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For each of the following elements, which set of orbitals is last to fill in the ground state? a. radium b. iodine c. gold d. uranium

Short Answer

Expert verified
In the ground state, the last set of orbitals to fill for each element is: a. Radium: \(7s\) b. Iodine: \(5p\) c. Gold: \(6s\) d. Uranium: \(5f\)

Step by step solution

01

Find the atomic number

For each element, we will find the atomic number, which is equal to the number of protons and electrons in a neutral atom. We can find atomic numbers on the periodic table: a. Radium (Ra): Atomic number = 88 b. Iodine (I): Atomic number = 53 c. Gold (Au): Atomic number = 79 d. Uranium (U): Atomic number = 92
02

Write the electron configuration

Using the Aufbau Principle, we will build the electron configuration for each element. The Aufbau Principle states that electrons are added to the lowest energy orbitals first before filling the higher energy ones. a. Radium (Ra): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^{10} 6p^6 7s^2\) b. Iodine (I): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^5\) c. Gold (Au): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^1 4f^{14} 5d^{10}\) d. Uranium (U): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^{10} 6p^6 7s^2 5f^{3}\)
03

Identify the last set of orbitals to fill

Finally, we will find the last set of orbitals in the electron configuration for each element as this will be the orbitals that fill up last in the ground state. a. Radium (Ra): \(\underline{7s^2}\) b. Iodine (I): \(\underline{5p^5}\) c. Gold (Au): \(\underline{6s^1}\) d. Uranium (U): \(\underline{5f^{3}}\) In the ground state, the last set of orbitals to fill for each element is: a. Radium: \(7s\) b. Iodine: \(5p\) c. Gold: \(6s\) d. Uranium: \(5f\)

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Key Concepts

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

Atomic Number
The atomic number is a fundamental property of an element. It tells us the number of protons in an atom's nucleus. This is crucial because it defines the element. For a neutral atom, the atomic number also equals the number of electrons.
The atomic number is used to determine where an element is on the periodic table. Each element has a unique atomic number that increases as you move across a period (row) or down a group (column).
For example,
  • Radium (Ra) has an atomic number of 88.
  • Iodine (I) has an atomic number of 53.
  • Gold (Au) has an atomic number of 79.
  • Uranium (U) has an atomic number of 92.
This atomic number sequence helps us understand the electron configuration and the chemical behavior of these elements.
Aufbau Principle
The Aufbau Principle is key to understanding electron configuration. "Aufbau" is a German word meaning "building up," and this principle relates to how electrons fill up orbitals in an atom.
According to this principle, electrons occupy the lowest energy orbitals available, gradually moving to higher ones as needed.
Here's a simple summary:
  • Electrons start by filling the 1s orbital.
  • Once 1s is full, electrons fill the 2s, then the 2p, and so on.
By following this filling order dictated by the Aufbau chart, we can create the electron configuration, which indicates the distribution of electrons in an atom's orbitals.
For example, in Gold (Au), the final electron fills the 6s orbital after passing through all preceding lower-energy orbitals according to the Aufbau sequence.
Ground State Orbitals
The ground state of an atom is the lowest energy state with all electrons in the lowest possible orbitals. The position of these electrons is expressed through electron configuration.
When an atom is in its ground state, it is stable and does not require additional energy to maintain its electron arrangement.
In this state, electrons fill the orbitals in a specific order, making use of the lowest available energy levels first.
For instance:
  • In Radium's configuration, the final orbital filled is 7s.
  • In Iodine, the last to fill is 5p.
These final orbitals indicate the atom's normalized low-energy configuration when it is not excited.
Periodic Table Elements
The periodic table is an essential tool for chemists and students alike. It arranges all known elements in a systematic way according to their atomic number.
This organization reflects periodic trends and helps predict various chemical properties.
Elements are arranged in rows (periods) and columns (groups) that highlight repeating patterns of properties.
Each element's position can give insight into its electron configuration. For example,
  • Elements in the same group often have similar valence electron configurations.
  • Radium, located in group 2, is an alkaline earth metal with its valence electrons in the 7s orbital.
  • Uranium, found in the actinide series, has its outer electrons in 5f orbital, indicating its placement in the f-block.
Understanding this layout allows us to predict chemical reactivity, physical properties, and even electron configuration patterns.

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Most popular questions from this chapter

Which of the following statements is(are) true? a. F has a larger first ionization energy than does Li. b. Cations are larger than their parent atoms. c. The removal of the first electron from a lithium atom (electron configuration is 1\(s^{2} 2 s^{1} )\) is exothermic - that is, removing this electron gives off energy. d. The He atom is larger than the \(\mathrm{H}^{+}\) ion. e. The Al atom is smaller than the Li atom.

A certain oxygen atom has the electron configuration 1\(s^{2} 2 s^{2} 2 p_{x}^{2} 2 p_{y}^{2} .\) How many unpaired electrons are present? Is this an excited state of oxygen? In going from this state to the ground state, would energy be released or absorbed?

For each of the following pairs of elements $$(\mathrm{Mg} \text { and } \mathrm{K}) \quad(\mathrm{F} \text { and } \mathrm{Cl})$$ pick the atom with a. more favorable (exothermic) electron affinity. b. higher ionization energy. c. larger size.

Consider an electron for a hydrogen atom in an excited state. The maximum wavelength of electromagnetic radiation that can completely remove (ionize) the electron from the H atom is 1460 \(\mathrm{nm}\) . What is the initial excited state for the electron \((n=?) ?\)

Identify the following three elements. a. The ground-state electron configuration is \([\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{4}\) b. The ground-state electron configuration is \([\mathrm{Ar}] 4 s^{2} 3 d^{10} 4 p^{2}\) c. An excited state of this element has the electron configuration 1\(s^{2} 2 s^{2} 2 p^{4} 3 s^{1}\)

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