Question

(a) Write the electron configuration for \(\mathrm{Li}\), and estimate the effective nuclear charge experienced by the valence electron. (b) The energy of an electron in a one-electron atom or ion equals \(\left(-2.18 \times 10^{-18} \mathrm{~J}\right)\left(\frac{\mathrm{Z}^{2}}{n^{2}}\right)\) where \(\mathrm{Z}\) is the nuclear charge and \(n\) is the principal quantum number of the electron. Estimate the first ionization energy of Li. (c) Compare the result of your calculation with the value reported in table \(7.4\), and explain the difference. (d) What value of the effective nuclear charge gives the proper value for the ionization energy? Does this agree with your explanation in (c)?

Short answer

The electron configuration of Li is 1s² 2s¹. The effective nuclear charge experienced by the valence electron in Li is approximately 1. The estimated first ionization energy of Li is -4.905×10⁻¹⁸ J, which differs from the -5.392×10⁻¹⁸ J reported in Table 7.4. This difference is due to the shielding effect. The value of the effective nuclear charge that gives the correct ionization energy is around 3.24, which agrees with our explanation for the discrepancy.

Step-by-step solution

Electron configuration of Li

To write the electron configuration of Li, we must first find the number of electrons in the element. Lithium has an atomic number of 3, which means it has 3 electrons in its neutral state. The electron configuration tells us how these electrons fill its atomic orbitals: 1s² 2s¹

Estimating the effective nuclear charge for the valence electron

The effective nuclear charge (\(Z_{eff}\)) is the net positive charge experienced by an electron in an atom due to protons in the nucleus and the shielding effect of inner electrons. The formula for effective nuclear charge is: \(Z_{eff} = Z - S\) where Z is the atomic number (number of protons) and S is the shielding constant (number of core electrons). For Li, Z=3 and S=2 (since there are two 1s electrons). Thus, the effective nuclear charge experienced by the valence electron in Li is: \(Z_{eff} = 3 -2 = 1\)

Estimating the first ionization energy of Li

The first ionization energy of Li can be calculated using the provided formula: \(E = \left(-2.18 \times 10^{-18} \mathrm{~J}\right)\left(\frac{\mathrm{Z}^{2}}{n^{2}}\right)\) We know Z=3 (for Li) and the principal quantum number n=2 (for 2s valence electron). Plugging in these values, we get: \(E_{Li} = (-2.18 \times 10^{-18} \mathrm{~J})\left(\frac{3^{2}}{2^{2}}\right) = -4.905 \times 10^{-18} \mathrm{~J}\)

Comparing the calculated ionization energy with Table 7.4

The calculated ionization energy is -4.905×10⁻¹⁸ J, and the ionization energy reported in Table 7.4 is -5.392×10⁻¹⁸ J. The difference between these values can be due to the shielding effect. We have only an approximate value for the effective nuclear charge.

Finding the effective nuclear charge that gives the correct ionization energy

We can use the provided formula with the given ionization energy to find the value of Z that gives the correct ionization energy: \((-5.392 \times 10^{-18} \mathrm{~J})= \left(-2.18 \times 10^{-18} \mathrm{~J}\right)\left(\frac{\mathrm{Z}^{2}}{n^{2}}\right)\) Solving for Z, we get: \(Z=\sqrt{\frac{-5.392 \times 10^{-18} \mathrm{~J}}{-2.18\times 10^{-18} \mathrm{~J}}\times 2^{2}}\approx 3.24\) The obtained Z value of 3.24 gives a slightly higher effective nuclear charge experienced by the valence electron. This agrees with our explanation in step 4 since we overestimated the shielding effect which then underestimates the ionization energy calculated.

Key concepts

Electron Configuration

Lithium (Li) is an element with an atomic number of 3, which means it has 3 electrons when it is neutral. The electron configuration gives us a way to understand how these electrons are distributed among the atomic orbitals. For Li, the electron configuration is represented as 1s² 2s¹.

• The "1s²" indicates that two electrons are in the first energy level, or "shell," specifically in the 's' orbital of that shell.
• The "2s¹" means there is one electron in the second energy level, also in an 's' orbital, and this electron is the valence electron.

Understanding electron configuration helps us predict how an element like lithium will engage in chemical bonding and what kind of ionization energy it might have.

Ionization Energy

Ionization energy is the energy required to remove an electron from an atom. For lithium, its first ionization energy can be calculated using the equation:
\[E = (-2.18 \times 10^{-18} \, \mathrm{J})\left(\frac{Z^2}{n^2}\right)\]
where \(Z\) is the atomic number, and \(n\) is the principal quantum number for the electron in question. For Li:

• \(Z = 3\), as lithium has 3 protons.
• The outside electron has \(n = 2\) since it's in the second shell.

The calculation gives us the energy for removing the outermost electron efficiently. However, the discrepancy in calculated and tabulated data points to factors not captured perfectly by simple equations, such as the shielding effect.

Shielding Effect

The shielding effect refers to the reduction in the effective nuclear charge on the electron cloud, due to a difference in the attraction forces of the electrons in different orbitals.

In lithium:

• The 1s electrons (inner electrons) act as a shield for the 2s (valence) electron, reducing the nuclear pull it experiences.
• Effective nuclear charge \(Z_{eff}\) is calculated as \(Z - S\), where \(S\) is the shielding constant, representing core (inner) electrons.

For lithium, \(Z = 3\) and \(S = 2\), this means the outer electron experiences a \(Z_{eff}\) of 1. Understanding how well an electron is shielded helps explain its ionization energy differences.

Principal Quantum Number

The principal quantum number, denoted as \(n\), is a crucial quantum mechanical concept that indicates the main energy level occupied by an electron in an atom.

• For lithium, the valence electron (in 2s¹) has a principal quantum number of \(n=2\).
• This number tells us how far the electron is from the nucleus; the higher the number, the farther away it is, typically reducing the electron's stability.

The principal quantum number is a foundational concept for understanding electron configurations, affecting an electron's energy state, and influencing ionization energy by determining how tightly an electron is held by an atom.