Trends in Ionisation Energy

9 Key excerpts on "Trends in Ionisation Energy"

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  eBook - PDF

  Chemistry

  The Molecular Nature of Matter

  • Neil D. Jespersen, Alison Hyslop(Authors)
  • 2021(Publication Date)
  • Wiley

    (Publisher)

  O Fe Mg P S K Ca 5 10 Cr Mn Fe Ni Cu Ni Zn Zn Fe AlSi N Zn S Ca 352 CHAPTER 7 The Quantum Mechanical Atom 352 CHAPTER 7 The Quantum Mechanical Atom Periodic Trends in IE Within the periodic table there are trends in the way IE varies that are useful to know and to which we will refer in later discussions. We can see these by examining Figure 7.34, which shows a graph of how the first ionization energy varies with an element’s position in the periodic table. Notice that the elements with the largest ionization energies are the nonmetals in the upper right of the periodic table, and that those with the smallest ionization energies are the metals in the lower left of the table. In general, then, the following trends are observed. Ionization energy generally increases from bottom to top within a group and increases from left to right within a period. Overall the ionization energy increases from the lower left corner of the periodic table to the upper right corner. This is usually referred to as a diagonal trend. The same factors that affect atomic size also affect ionization energy. As the value of n increases going down a group, the orbitals become larger and the outer electrons are farther from the nucleus. Electrons farther from the nucleus are bound less tightly, so IE decreases from top to bottom. As you can see, there is a gradual overall increase in IE as we move from left to right across a period, although the horizontal variation of IE is somewhat irregular. The reason for the overall trend is the increase in effective nuclear charge felt by the valence electrons as we move across a period. As we’ve seen, this draws the valence electrons closer to the nucleus and leads to a decrease in atomic size as we move from left to right. However, the increasing effective nuclear charge also causes the valence electrons to be held more tightly, which makes it more difficult to remove them.

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  Foundations of Chemistry

  An Introductory Course for Science Students

  • Philippa B. Cranwell, Elizabeth M. Page(Authors)
  • 2021(Publication Date)
  • Wiley

    (Publisher)

  The first ionisation energy, IE 1 , is the energy required to remove the first electron from its outer (valence) shell and is represented by the following equation, where M is any element: M g M g + + e -IE 1 The first ionisation energies of the elements decrease on going down a group. The size of the first ionisation energy of an element depends upon the atomic radius of the element and the effective nuclear charge: • The larger the atomic radius, the greater the distance between the nucleus and the outer electrons, and so less energy is required to remove an elec-tron and the ionisation energy therefore is smaller. • The weaker effective nuclear charge exerts a smaller pull on the valence electrons, and so the ionisation energy is lower. On going down a group in the periodic table the effective nuclear charge gets smaller and less energy is required to remove an outer electron. Figure 10.5 shows a plot of the first ionisation energies of the elements in Group 2 and illus-trates how the first ionisation energy decreases on going down the group. 400 500 600 700 800 900 1000 Be M g Ca Sr Ba IE 1 / kJ mol –1 Figure 10.5 First ionisation energy of Group 2 elements. On descending a group, the atomic radius increases. The first ionisation energy decreases on going down a group in the peri-odic table. 10.2 Trends in properties of elements in the same vertical group of the periodic table 333 10.2.5 Electronegativity The electronegativity of an element is the power of an atom to pull a pair of bonded electrons towards itself. Electronegativity arises through the positive charge of the nucleus attracting electrons from surrounding atoms. The more electronegative the element, the stronger the pull on bonded electrons. On going down a group in the periodic table the atoms get larger and there is a bigger dis-tance between the nucleus and any bonded electrons. The electronegativity of elements therefore decreases on going down a group.

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  Chemical Fundamentals of Geology and Environmental Geoscience

  • Robin Gill(Author)
  • 2014(Publication Date)
  • Wiley-Blackwell

    (Publisher)

  Although the architecture of the Periodic Table can be thought of as an outcome of wave-mechanical theory, it was originally worked out from chemical observation. It was first published in its modern form by the Russian chemist Dimitri Mendeleev in 1869, almost 60 years before Schrödinger published his paper on wave mechanics.

  Ionization energy

  The bonds formed by an atom involve the transfer or sharing of electrons. It therefore makes sense to illustrate the periodicity of chemical properties by looking at a parameter that expresses how easy or difficult it is to remove an electron from an atom. The ionization energy of an element is the energy input (expressed in J mo1−1 ) required to detach the loosest electron from atoms of that element (in its ground state). It is the energy difference between the ‘free electron at rest' state (the zero on the scale of electron energy levels) and the highest occupied energy level in the atom concerned. What this means in the simplest case, the hydrogen atom, is shown in Figure 5.6 . A low ionization energy denotes an easily removed electron, a high value a strongly held one.

  We can picture how ionization energy will vary with atomic number by considering the highest occupied energy level in each type of atom (Figure 5.7 ). In lithium (Li; Z = 3, electronic configuration = ls2 2s1 ) and beryllium (Be; Z = 4, 1s2 2s2 ) it is the 2 s level; in boron (B; Z = 5, ls2 2s2 2p1 ) it is the 2p level; and so on. If we were to disregard the increasing nuclear charge, we would predict that the energy needed to strip an electron from this ‘outermost' level would vary with atomic number as shown in Figure 6.1 a. One would expect a general decline in ionization energy with increasing Z, punctuated by sudden drops marking the large energy gaps between one ‘shell' and the next one up (Figure 5.6 ); the downward series of steps in Figure 6.1 a thus reflects the occupation of progressively higher energy levels in Figure 5.6 . There is no suggestion of periodicity.

  Figure 6.1

  (a) A notional plot of ionization energy against atomic number, predicted without regard to the effect of increasing nuclear charge. (b) The variation of measured ionization energy with atomic number Z among the first 20 elements. (The whole Z

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  Chemistry, 5th Edition

  • Allan Blackman, Steven E. Bottle, Siegbert Schmid, Mauro Mocerino, Uta Wille(Authors)
  • 2022(Publication Date)
  • Wiley

    (Publisher)

  As with atomic radius, ionisation energy changes less for elements in the d and f blocks, because increased shielding from the d and f orbitals offsets increases in Z. As a rule of thumb, the trend in the ionisation energies is inverse to that of the atomic radii, that is, smaller atoms have higher ionisation energies. Therefore, the trend in the ionisation energies can be rationalised in similar fashion. FIGURE 4.41 The first ionisation energy generally increases from left to right across a period (blue arrow) and decreases from top to bottom down a group (red arrow) of the periodic table. 0 10 20 30 40 50 60 70 80 90 First ionisation energy, E i1 (kJ mol –1 ) Atomic number, Z 2000 1500 1000 500 0 Rb Xe Rn Ar H Li Na K 2500 He Ne Kr Cs As Z increases, E i1 increases. As n increases, E i1 decreases. 6 5 4 3 2 1 Row He 2372 H 1312 1 2 2000 1500 1000 2500 First ionisation energy (kJ mol –1 ) 500 2000 1500 1000 2500 500 0 0 Group 18 13 14 15 16 17 Ne 2080 F 1681 O 1314 N 1402 C 1086 B 800 Be 899 Li 520 Ar 1520 Cl 1256 S 999 P 1012 Si 786 Al 577 Mg 738 Na 496 Kr 1351 Br 1143 Se 941 As 947 Ge 761 Ga 579 Ca 590 K 419 Xe 1170 I 1009 Te 869 Sb 834 Sn 708 In 558 Sr 549 Rb 403 Rn 1037 At (926) Po 813 Bi 703 Pb 715 Tl 589 Ba 503 Cs 376 Higher ionisations A multi-electron atom can lose more than one electron, but ionisation becomes more difficult as positive charge increases. The first three ionisation energies for a magnesium atom in the gas phase provide an illustration. Process Configurations E i Mg(g) ← → Mg + (g) + e - [ Ne ] 3s 2 ← → [ Ne ] 3s 1 738 kJ mol - 1 Mg + (g) ← → Mg 2+ (g) + e - [ Ne ] 3s 1 ← → [ Ne ] 1450 kJ mol - 1 Mg 2+ (g) ← → Mg 3+ (g) + e - [ Ne ] ← → [ He ] 2s 2 2p 5 7730 kJ mol - 1 178 Chemistry Notice that the second ionisation energy of magnesium is almost twice as large as the first, even though each electron is removed from a 3s orbital. This is because Z eff increases as the number of electrons decreases.

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  Introduction to Modern Inorganic Chemistry, 6th edition

  • R.A. Mackay, W. Henderson(Authors)
  • 2017(Publication Date)
  • CRC Press

    (Publisher)

  IG . 8.4 Variations of first ionization potential across the first short Period

  As the ionization potentials measure the energy required to remove the least tightly bound electron from an atom or ion, values reflect the stability of the configuration from which the electron is being removed. Table. 2.8 gives the ionization potentials of the elements.

  The energy gaps between successive levels with the same l value decrease as the n values increase, so that all the atomic orbitals get closer in energy as the atomic number increases. This trend is not completely regular, and larger than average energy gaps occur between the 4p and 5s levels where the first set of d orbitals has been filled, and between the 6s and 7s levels where the first of the f levels comes. These energy jumps reflect the poorer-than-average shielding powers of d and f electrons.

  Apart from the major discontinuities at the rare gases, there is also a gap in energy wherever the outermost electron enters a new atomic orbital. These gaps correspond to stabilization of the filled shell configurations, s2 , d10 s2 and f14 d10 s2 , before the p orbitals are occupied, and also suggest the possibility of transfer of s electrons into the d shell to give the d10 configuration, or of d electrons into the f shell to give the f14 arrangement, which was discussed in the previous section.

  The stability of the rare gas configurations can be seen, both from the high energies required to remove an electron from the rare gases themselves, and from the leap in the values of the potential when the rare gas configuration has to be broken (i.e. when the second electron is removed from an alkali metal, the third electron is removed from an alkaline earth, the fourth electron from a boron Group element, etc.). The very low first ionization potentials of the alkali metals, and, to a lesser extent, the low first and second potentials of the alkaline earths, show how loosely held are the first one or two electrons outside the rare gas configuration.

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  Chemistry: Atoms First 2e

  • Edward J. Neth, Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson(Authors)
  • 2019(Publication Date)
  • Openstax

    (Publisher)

  Removing an electron from a cation is more difficult than removing an electron from a neutral atom because of the greater electrostatic attraction to the cation. Likewise, removing an electron from a cation with a higher positive charge is more difficult than removing an electron from an ion with a lower charge. Thus, successive ionization energies for one element always increase. As seen in Table 3.3, there is a large increase in the ionization energies for each element. This jump corresponds to removal of the core electrons, which are harder to remove than the valence electrons. For example, Sc and Ga both have three valence electrons, so the rapid increase in ionization energy occurs after the third ionization. Successive Ionization Energies for Selected Elements (kJ/mol) Element IE 1 IE 2 IE 3 IE 4 IE 5 IE 6 IE 7 K 418.8 3051.8 4419.6 5876.9 7975.5 9590.6 11343 Ca 589.8 1145.4 4912.4 6490.6 8153.0 10495.7 12272.9 Sc 633.1 1235.0 2388.7 7090.6 8842.9 10679.0 13315.0 Ga 578.8 1979.4 2964.6 6180 8298.7 10873.9 13594.8 TABLE 3.3 3.5 • Periodic Variations in Element Properties 147 Element IE 1 IE 2 IE 3 IE 4 IE 5 IE 6 IE 7 Ge 762.2 1537.5 3302.1 4410.6 9021.4 Not available Not available As 944.5 1793.6 2735.5 4836.8 6042.9 12311.5 Not available TABLE 3.3 EXAMPLE 3.13 Ranking Ionization Energies Predict the order of increasing energy for the following processes: IE 1 for Al, IE 1 for Tl, IE 2 for Na, IE 3 for Al. Solution Removing the 6p 1 electron from Tl is easier than removing the 3p 1 electron from Al because the higher n orbital is farther from the nucleus, so IE 1 (Tl) < IE 1 (Al). Ionizing the third electron from requires more energy because the cation Al 2+ exerts a stronger pull on the electron than the neutral Al atom, so IE 1 (Al) < IE 3 (Al). The second ionization energy for sodium removes a core electron, which is a much higher energy process than removing valence electrons.

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  Principles of Inorganic Chemistry

  • Brian W. Pfennig(Author)
  • 2021(Publication Date)
  • Wiley

    (Publisher)

  The ionization energy (IE) is defined as the energy that it takes to remove an elec- tron from a gaseous atom to an infinite distance from the nucleus, as shown in Equation (3.23). The ionization energy is approximately equal in magnitude to the absolute value of the difference in energy of the electron configurations of M and M + and can be calculated from Equation (3.24). M g M + g + e - ∞ ionization energy definition (3.23) IE = E el M + - E el M , where E el = - 2 178 × 10 - 18 J Z ∗ n ∗ 2 = - 13 6 eV Z ∗ n ∗ 2 (3.24) TABLE 3.2 The relationship between the principle quantum number n and Slater’s effective quantum number n . N n ∗ 1 1 2 2 3 3 4 3.7 5 4.0 6 4.2 7 4.3 a a extrapolated value. 104 3 THE PERIODICITY OF THE ELEMENTS Example 3-4. Calculate the first ionization energy for a 4s electron in K. Solution. Using Slater’s rules, the shielding parameter for K is calculated as: 1s 2 2s, 2p 8 3s, 3p 8 3d 0 4s, 4p 1 σ = 8 0 85 + 10 1 00 = 16 8 Z ∗ = Z – σ = 19 0 - 16 8 = 2 2 Since all 18 of the core electrons are the same configuration in both K and K + , we only need to calculate the energy of the 19th electron in order to deter- mine the total difference in the energies of the electrons between K and K + . Using Equation (3.24), the ionization energy is calculated as follows: IE = - 13 6 eV 2 2 3 7 2 = 4 81 eV the actual value is 4 34 eV Example 3-5. Calculate the ionization energy of a 2p electron in N. Solution. The electron configuration of N using Slater’s rules is (1s) 2 (2s, 2p) 5 . Again, we can ignore all the core electrons in the filled 1s subshell of N since their energies will be identical in both N and N + .

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  Theoretical Aspects of Chemical Reactivity

  • (Author)
  • 2006(Publication Date)
  • Elsevier Science

    (Publisher)

  Theoretical Aspects of Chemical Reactivity A. Toro-Labbé (Editor) © 2007 Published by Elsevier B.V. Chapter 8 The average local ionization energy: concepts and applications Peter Politzer and Jane S. Murray Department of Chemistry, University of New Orleans, New Orleans, LA 70148, USA 1. Orbital ionization energies The ionization energy I of an N-electron atom or a molecule having energy E is a well-defined property: I = EN − 1 − EN (1) I has been measured experimentally at a high level of accuracy for a large number of atoms and molecules [1]. For molecules, it can be important to distinguish between the adiabatic and the vertical ionization energies. The former corresponds to the neutral molecule and the ion, both being in the ground states, which may or may not have similar geometries. The latter refers to the situation in which the positions of the nuclei are the same in the ion, as in the neutral molecule, which may not be the ion’s ground vibrational state. Our interest shall be in vertical ionization energies. Computationally, either I can be obtained via (1) by evaluating the appropriate EN − 1 and EN . However, another approach is used very frequently. In Hartree– Fock (HF) theory, it follows directly from the formalism that the vertical ionization energy I i of any electron i would equal the negative of its orbital energy i if all of the orbitals of the system were unaffected by the loss of the electron. Koopmans’ theorem assures the stability of the one from which the electron is lost [2,3], and thus the approximation I i ≈ − i (2) is a common one at the HF level. Then, the molecular vertical ionization energy I corresponds to i for the highest occupied orbital. 119 120 Average local ionization energy In reality, the remaining orbitals do undergo some changes when an electron is removed from one of them. By ignoring this, which stabilizes the positive ion, (2) overestimates I i .

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  Chemistry 2e

  • Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson(Authors)
  • 2019(Publication Date)
  • Openstax

    (Publisher)

  Removing an electron from a cation is more difficult than removing an electron from a neutral atom because of the greater electrostatic attraction to the cation. Likewise, removing an electron from a cation with a higher positive charge is more difficult than removing an electron from an ion with a lower charge. Thus, successive ionization energies for one element always increase. As seen in Table 6.3, there is a large increase in the ionization energies for each element. This jump corresponds to removal of the core electrons, which are harder to remove than the valence electrons. For example, Sc and Ga both have three valence electrons, so the rapid increase in ionization energy occurs after the third ionization. Successive Ionization Energies for Selected Elements (kJ/mol) Element IE 1 IE 2 IE 3 IE 4 IE 5 IE 6 IE 7 K 418.8 3051.8 4419.6 5876.9 7975.5 9590.6 11343 Ca 589.8 1145.4 4912.4 6490.6 8153.0 10495.7 12272.9 Sc 633.1 1235.0 2388.7 7090.6 8842.9 10679.0 13315.0 Ga 578.8 1979.4 2964.6 6180 8298.7 10873.9 13594.8 TABLE 6.3 6.5 • Periodic Variations in Element Properties 301 Element IE 1 IE 2 IE 3 IE 4 IE 5 IE 6 IE 7 Ge 762.2 1537.5 3302.1 4410.6 9021.4 Not available Not available As 944.5 1793.6 2735.5 4836.8 6042.9 12311.5 Not available TABLE 6.3 EXAMPLE 6.13 Ranking Ionization Energies Predict the order of increasing energy for the following processes: IE 1 for Al, IE 1 for Tl, IE 2 for Na, IE 3 for Al. Solution Removing the 6p 1 electron from Tl is easier than removing the 3p 1 electron from Al because the higher n orbital is farther from the nucleus, so IE 1 (Tl) < IE 1 (Al). Ionizing the third electron from requires more energy because the cation Al 2+ exerts a stronger pull on the electron than the neutral Al atom, so IE 1 (Al) < IE 3 (Al). The second ionization energy for sodium removes a core electron, which is a much higher energy process than removing valence electrons.

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