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Chapter 6: Problem 76

The energy of any one-electron species in its nth state \((\mathbf{n}=\) principal quantum number) is given by \(E=-B Z^{2} / \mathbf{n}^{2}\), where \(Z\) is the charge on the nucleus and \(B\) is \(2.180 \times 10^{-18} \mathrm{~J}\). Find the ionization energy of the \(\mathrm{Li}^{2+}\) ion in its first excited state in kilojoules per mole.

Short Answer

Expert verified
Answer: The ionization energy of the Li²⁺ ion in its first excited state is approximately -267.0 kJ/mol.

Step by step solution

01

Identify the values for the Li²⁺ ion

The Li²⁺ ion has a charge on the nucleus (Z) of 3, as it is lithium with two electrons removed. The first excited state means that the principal quantum number (n) is 2. The constant B is given as 2.180 × 10⁻¹⁸ J.
02

Calculate the energy in the n=1 state

Using the given formula, we plug in the values for Z, n, and B: E₁ = -B * Z² / n² E₁ = -2.180 × 10⁻¹⁸ J * (3²) / (1²) E₁ = -5.914 × 10⁻¹⁸ J
03

Calculate the energy in the n=2 state

Now we find the energy in the n=2 state: E₂ = -B * Z² / n² E₂ = -2.180 × 10⁻¹⁸ J * (3²) / (2²) E₂ = -1.479 × 10⁻¹⁸ J
04

Calculate the ionization energy

The ionization energy is the energy required to remove the electron, which is the difference between the energy in the n=2 state and n=1 state: Ionization energy = E₁ - E₂ Ionization energy = (-5.914 × 10⁻¹⁸ J) - (-1.479 × 10⁻¹⁸ J) Ionization energy = -4.435 × 10⁻¹⁸ J
05

Convert joules to kilojoules per mole

To convert the ionization energy in joules to kilojoules per mole, we first divide by 1000 to get the energy in kilojoules: Ionization energy = -4.435 × 10⁻²¹ kJ Next, we multiply by Avogadro's number (6.022 × 10²³) to convert to kilojoules per mole: Ionization energy = -4.435 × 10⁻²¹ kJ * (6.022 × 10²³) Ionization energy = -267.0 kJ/mol Thus, the ionization energy of the Li²⁺ ion in its first excited state is approximately -267.0 kJ/mol.

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Most popular questions from this chapter

Suppose that the spin quantum number could have the values \(\frac{1}{2}, 0\), and \(-\frac{1}{2} .\) Assuming that the rules governing the values of the other quantum numbers and the order of filling sublevels were unchanged, (a) what would be the electron capacity of an s sublevel? A p sublevel? A d sublevel? (b) how many electrons could fit in the \(\mathbf{n}=3\) level? (c) what would be the electron configuration of the element with atomic number 8? \(17 ?\)

Explain the difference between (a) the Bohr model of the atom and the quantum mechanical model. (b) wavelength and frequency. (c) the geometries of the three different p orbitals.

For the following pairs of orbitals, indicate which is lower in energy in a many-electron atom. (a) \(3 \mathrm{~d}\) or \(4 \mathrm{~s}\) (b) \(4 \mathrm{f}\) or \(3 \mathrm{~d}\) (c) \(2 \mathrm{~s}\) or \(2 \mathrm{p}\) (d) \(4 \mathrm{f}\) or \(4 \mathrm{~d}\)

A bar code scanner in a grocery store is a He-Ne laser with a wavelength of \(633 \mathrm{~nm}\). If \(12 \mathrm{~kJ}\) of energy is given off while the scanner is "reading" bar codes, how many photons are emitted?

Name and give the symbol for the element with the characteristic given below: (a) Its electron configuration is \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}\). (b) Lowest ionization energy in Group 14. (c) Its \(+2\) ion has the configuration \(\left[{ }_{18} \mathrm{Ar}\right] 3 \mathrm{~d}^{5}\). (d) It is the alkali metal with the smallest atomic radius. (e) Largest ionization energy in the fourth period.
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