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In studying electron screening in multielectron atoms, you begin with the alkali metals. You look up experimental data and find the results given in the table. $$ \begin{array}{l|cccccc} \text { Element } & \mathrm{Li} & \mathrm{Na} & \mathrm{K} & \mathrm{Rb} & \mathrm{Cs} & \mathrm{Fr} \\ \hline \begin{array}{l} \text { Ionization energy } \\ (\mathbf{k} \mathbf{J} / \mathrm{mol}) \end{array} & 520.2 & 495.8 & 418.8 & 403.0 & 375.7 & 380 \\ & & & & & & \end{array} $$ The ionization energy is the minimum energy required to remove the least-bound electron from a ground-state atom. (a) The units \(\mathrm{kJ} / \mathrm{mol}\) given in the table are the minimum energy in kJ required to ionize 1 mol of atoms. Convert the given values for ionization energy to the energy in eV required to ionize one atom. (b) What is the value of the nuclear charge \(Z\) for each element in the table? What is the \(n\) quantum number for the least-bound electron in the ground state? (c) Calculate \(Z_{\text {cff }}\) for this electron in each alkali-metal atom. (d) The ionization energies decrease as \(Z\) increases. Does \(Z_{\text {eff }}\) increase or decrease as \(Z\) increases? Why does \(Z_{\text {eff }}\) have this behavior?

Step by step solution

01

Conversion of Energy from kJ/mol to eV/atom

The conversion can be achieved using the relationship 1 eV = 96.485 kJ/mol. The conversion involves dividing the given energy by Avogadro's number and converting joules to eV. The formula would be \(Energy_{atom} = Energy_{mol} / (6.022*10^{23} * 96.485)\) in eV.

02

Determine Nuclear Charge Z and Quantum Number n

For alkali metals, the atomic number Z corresponds to the number of protons in the atom and equals the position in the periodic table. For example, Li has Z = 3. The quantum number n for the least tightly bound electron (valence electron) in the ground state equals the period number. For Li, n = 2.

03

Calculate the Effective Nuclear Charge

The effective nuclear charge \(Z_{eff}\) can be approximated using the formula \(Z_{eff} = Z - S\), where Z is the atomic number and S is the screening constant. For alkali metals, S can be approximated as \(n - 1\). So, \(Z_{eff} = Z - (n - 1)\). Calculate \(Z_{eff}\) for each element.

04

Determine the Trend of Effective Nuclear Charge

As Z increases, the ionization energies decrease. This is because as Z increases, the distance between outermost electron and nucleus also increases and electron experiences less nuclear charge and it can be removed more easily. Thus, although Z is increasing, \(Z_{eff}\) essentially remains the same for alkali metals because the additional positive charge is offset by the additional electron.

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Key Concepts

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

Ionization Energy

Ionization energy is the amount of energy required to remove the most loosely bound electron from an atom in its ground state. It is measured in kilojoules per mole (kJ/mol) or electronvolts (eV) per atom, depending on the scale. By understanding ionization energy, we gain insight into how strongly an electron is bound to an atom.
For alkali metals, ionization energy tends to be lower compared to other elements. This is because they have only one electron in their outermost shell, which is relatively easy to remove. As we go down the group in the periodic table—from lithium (Li) to francium (Fr)—the ionization energy decreases. This trend occurs due to the increasing atomic size and shielding effect, which makes the outermost electron easier to remove.

Nuclear Charge

Nuclear charge, often represented by the symbol \( Z \), represents the number of protons present in the nucleus of an atom. This positive charge attracts the negatively charged electrons. As the number of protons increases, the nuclear charge also increases, pulling electrons closer to the nucleus.
In alkali metals, the nuclear charge increases down the group (for Li, Na, K, etc.), as each succeeding element has one extra proton compared to the previous one. Despite the increasing nuclear charge, the outermost electron in these elements is less affected due to greater distance from the nucleus and increased electron shielding.

Quantum Number

The quantum number \( n \) acts as a shell indicator and is crucial in determining an electron's energy level within an atom. It is the principal quantum number associated with the main electron shell or energy level in which the electron resides.
For alkali metals, the least-bound electron, or valence electron, is in the outermost shell, with the principal quantum number \( n \) corresponding to the period number in the periodic table. For example, lithium (Li) has \( n = 2 \), sodium (Na) has \( n = 3 \), and potassum has \( n = 4 \), and so on, as you move down the group.

Effective Nuclear Charge

Effective nuclear charge, denoted \( Z_{eff} \), accounts for the net positive charge experienced by an electron in the outermost shell. It takes into account both the attraction from the nucleus (protons) and the repulsion from electrons in inner shells. This concept is crucial for understanding electron screening.
The formula for effective nuclear charge is \( Z_{eff} = Z - S \), where \( S \) is the screening constant approximated as \( n - 1 \) for alkali metals. Despite the increase in nuclear charge (\( Z \)) as we move down the group, \( Z_{eff} \) tends to remain relatively constant for these elements because the additional valence electron is also shielded more by the increase in inner-shell electrons.

Alkali Metals

Alkali metals are a group of elements found in Group 1 of the periodic table. They are characterized by having a single electron in their outermost shell, which makes them highly reactive. This single valence electron is easily lost, making them great conductors and often used in various applications due to their reactivity.
Properties of alkali metals include:

• Softness and tendency to react with water
• Low melting points compared to most other metals
• Forming strong bases (alkalis) upon reacting with water

As we move down this group from lithium (Li) to francium (Fr), these properties intensify due to the gradual increase in atomic and ionic size.