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Give the formula of a negative ion that would have the same number of electrons as each of the following positive ions. a. \(\mathrm{Na}^{+} \quad\) c. \(\mathrm{Al}^{3+}\) b. \(\mathrm{Ca}^{2+} \quad\) d. \(\mathrm{Rb}^{+}\)

Short Answer

Expert verified
a. \(\mathrm{F}^-\) b. \(\mathrm{S}^{2-}\) c. No suitable ion found d. \(\mathrm{Br}^-\)

Step by step solution

01

Determine the number of electrons in the positive ions

To determine the number of electrons in each positive ion, we will subtract the charge of each ion from the atomic number of the element. a. \(\mathrm{Na}^{+}\) - Sodium (Na) has an atomic number of 11. Therefore, the number of electrons in \(\mathrm{Na}^+\) is 11 - 1 = 10 electrons. b. \(\mathrm{Ca}^{2+}\) - Calcium (Ca) has an atomic number of 20. Therefore, the number of electrons in \(\mathrm{Ca}^{2+}\) is 20 - 2 = 18 electrons. c. \(\mathrm{Al}^{3+}\) - Aluminum (Al) has an atomic number of 13. Therefore, the number of electrons in \(\mathrm{Al}^{3+}\) is 13 - 3 = 10 electrons. d. \(\mathrm{Rb}^{+}\) - Rubidium (Rb) has an atomic number of 37. Therefore, the number of electrons in \(\mathrm{Rb}^+\) is 37 - 1 = 36 electrons.
02

Find elements with same proton number but negative charge

In this step, we will find elements that match the number of protons as the positive ions, but have a negative charge. a. Fluorine (F) has an atomic number of 9. To have 10 electrons, the charge on the fluorine ion will be -1. Hence, the negative ion is \(\mathrm{F}^-\). b. Sulfur (S) has an atomic number of 16. To have 18 electrons, the charge on the sulfur ion will be -2. Hence, the negative ion is \(\mathrm{S}^{2-}\). c. Neon (Ne) has an atomic number of 10. To have 10 electrons, the charge on the neon ion will be 0. Since neon is a noble gas, it does not form ions. Therefore, there is no negative ion that matches this condition. d. Bromine (Br) has an atomic number of 35. To have 36 electrons, the charge on the bromine ion will be -1. Hence, the negative ion is \(\mathrm{Br}^-\).
03

Write the ion's formula

Finally, we will write the ion formulas for each of the elements we found in Step 2. a. \(\mathrm{F}^-\) b. \(\mathrm{S}^{2-}\) c. No suitable ion found d. \(\mathrm{Br}^-\)

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Key Concepts

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

Electron Configuration
Electron configuration is the arrangement of electrons in an atom or ion. This configuration determines the chemical properties of an element. Electrons are filled into orbitals in a sequence determined by the principles of quantum mechanics.

In a neutral atom, the number of electrons equals the atomic number. They occupy orbitals starting from the lowest energy level, like the 1s, 2s, and 2p orbitals for the first few elements.
  • 1s orbital can hold up to 2 electrons.
  • 2s orbital can hold up to 2 electrons.
  • 2p orbitals can hold up to 6 electrons (3 orbitals, 2 electrons each).
When electrons are removed or added in the case of ions, the electron configuration changes. For example, when sodium ( Na ) loses one electron to become Na ^+ , its electron configuration changes from 1s² 2s² 2p⁶ 3s¹ to 1s² 2s² 2p⁶.
Atomic Number
The atomic number is a fundamental property of an element. It represents the number of protons in the nucleus of an atom.

This number is crucial because it defines the identity of an element, determining its place in the periodic table. For example, sodium has an atomic number of 11, meaning every sodium atom has 11 protons.

The atomic number also equals the number of electrons in a neutral atom, which defines its electron configuration and chemical behavior. However, when ions form, electrons are gained or lost without affecting the atomic number.
  • Atomic number dictates the element’s identity.
  • In a neutral atom, protons = electrons = atomic number.
  • Atomic number remains unchanged when an atom becomes an ion.
Negative Ions
Negative ions, also called anions, form when an atom gains electrons. This usually occurs because an atom is trying to complete its valence shell to achieve a stable electron configuration.

For example, a fluorine atom (atomic number 9) gains one electron to achieve the same electron configuration as neon, resulting in the fluoride ion ( F^− ). The additional electron gives the anion a negative charge, equal to the number of gained electrons.
  • Anions have more electrons than protons.
  • Commonly formed by nonmetals, which gain electrons to complete their outer shell.
  • Anion charge is represented with a minus sign (e.g., Cl^−).
Positive Ions
Positive ions, or cations, form when an atom loses electrons. This leads to a reduced electron count compared to protons, giving the ion a positive charge.

Typically, metals form cations to achieve a stable electron configuration. For instance, a calcium atom (atomic number 20) loses two electrons to become Ca^{2+} , matching the electron configuration of argon.
  • Cations have fewer electrons than protons.
  • Commonly formed by metals, which lose electrons to achieve stability.
  • Cation charge is indicated with a positive sign (e.g., Na^+).

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Most popular questions from this chapter

Two different compounds exist having the formula \(\mathrm{N}_{2} \mathrm{F}_{2}\) . One compound is polar whereas the other is nonpolar. Draw Lewis structures for \(\mathrm{N}_{2} \mathrm{F}_{2}\) consistent with these observations.

Use the following data to estimate \(\Delta H_{\mathrm{f}}^{\circ}\) for magnesium fluoride. $$\mathrm{Mg}(s)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{MgF}_{2}(s)$$ \(\begin{array}{l}{\text { Lattice energy }} & {-22913 . \mathrm{kJ} / \mathrm{mol}} \\ {\text { First ionization energy of } \mathrm{Mg}} & \quad{735 \mathrm{kJ} / \mathrm{mol}} \\ {\text {Second ionization energy of } \mathrm{Mg}} & \quad {1445 \mathrm{kJ} / \mathrm{mol}}\\\\{\text { Electron affinity of } \mathrm{F}} & {-328 \mathrm{kJ} / \mathrm{mol}} \\ {\text { Bond energy of } \mathrm{F}_{2}} & \quad {154 \mathrm{kJ} / \mathrm{mol}} \\\ {\text { Enthalpy of sublimation for } \mathrm{Mg}} & \quad {150 . \mathrm{kJ} / \mathrm{mol}} \end{array}\)

Although both \(\mathrm{Br}_{3}^{-}\) and \(\mathrm{I}_{3}^{-}\) ions are known, the \(\mathrm{F}_{3}-\) ion has not been observed. Explain.

Draw a Lewis structure for the \(N, N\) -dimethylformamide molecule. The skeletal structure is Various types of evidence lead to the conclusion that there is some double bond character to the C-N bond. Draw one or more resonance structures that support this observation.

Arrange the atoms and/or ions in the following groups in order of decreasing size. a. \(\mathrm{O}, \mathrm{O}^{-}, \mathrm{O}^{2-}\) b. \(\mathrm{Fe}^{2+}, \mathrm{Ni}^{2+}, \mathrm{Zn}^{2+}\) c. \(\mathrm{Ca}^{2+}, \mathrm{K}^{+}, \mathrm{Cl}^{-}\)

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