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What is the identity of the element \(\mathrm{X}\) in the following ions? (a) \(\mathrm{X}^{2+}\), a cation that has 36 electrons (b) \(\mathrm{X}\), an anion that has 36 electrons

Short Answer

Expert verified
(a) Strontium (Sr); (b) Bromine (Br).

Step by step solution

01

Understanding the Problem for Part (a)

We are given a cation \(\mathrm{X}^{2+}\) that has 36 electrons. Our task is to find the atomic number of element \(\mathrm{X}\) when it was neutral.
02

Calculating Atomic Number for Part (a)

In a cation \(X^{2+}\), the atom has lost 2 electrons. Thus the neutral atom \(\mathrm{X}\) has 2 more electrons than the cation. Hence, \(\mathrm{X}\) must have \(36 + 2 = 38\) electrons in its neutral state, meaning its atomic number is 38.
03

Identifying Element for Part (a)

The periodic table shows that the element with atomic number 38 is Strontium (Sr). Therefore, the cation \(\mathrm{X}^{2+}\) is Strontium.
04

Understanding the Problem for Part (b)

We are given an anion \(\mathrm{X}^{-}\) where \(X\) has 36 electrons. We need to determine the atomic number of the neutral element \(\mathrm{X}\).
05

Calculating Atomic Number for Part (b)

In an anion \(X^{-}\), the atom has gained 1 electron. Thus, the neutral atom \(\mathrm{X}\) has 1 less electron than the anion. Consequently, \(\mathrm{X}\) must have \(36 - 1 = 35\) electrons in its neutral state, meaning its atomic number is 35.
06

Identifying Element for Part (b)

The periodic table shows that the element with atomic number 35 is Bromine (Br). Therefore, the anion \(\mathrm{X}^{-}\) is Bromine.

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Key Concepts

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

Cations and Anions
Cations and anions are types of ions, which are atoms or molecules that have gained or lost electrons and have an electric charge. When an atom loses electrons, it becomes a cation, which has a positive charge due to having more protons than electrons. Conversely, when an atom gains electrons, it becomes an anion, carrying a negative charge because it now has more electrons than protons.

In our exercise, the element X forms a cation, denoted as \( \mathrm{X}^{2+} \), which means it has lost 2 electrons. Originally having 36 electrons in the cation form, X would have 38 electrons in its neutral state. For the anion \( \mathrm{X}^{-} \), X has gained one additional electron, indicating it had 35 electrons in the neutral state since it shows 36 as the anion. Identifying these changes in electrons helps determine the chemical nature and identity of the ions.

Understanding cations and anions is fundamental in chemistry as these charged species are significant in many chemical reactions, governing the interactions between different materials and elements.
Periodic Table
The periodic table is a comprehensive chart that organizes chemical elements according to their atomic number, electron configurations, and recurring chemical properties. This arrangement makes it easier to predict characteristics and behaviors of elements.

For instance, in our exercise, the periodic table assists in identifying the elements based on the atomic numbers derived from their electron configurations. The element with atomic number 38 is Strontium (\( \mathrm{Sr} \)), and the element with atomic number 35 is Bromine (\( \mathrm{Br} \)). The periodic table is pivotal in such identifications as it lays out all known elements, providing easy reference and insight into the nature of the elements like their availability, common compounds, and general chemical behavior.
  • Understanding Groups and Periods: Elements are arranged in rows (periods) and columns (groups) which reflect increments in electron configurations.
  • Elemental Trends: Trends such as electronegativity and atomic radius are consistent across periods and groups, offering the ability to predict element behavior.
The periodic table's design highlights its utility as a fundamental tool for chemists and students alike.
Electron Configuration
Electron configuration refers to the distribution of electrons of an atom or molecule in atomic or molecular orbitals. Understanding how electrons are arranged helps predict how an element might interact with others.

In our exercise, electron configuration is vital. For example, the cation \( \mathrm{X}^{2+} \) with 36 electrons suggests its neutral state, based on periodic table, would involve filling orbitals up to 38 electrons, typical for Strontium. Similarly, an anion \( \mathrm{X}^{-} \) with 36 electrons reveals its neutral form corresponds to 35 electrons, suited for Bromine.
  • Why Electrons Matter: Electrons in outer shells determine bonding and reactivity, making their configuration essential in predicting chemical behavior.
  • Orbital Filling Order: Electron shells fill in a specific order (1s, 2s, 2p, etc.), crucial for determining configuration accurately.
  • Application in Problem Solving: Recognizing configurations allows identification of elements, understanding ion formation, and chemical bonding.
The mastery of electron configuration equips students with a clearer insight into not just elemental identity, but the possibilities of chemical transformations and reactions.

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Most popular questions from this chapter

An ionic liquid consisting of a bulky amine cation and chloride anion has a melting point of \(41^{\circ} \mathrm{C}\). For a green chemistry solvent application, it is desirable to have a melting point below room temperature \(\left(25^{\circ} \mathrm{C}\right)\) so energy is not required to heat the compound. The melting point of the ionic liquid can be altered by replacing chloride with a different anion. The anion possibilities are: \(\mathrm{F}^{-}, \mathrm{Se}^{2-}, \mathrm{O}^{2-}\), and \(\mathrm{Br}^{-}\) (a) Write the electron configuration for each of these ions. (b) Which ions are isoelectronic? (c) Which ion is the best choice for replacing the chloride ion to make an ionic liquid with a lower melting temperature?

The following ions all have the same number of electrons: \(\mathrm{Ti}^{4+}, \mathrm{Sc}^{3+}, \mathrm{Ca}^{2+}, \mathrm{S}^{2-} .\) Order them according to their expected sizes, and explain your answer. 6.88 Calculate overall energy changes in kilojoules per mole for the formation of \(\mathrm{MgF}(s)\) and \(\mathrm{MgF}_{2}(s)\) from their elements. In light of your answers, which compound is more likely to form in the reaction of magnesium with fluorine, \(\mathrm{MgF}\) or \(\mathrm{MgF}_{2}\) ? The following data are needed: \(E_{\mathrm{ea}}\) for \(\mathrm{F}(\mathrm{g})=-328 \mathrm{~kJ} / \mathrm{mol}\) \(E_{i 1}\) for \(\mathrm{Mg}(g)=+737.7 \mathrm{~kJ} / \mathrm{mol}\) \(E_{i 2}\) for \(\mathrm{Mg}(g)=+1450.7 \mathrm{~kJ} / \mathrm{mol}\) Heat of sublimation for \(\mathrm{Mg}(s)=+147.7 \mathrm{~kJ} / \mathrm{mol}\)Bond dissociation energy for \(\mathrm{F}_{2}(\mathrm{~g})=+158 \mathrm{~kJ} / \mathrm{mol}\) Lattice energy for \(\mathrm{MgF}_{2}(s)=+2952 \mathrm{~kJ} / \mathrm{mol}\) Lattice energy for \(\mathrm{MgF}(s)=930 \mathrm{~kJ} / \mathrm{mol}\) (estimated)

Iron is commonly found as \(\mathrm{Fe}, \mathrm{Fe}^{2+}\), and \(\mathrm{Fe}^{3+}\). (a) Write electron configurations for each of the three. (b) What are the \(n\) and \(l\) quantum numbers of the electron removed on going from \(\mathrm{Fe}^{2+}\) to \(\mathrm{Fe}^{3+}\) ? (c) The third ionization energy of Fe is \(E_{i 3}=+2952 \mathrm{~kJ} / \mathrm{mol}\). What is the longest wavelength of light that could ionize \(\mathrm{Fe}^{2+}(g)\) to \(\mathrm{Fe}^{3+}(g) ?\) (d) The third ionization energy of \(\mathrm{Ru}\) is less than the third ionization energy of Fe. Explain.

What is the electron configuration of \(\mathrm{Ca}^{2+}\) ? What is the electron configuration of \(\mathrm{Ti}^{2+}\) ?

Predict the ground-state electron configuration for each of the following ions: (a) \(\mathrm{Ra}^{2+}\) (b) \(\mathrm{Ni}^{2+}\) (c) \(\mathrm{N}^{3-}\)

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