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Define ionization energy. Explain why ionization energy measurements are usually made when atoms are in the gaseous state. Why is the second ionization energy always greater than the first ionization energy for any element?

Short Answer

Expert verified
Ionization energy refers to the energy needed to remove an electron from an atom. It is measured in the gaseous state for accuracy, and the second ionization energy is higher due to increased nuclear attraction after the first electron is removed.

Step by step solution

01

Defining Ionization Energy

Ionization energy is the amount of energy required to remove an electron from an isolated atom or ion in its gaseous state. This energy is needed to overcome the attractive forces between the electron and the nucleus.
02

Gaseous State for Ionization Energy Measurements

Ionization energy is typically measured in the gaseous state to ensure that atoms are isolated. In a gaseous state, atoms have minimal interactions with each other, providing a clearer and more accurate measurement of the energy required to remove an electron.
03

Understanding the Second Ionization Energy

The second ionization energy is always greater than the first because after the first electron is removed, there is less electron-electron repulsion and the remaining electrons are drawn closer to the nucleus. This increased attraction means more energy is required to remove the next electron.

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Key Concepts

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

Understanding the Gaseous State
When we measure ionization energy, it is crucial to work with atoms in their gaseous state. In this form, atoms exist as individual and isolated entities. There is minimal interaction with other particles. This is important because any interactions could influence the energy measurements, leading to inaccurate results.
  • In the gaseous state, atoms are free from external forces like van der Waals forces that may be present in solid or liquid states.
  • This makes isolating an atom easier, providing a pure measurement of the energy needed to detach an electron.
  • It ensures the results are solely due to the intrinsic properties of the atom being studied.
So, by using the gaseous state, scientists can measure ionization energy with high precision.
The Process of Electron Removal
Ionization energy is fundamentally about removing electrons from an atom. This process involves overcoming the attractive force that the positively-charged nucleus exerts on the negatively-charged electrons. Removing an electron requires a specific amount of energy because you're essentially breaking this attraction.
  • Each electron experiences a pull toward the nucleus due to opposite electrical charges.
  • For the first ionization energy, the outermost electron is removed, which is often the easiest due to its distance from the nucleus.
  • This represents the minimum energy needed to turn an atom into a positively charged ion.
It's important to note that once an electron is removed, the atom undergoes changes that affect further electron removal, impacting the energy required for subsequent electrons.
Why Second Ionization Energy is Higher
Once the first electron is removed, a few things change within the atom. Primarily, with one fewer negatively charged electron to balance the positively charged nucleus, the remaining electrons are held more tightly.
  • The electron cloud is now smaller, increasing the effective nuclear charge felt by the remaining electrons.
  • This leads to stronger attraction between the nucleus and the remaining electrons, making it harder to remove another electron.
  • Consequently, the second ionization energy is the energy needed to remove a second electron from an already positively charged ion, which is always higher than the first ionization energy.
Thus, the second ionization energy is greater because it requires additional energy to overcome the increased attraction and remove another electron.

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Most popular questions from this chapter

Arrange the elements in each of the following groups in (b) F, order of increasing electron affinity: (a) \(\mathrm{Li}, \mathrm{Na}, \mathrm{K}\) Cl. Br. I.

The ionization energies of sodium (in \(\mathrm{kJ} / \mathrm{mol}\) ), starting with the first and ending with the eleventh, are 496 , 4562,6910,9543,13,354,16,613,20,117,25,496 \(28,932,141,362,159,075 .\) Plot the log of ionization energy \((y\) axis \()\) versus the number of ionization \((x\) axis); for example, \(\log 496\) is plotted versus 1 (labeled \(I E_{1}\), the first ionization energy), \(\log 4562\) is plotted versus 2 (labeled \(I E_{2},\) the second ionization energy \(),\) and so on. (a) Label \(I E_{1}\) through \(I E_{11}\) with the electrons in orbitals such as \(1 s, 2 s, 2 p,\) and \(3 s\) (b) What can you deduce about electron shells from the breaks in the curve?

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A technique called photoelectron spectroscopy is used to measure the ionization energy of atoms. A gaseous sample is irradiated with UV light, and electrons are ejected from the valence shell. The kinetic energies of the ejected electrons are measured. Because the energy of the UV photon and the kinetic energy of the ejected electron are known, we can write $$ h v=I E+\frac{1}{2} m u^{2} $$ where \(v\) is the frequency of the UV light, and \(m\) and \(u\) are the mass and velocity of the electron, respectively. In one experiment the kinetic energy of the ejected electron from potassium is found to be \(5.34 \times 10^{-19} \mathrm{~J}\) using a UV source of wavelength \(162 \mathrm{nm}\). Calculate the ionization energy of potassium. How can you be sure that this ionization energy corresponds to the electron in the valence shell (i.e., the most loosely held electron)?

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