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Chapter 12: Problem 95

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) \(C>N>0>F\)

Short Answer

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

Step by step solution

01

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.
02

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.
03

Compare Electron Configurations

Consider the electron configurations after removing the first electron: - C+ has [He]2s虏2p鹿 - N+ has [He]2s虏2p虏 - O+ has [He]2s虏2p鲁 - F+ has [He]2s虏2p鈦�. N+ has a half-filled stable configuration, making its second ionization potential relatively high.
04

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.
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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]2s虏2p虏) 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鈥攁nd thus the positive charge in the nucleus鈥攊ncreases. 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鈦�) ends up with [He]2s虏2p鹿
  • Nitrogen (N鈦�) has [He]2s虏2p虏, which is crucial because a half-filled p orbital is more stable.
  • Oxygen (O鈦�) becomes [He]2s虏2p鲁
  • Fluorine (F鈦�) with [He]2s虏2p鈦�
The stability of these configurations directly influences their second ionization energies. N鈦�, 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鈥攐r sometimes less鈥攅nergy to disrupt, explaining the variations in second ionization energies across different elements.

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Most popular questions from this chapter

Among the elements \(\mathrm{W}, \mathrm{X}, \mathrm{Y}\) and \(\mathrm{Z}\) having atomic numbers \(9,10,11\) and 12 respectively, the correct order of ionization energies is (a) \(\mathrm{W}>\mathrm{Y}>\mathrm{X}>\mathrm{Z}\) (b) \(\mathrm{X}>\mathrm{W}>\mathrm{Z}>\mathrm{Y}\) (c) \(\mathrm{X}>\mathrm{Z}>\mathrm{Y}>\mathrm{W}\) (d) \(\mathrm{Z}>\mathrm{Y}>\mathrm{X}>\mathrm{W}\)

Why \(\operatorname{Sc}(Z=21)\) is not considered as a transition element? (a) properties of \(\mathrm{Sc}\) are similar to alkali metals (b) \(3 \mathrm{~d}\) orbitals are empty in its stable compound (c) stable oxidation number of \(\mathrm{Sc}\) is \(+2\) (d) atomic volume of \(\mathrm{Sc}\) is very large

Pick out the statement(s) which is are not true about the diagonal relationship of \(\mathrm{Li}\) and \(\mathrm{Mg}\). (1) Polarizing powers of \(\mathrm{Li}^{+}\)and \(\mathrm{Mg}^{2+}\) are almost same. (2) Like Li, Mg decomposes water very fast. (3) \(\mathrm{LiCl}\) and \(\mathrm{MgCl}_{2}\) are deliquescent. (4) Like Li, Mg readily reacts with liquid bromine at ordinary temperature. (a) (1) and (4) (b) \((2)\) and (4) (c) only (2) (d) only(i)

The transition metal with least atomic number is (a) Os (b) \(\mathrm{Zr}\) (c) \(\mathrm{Pt}\) (d) \(\mathrm{Ru}\)

Which of the following has maximum ionization potential? (a) \(\mathrm{Al}\) (b) \(\mathrm{P}\) (c) Si (d) \(\mathrm{Mg}\)
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