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The energies of the \(4 s, 4 p,\) and \(4 d\) states of potassium are given in Example \(41.10 .\) Calculate \(Z_{\text {eff }}\) for each state. What trend do your results show? How can you explain this trend?

Short Answer

Expert verified
The calculated effective nuclear charge (\(Z_{eff}\)) for potassium in 4s, 4p and 4d states are 0.5, 1.5 and 1.65 respectively. The trend shows that \(Z_{eff}\) increases from 4s to 4d state. This is due to the fact that as you move from 4s to 4d, penetration of those orbitals decreases, meaning, the 4s electrons are closer to the nucleus and screen the 4p and 4d electrons more than they screen each other.

Step by step solution

01

Calculate \(Z_{eff}\) for 4s state

According to Slater’s Rules, for 'ns' or 'np' electrons: \(Z_{eff} = Z - S\) where Z is the atomic number and S is the shielding constant. For 4s state, based on the rule, every electron in the 4s, 4p, 4d and 4f sub-shells contributes 0.35 to S except the one electron being considered, in addition every electron in the n=3 shell contributes 0.85 and every electron in the n=1 or n=2 shells contributes 1 to S. So, S for 4s state will be: \(S = (8*1) + (10*0.85) + (0.35*0) = 18.5\) And, \(Z_{eff} = Z - S = 19 - 18.5 = 0.5\) for 4s state.
02

Calculate \(Z_{eff}\) for 4p state

For 4p state, based on Slater’s Rule, every other electron in the 4p, 4d and 4f sub-shells contributes 0.35 to S, every electron in the 4s, 3d, 3p and 3s sub-shells contributes 0.85 and every electron in the n=1 or n=2 shells contributes 1 to S. So, \(S = (8*1) + (10*0.85) + (1*0.85) + (0.35*0) = 17.5\) And, \(Z_{eff} = Z - S = 19 - 17.5 = 1.5\) for 4p state.
03

Calculate \(Z_{eff}\) for 4d state

For 4d state, every electron on the same level contributes 0.35 to S, every electron in the 4p, 4s, 3d, 3p and 3s sub-shells contributes 0.85 and every electron in the n=1 and n=2 shells contributes 1 So, \(S = (8*1) + (11*0.85) + (0.35*0) = 17.35\) And, \(Z_{eff} = Z - S = 19 - 17.35 = 1.65\) for 4d state.
04

Explain the trend

From the calculated values of \(Z_{eff}\) for 4s, 4p and 4d states, the trend shows that \(Z_{eff}\) increases from 4s to 4d state. This is because as you move from 4s to 4d, penetration decreases. The 4s electrons are closer to the nucleus and screen the 4p and 4d electrons more than they screen each other.

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Key Concepts

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

Slater's Rules
Understanding the concept of effective nuclear charge, or \(Z_{eff}\), is essential in grasping how electrons behave in an atom. Slater's Rules are a set of guidelines that help us predict \(Z_{eff}\) for electrons in different orbitals. They provide a way to calculate a shielding constant, \(S\), which reflects the extent to which an electron is shielded from the nucleus by other electrons. The rule takes into account the electron configuration of an atom and assigns values based on the distribution of electrons across different shells and subshells.

To apply these rules, you first identify the electron of interest and then consider the repulsion from other electrons. Contributions to \(S\) vary; electrons in the same group contribute less to shielding than those closer to the nucleus. In the exercise, Slater's Rules help us see why different orbitals, such as 4s, 4p, and 4d, have varying \(Z_{eff}\). These calculations are crucial as they influence properties like atomic radius, ionization energy, and electron affinity, making Slater's Rules a fundamental concept in atomic and quantum chemistry.
Atomic Orbital Penetration
Atomic orbital penetration is a concept that describes how close an electron in a given orbital can get to the nucleus. It's a direct factor affecting the shielding effect and, ultimately, the effective nuclear charge. Electrons in orbitals with higher penetration (such as s orbitals) experience a greater nuclear charge, since they can penetrate the electron cloud more deeply and approach the nucleus more closely.

This phenomenon is precisely what we observe in the exercise when comparing the 4s, 4p, and 4d states of potassium. The 4s orbital has a higher penetrating ability than 4p and 4d. Consequently, electrons in the 4s orbital are more effective at shielding (thus reducing \(Z_{eff}\)) than those in the 4p or 4d orbitals, which are further from the nucleus and have lower penetration. Understanding orbital penetration helps explain trends across the periodic table, such as variations in atomic size and ionization energies within a given period.
Electron Shielding
Electron shielding, also known as screening, is the process by which inner electrons reduce the nuclear charge felt by outer electrons. This creates an effective nuclear charge that is less than the actual charge of the nucleus due to the partial cancellation of the nuclear charge by the negative charge of the inner electrons.

In the context of our exercise, electron shielding is what leads to the calculation of different \(S\) values for the 4s, 4p, and 4d electrons. Each electron cloud between the nucleus and the electron of interest contributes to the shielding effect, thereby reducing the nuclear charge experienced by outer electrons. The exercise improvement advice suggests emphasizing this point to strengthen the understanding that the differences in the calculated \(Z_{eff}\) for each state are due to the varying degrees of electron shielding. The stronger the shielding, the lower the \(Z_{eff}\), and vice versa. Electron shielding plays a vital role in many chemical properties and behaviors, making it crucial for students to understand its impact on atomic structure.

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Most popular questions from this chapter

A hydrogen atom is in a state that has \(L_{z}=2 h .\) In the semiclassical vector model, the angular momentum vector \(\overrightarrow{\boldsymbol{L}}\) for this state makes an angle \(\theta_{L}=63.4^{\circ}\) with the \(+z\) -axis. (a) What is the \(l\) quantum number for this state? (b) What is the smallest possible \(n\) quantum number for this state?

An electron is in the hydrogen atom with \(n=5 .\) (a) Find the possible values of \(L\) and \(L_{z}\) for this electron, in units of \(h .\) (b) For each value of \(L\), find all the possible angles between \(\overrightarrow{\boldsymbol{L}}\) and the z-axis. (c) What are the maximum and minimum values of the magnitude of the angle between \(\vec{L}\) and the \(z\) -axis?

In a particular state of the hydrogen atom, the angle between the angular momentum vector \(\vec{L}\) and the \(z\) -axis is \(\theta=26.6^{\circ}\). If this is the smallest angle for this particular value of the orbital quantum number \(l\). what is \(l ?\)

Consider an electron in the \(N\) shell. (a) What is the smallest orbital angular momentum it could have? (b) What is the largest orbital angular momentum it could have? Express your answers in terms of \(\hbar\) and in SI units. (c) What is the largest orbital angular momentum this electron could have in any chosen direction? Express your answers in terms of \(\hbar\) and in SI units. (d) What is the largest spin angular momentum this electron could have in any chosen direction? Express your answers in terms of \(\hbar\) and in \(\mathrm{SI}\) units. (e) For the electron in part (c), what is the ratio of its spin angular momentum in the \(z\) -direction to its orbital angular momentum in the \(z\) -direction?

Show that the total number of atomic states (including different spin states) in a shell of principal quantum number \(n\) is \(2 n^{2}\). [Hint: The sum of the first \(N\) integers \(1+2+3+\cdots+N\) is equal to \(N(N+1) / 2 .]\) (b) Which shell has 50 states?

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