

Step 1: Understand Ionization Potential

Ionization potential is the energy required to remove an electron from an atom. The second ionization potential refers to removing a second electron after the first has been removed. It generally increases across a period as atomic size decreases and nuclear charge increases.

Step 2: Remove Initial Electron

For the elements carbon (C), nitrogen (N), oxygen (O), and fluorine (F), initially one electron is removed to give C+, N+, O+, and F+. The second ionization potential is concerned with removing another electron from these cations.

Step 3: Compare Electron Configurations

Consider the electron configurations after removing the first electron: - C+ has [He]2s²2p¹ - N+ has [He]2s²2p² - O+ has [He]2s²2p³ - F+ has [He]2s²2p⁴. N+ has a half-filled stable configuration, making its second ionization potential relatively high.

Step 4: Determine Ionization Energy Trends

Moving across the period, second ionization potential generally increases. Fluorine, being the smallest with the highest nuclear charge after removing one electron, is expected to have the highest second ionization energy.

Step 5: Logical Deduction Based on Trends

Arrange by decreasing second ionization energies: - Fluorine will have the highest second ionization potential. - Oxygen and nitrogen will follow, with oxygen being slightly higher due to increased nuclear charge and electron removal from a paired state. - Carbon will have the lowest due to its larger size and less nuclear charge.

Key Concepts

Second Ionization Energy

Second ionization energy is all about what it takes to remove a second electron from an already positive ion. Picture this: you've already removed one electron from an atom, and now you want to snatch another one. This second electron removal requires even more energy than the first one. Here's why:

• The positive ion left after the first electron has been removed, is even more attracted to its remaining electrons. The reason is simple: fewer electrons means a stronger pull from the protons in the nucleus.
• Removing a second electron often involves breaking up stable electronic configurations, like half-filled or fully filled subshells, which naturally require more energy.

So, for carbon (C), nitrogen (N), oxygen (O), and fluorine (F), their second ionization energies reflect their tendency towards stability. Notably, nitrogen's half-filled electronic configuration ([He]2s²2p²) makes its second ionization energy notably high.

Periodic Trends

When it comes to periodic trends, the second ionization energy tends to increase as you move across a period from left to right. Why does this happen? Simple! The atomic number—and thus the positive charge in the nucleus—increases. This increasing nuclear charge pulls electrons closer, making them harder to remove:

• The atomic size decreases across a period, which means electrons are closer to the nucleus and more tightly held.
• Elements like fluorine, to the right of the period, have a smaller atomic radius and a higher effective nuclear charge. This contributes to fluorine having a high second ionization energy.

Indicators like these help you predict the behavior of elements and their tendency to hold onto their electrons more aggressively as you move across a period.

Electron Configuration

Electron configuration is the arrangement of electrons in an atom's orbitals and tells us about stability. Consider these ions after the first electron removal:

• Carbon (C⁺) ends up with [He]2s²2p¹
• Nitrogen (N⁺) has [He]2s²2p², which is crucial because a half-filled p orbital is more stable.
• Oxygen (O⁺) becomes [He]2s²2p³
• Fluorine (F⁺) with [He]2s²2p⁴

The stability of these configurations directly influences their second ionization energies. N⁺, with its half-filled stability, is less willing to give up another electron, increasing its ionization energy. Electron configurations that bring atoms to a more stable state can require more—or sometimes less—energy to disrupt, explaining the variations in second ionization energies across different elements.