Anatomy of Electron Clouds

TP The energy required to remove a second electron from nitrogen, 𝑒 𝑔 , is called the second ionization energy, 𝐼 . Compared N 𝑔 →N 𝑔 to the first ionization energy, 𝐼 , the second ionization energy 1. 2. 3. 4. 5. must be smaller must be about the same must be larger Further information required Not sure Lecture • • • • • • Anatomy of electron clouds 𝐼𝐸 → electron cloud expansion Electron affinity: X 𝑔 → X 𝑔 𝑒 Electronegativity: 𝐸𝑁 ~ 𝐼𝐸 𝐸𝐴 Dipole moment and ionic character: ~ Δ𝐸𝑁 Lewis diagrams Anatomy of electron clouds Anatomy of electron clouds The volume of atoms is due to their electron clouds. Let’s do this for the nitrogen atom. Studying successive ionization energies how much energy is needed to remove an electron from the cloud reveals the anatomy of the cloud. Here is the equation describing removal of one electron from a neutral N atom. N 𝑔 →N 𝑔 𝑒 𝑔 ,𝐼 2.32 aJ 2.32 𝐼 is called the first ionization energy, and 1 aJ 10 1 J 10 J attojoule 1 TP The energy required to remove a second electron from nitrogen, 𝑒 𝑔 , is called the second ionization energy, 𝐼 . Compared N 𝑔 →N 𝑔 to the first ionization energy, 𝐼 , the second ionization energy 1. 2. 3. 4. 5. must be smaller must be about the same must be larger Further information required Not sure Anatomy of electron clouds The key in comparing ionization energies, the ionizations N 𝑔 →N N 𝑔 𝑔 →N 𝑒 𝑔 𝑔 ,𝐼 𝑒 𝑔 ,𝐼 is Coulomb’s law, 𝐸 ~ In words, Coulomb’s law says that opposite charges attract, the more so the greater the charges, 𝑄 and 𝑄 , and the more so the closer, 𝑑, they are. Structure of electron clouds The key in comparing ionization energies, the ionizations N 𝑔 →N N 𝑔 →N 𝑔 𝑒 𝑔 𝑔 𝑒 𝐼 𝑔 𝐼 The greater the product of the charges, the more energy required to separate them. Therefore, 𝐼 must be larger than 𝐼 , because for 𝐼 , 𝑄 𝑄 2 1 2 for 𝐼 , 𝑄 𝑄 1 1 1 TP For the 5th ionization energy of N atom, 𝐼 , what are the values of the Coulomb’s law charges 𝑄 and 𝑄 ? 1 and 𝑄 1. 𝑄 2 and 𝑄 2. 𝑄 4 and 𝑄 3. 𝑄 5 and 𝑄 4. 𝑄 6 and 𝑄 5. 𝑄 6. Something else 7. Not sure 1 1 1 1 1 2 TP The energy required to remove a third electron from nitrogen, 𝑒 𝑔 , is called the third ionization energy, 𝐼 . How do the N 𝑔 →N 𝑔 first three ionization energies compare? Tell your classmates why. 1. 2. 3. 4. Successive ionization energies 𝐼 𝐼 𝐼 𝐼 𝐼 𝐼 𝐼 𝐼 𝐼 Not sure Plot log(IEn) versus n for Mg Plot log(IEn) versus n for Mg You will need the final two ionization energies, 𝐼𝐸 11 and 𝐼𝐸 12 . Physically in terms of Coulomb’s law , what accounts for the discontinuity between 𝑛 2 and 𝑛 3? https://en.wikipedia.org/wiki/Ionization_energies_of_the_elements_ data_page 𝐼𝐸 1762 eV 𝐼𝐸 1962 eV 3 Successive IEn → electron cloud contraction 𝐼𝐸 𝐼𝐸 : Mg2 Mg increase in residual ion charge 𝐼𝐸 ≫ 𝐼𝐸 : Mg3 ≫ Mg2 large contraction of electron cloud 𝐼𝐸 𝐼𝐼𝐸 𝐼𝐸 ≫ 𝐼𝐸 𝐼𝐸 𝐼𝐸 ⋯ Structure of electron clouds Evidence of charge cloud contraction is seen in every element other than H and He. For example, 𝐼𝐸 : increase in residual ion charge : Mg11 ≫ Mg10 large contraction of electron cloud : Mg12 Mg11 integer increase in residual ion charge Structure of electron clouds Structure of electron clouds The “easier to remove” electrons are called valence electrons. The remaining ”harder to remove” electrons are called core electrons. The “easy-to-remove” part of electron clouds account for the periodicity of atom properties. The Lewis electron-dot formula summarizes the number of easily removed valence electrons until the much smaller cloud of inner-core electrons remains. 4 IE1 → electron cloud expansion IE1 → electron cloud expansion Elements in successive rows of the same column of the periodic table contain the same number of valence electrons but increasing numbers of core electrons. This means the valence portion of the electron clouds of successive elements in the same column is larger and so farther from the nucleus. This in turn means it is increasingly easier to remove valence electrons going down a column of the periodic table. Ionization energies decrease due to this expansion of the valence region of the electron cloud. IE1 → electron cloud expansion Lewis dot picture summarizes effect of electron cloud expansion. Problem 3.9 TP Which is the correct order of first ionization energy of the atoms barium, beryllium, chlorine, and iodine. 1. Ba Be Cl I 2. I Cl Be Ba 3. I Ba Cl Be 4. Be Cl Ba I 5. Ba I Cl Be 6. Ba Be I Cl 7. Something else 8. Not sure Electron affinity, EA: X−(𝑔) → X(𝑔) + 𝑒 𝐼𝐸 of barium, beryllium, chlorine, and iodine Ba Be Ba 502.9 kJ/mol I Cl Be 899.5 kJ/mol Cl 1251.2 kJ/mol I 1008.5 kJ/mol Electron affinity, EA: X−(𝑔) → X(𝑔) + 𝑒 6 Electronegativity: 𝐸𝑁 ~ 𝐼𝐸 1 + 𝐸𝐴 Problem 3.23a 𝐼𝐸 𝐼𝐸 𝐸𝐴 𝐸𝐴 K Cl K Cl 4.34 eV/atom 12.97 eV/atom 48 kJ/mol 349 kJ/mol Dipole moment and ionic character: ~ Δ𝐸𝑁 μ charge separation Dipole moment of charges e and e separated by 0.2082 as one Debye, 1D 3.336 10 30 Dipole moment Typically, only a fraction, δ, of an electron charge is displaced 10 10 m is defined 𝑞 δ |e| Where δ is the fractional ionic character. Cm 7 Ionic character Ionic character The internuclear distance of HF is Re is 1.82 D. 0.917 10 10 m and its dipole moment Calculate dipole moment, assuming displacement of one full unit of charge, |e| . Answer: 1.602 10 C 0.917 10 m . 4.40 D The internuclear distance of HF is Re is 1.82 D. 0.917 10 10 m and its dipole moment Calculate what fraction of an electron charge is displaced, that is, the fractional ionic character of HF. . . Answer: 𝛿 0.41 Ionic character Ionic character The internuclear distance of HF is Re 0.917 10 10 m and its dipole moment is 1.82 D. The fractional ionic character of HF is 0.41. If the ionic character were 100 %, the molecule would be Percent ionic character is δ 100 % 41 % H . F . with 𝜇 𝜇 𝑒 𝑅 4.40 𝐷 . with one full unit of electron charge transferred. Since the ionic character is 41 %, the molecule actually is H . F . with 𝜇 𝜇 0.41𝑒 𝑅 . 1.82 D That is, only 0.41 of an electron charge is transferred. So the fractional charge transferred is 𝛿 0.41. 8 Percent ionic character Lewis diagram of acetic acid, CH3COOH 0. skeleton Lewis diagram of acetic acid, CH3COOH 0. 1. 2. 3. skeleton e’s we have 4 8 12 24 e’s needed: 8 32 40 e’s shared single needed – have 40 – 24 16 9