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Use the periodic table to write electron configurations for each element. (a) \(\mathrm{Sr}\) (b) \(Y\) (c) \(\mathrm{Ti}\) (d) \(\mathrm{Te}\)

Short Answer

Expert verified
\(\mathrm{Sr: } 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2\), \(Y: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^1\), \(Ti: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^2\), \(Te: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^4\).

Step by step solution

01

Identify the Atomic Number

First, find the atomic number of the element from the periodic table. This atomic number represents the number of protons and, in a neutral atom, also the number of electrons.
02

Determine the Electron Configuration for Strontium (Sr)

Strontium (Sr) has an atomic number 38. Follow the periodic table to fill orbitals: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s. Sr electron configuration is written as \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2\).
03

Determine the Electron Configuration for Yttrium (Y)

Yttrium (Y) has an atomic number 39. Following the order from Step 1, fill orbitals up to the 5s orbital, then the 4d. Y electron configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^1\).
04

Determine the Electron Configuration for Titanium (Ti)

Titanium (Ti) has an atomic number 22. Proceed with orbital filling: 1s, 2s, 2p, 3s, 3p, 4s, 3d. The electron configuration for Ti is \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^2\).
05

Determine the Electron Configuration for Tellurium (Te)

Tellurium (Te) has an atomic number 52. Follow the order of the periodic table: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p. Te electron configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^4\).

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Key Concepts

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

Periodic Table
The periodic table is a fundamental tool in chemistry that organizes all known chemical elements in an informative array. Elements are arranged from left to right and top to bottom in order of increasing atomic number, which is the total number of protons in an element's nucleus. Each row corresponds to a period, and each column forms a group with similar chemical properties.

As you move across a period, the atomic radius decreases, electron affinity increases, and ionization energy increases. Vertically, groups maintain similar electronic configurations, particularly in their valence shells, which explains their shared chemical characteristics. Understanding the layout of the periodic table is crucial for determining electron configurations, as it follows the order of the elements.
Atomic Number
The atomic number of an element, symbolized as 'Z', is the number of protons found in the nucleus of an atom. It is a unique identifier for each chemical element and is used to distinguish one element from another on the periodic table. Since atoms are electrically neutral, the atomic number also indicates the number of electrons in the orbitals of an atom.

As we have seen in the textbook exercise, identifying the atomic number is the first step in writing out the electron configuration of an element. This number directs us through the order of orbital filling, ensuring that we allocate the correct number of electrons for each element.
Orbital Filling Order
Orbital filling order, also known as electron configuration order, is the sequence in which electrons populate the atomic orbitals of an atom. Electrons fill orbitals starting from the lowest energy levels to the higher ones, following the Aufbau principle, which translates to 'building up'. The sequence is typically: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, and so on, where the numbers indicate the energy level and the letters stand for the type of orbital (s, p, d, or f).

Prior knowledge of orbital energy and the specific order is paramount when writing electron configurations, as incorrectly filling orbitals can lead to an inaccurate representation of an element's electrons.
Electron Configuration Notation
Electron configuration notation is a shorthand method of writing an atom's electron arrangement. It consists of a sequence of numbers, letters, and superscripts that concisely reflects how electrons are distributed among the orbitals and energy levels. In this notation, the number indicates the principal energy level, the letter (s, p, d, f) specifies the type of orbital, and the superscript denotes the number of electrons in that orbital.

For instance, the electron configuration of hydrogen is written as 1s¹, indicating one electron in the 1s orbital. As the exercises demonstrate, this notation provides a clear and organized way of representing the electron configurations that can easily be translated to the visual layout of the periodic table.

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Most popular questions from this chapter

Which electron is, on average, farther from the nucleus: an electron in a \(3 p\) orbital or an electron in a \(4 p\) orbital?

Write full electron configurations for each element. (a) \(\mathrm{Sr}\) (b) Ge (c) \(\mathrm{Li}\) (d) \(\mathrm{Kr}\)

List the quantum-mechanical orbitals through \(5 s\), in the correct energy order for multi-electron atoms.

You learned in this chapter that ionization generally increases as you move from left to right across the periodic table. However, consider the following data, which shows the ionization energies of the period 2 and 3 elements: $$ \begin{array}{ccccc} \text { Group } & \begin{array}{c} \text { Period 2 } \\ \text { Elements } \end{array} & \begin{array}{c} \text { lonization } \\ \text { Energy } \\ \text { (kJ/mol) } \end{array} & \begin{array}{c} \text { Period 3 } \\ \text { Elements } \end{array} & \begin{array}{c} \text { Ionization } \\ \text { Energy } \\ \text { (kJ/mol) } \end{array} \\ \text { 1A } & \text { Li } & 520 & \text { Na } & 496 \\ \text { 2A } & \text { Be } & 899 & \text { Mg } & 738 \\ \text { 3A } & \text { B } & 801 & \text { Al } & 578 \\ \text { 4A } & \text { C } & 1086 & \text { Si } & 786 \\ \text { 5A } & \text { N } & 1402 & \text { P } & 1012 \\ \text { 6A } & \text { 0 } & 1314 & \text { S } & 1000 \\ \text { 7A } & \text { F } & 1681 & \text { Cl } & 1251 \\ \text { 8A } & \text { Ne } & 2081 & \text { Ar } & 1521 \\ \hline \end{array} $$ Notice that the increase is not uniform. In fact, ionization energy actually decreases a bit in going from elements in group \(2 \mathrm{~A}\) to \(3 \mathrm{~A}\) and then again from \(5 \mathrm{~A}\) to \(6 \mathrm{~A}\). Use what you know about electron configurations to explain why these dips in ionization energy exist.

What is the maximum number of electrons that can occupy the \(n=4\) quantum shell?

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