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Using Lewis symbols, diagram the reaction between magnesium and oxygen atoms to give the ionic substance \(\mathrm{MgO}\).

Short Answer

Expert verified
Magnesium (Mg) has 2 valence electrons and oxygen (O) has 6 valence electrons. Magnesium transfers its 2 valence electrons to oxygen to achieve a full electron shell configuration: Mg^(2+) and O^(2-). This results in the formation of the ionic compound MgO: \(\mathrm{MgO}\).

Step by step solution

01

Identify the number of valence electrons for each atom

Magnesium (Mg) is in group 2 of the periodic table, which means that it has 2 valence electrons. Oxygen (O) is in group 16, which means that it has 6 valence electrons.
02

Represent the Lewis symbols for the atoms

A Lewis symbol represents an atom with the chemical symbol surrounded by dots representing its valence electrons. For magnesium, this would be: Mg: •• For oxygen, this would be: O: •• •• ••
03

Show the interaction between the atoms

Magnesium can achieve a full electron shell by losing its 2 valence electrons, while oxygen can achieve a stable electron configuration by gaining 2 electrons. In this reaction, magnesium will transfer its 2 valence electrons to oxygen to form the ionic compound.
04

Indicate the transfer of electrons

By transferring the 2 electrons from magnesium to oxygen, we obtain the following new Lewis symbols which represent a complete electron shell configuration: Mg: O: •• •• •• •• Here, the magnesium atom donates its two valence electrons to the oxygen atom. This changes both of them to ion forms. Magnesium loses electrons and becomes a cation and oxygen gains electrons and becomes an anion.
05

Show the formation of the ionic compound

After the interaction that occurred in the previous step, magnesium becomes a cation with a charge of \(+2\) and oxygen becomes an anion with a charge of \(-2\). The attractive force between these oppositely charged ions results in the formation of the ionic compound MgO. The final result of the reaction is: Mg^(2+) O^(2-): \(\mathrm{MgO}\) This shows the Lewis symbols representing the formation of the ionic substance MgO from magnesium and oxygen atoms.

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Key Concepts

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

Valence Electrons
Valence electrons are the electrons present in the outermost shell of an atom. They play a pivotal role in chemical bonding, as they are the electrons that are involved in the formation of chemical bonds. Understanding valence electrons is crucial when exploring ionic bonding and compound formation.

For example, magnesium (Mg), which falls under group 2 of the periodic table, has 2 valence electrons. These valence electrons are represented in Lewis symbols by dots around the element's symbol. Similarly, oxygen (O), from group 16, has 6 valence electrons, which are also depicted as dots in its Lewis symbol.

Why Valence Electrons Matter

These outer electrons dictate how an element will interact with other elements. The ultimate goal for an atom, based on the octet rule, is to have a full set of valence electrons, usually eight, to attain a stable electronic configuration similar to that of the noble gases.
Electron Transfer Process
During the formation of an ionic compound, atoms undergo an electron transfer process. This process involves the movement of electrons from one atom to another, enabling each to reach a state of electric stability.

In the interaction between magnesium and oxygen, magnesium donates its 2 valence electrons to oxygen. Such a transfer alters the electron configurations of both atoms: magnesium loses electrons and transforms into a positively charged ion, or cation, while oxygen gains electrons to become negatively charged, or an anion.

Understanding Electron Traffic

Depicted in Lewis symbols, we can visualize this traffic of electrons. Post transfer, magnesium’s Lewis symbol has no dots, symbolizing a full outer shell, whereas oxygen’s Lewis symbol shows eight dots, illustrating its new full shell attained through the gained electrons.
Ionic Compound Formation
The culmination of the electron transfer process is the formation of an ionic compound. When atoms like magnesium and oxygen undergo electron transfer, the resulting ions oppositely charge and attract one another, forming a strong electrostatic bond.

In our example, magnesium becomes Mg²⁺ and oxygen becomes O²⁻. The ionic bond forms because the positively charged magnesium cation is attracted to the negatively charged oxygen anion.

Final Compounds as Ionic Puzzles

This attraction results in the ionic compound MgO, where each magnesium ion pairs with an oxygen ion. The balance of charges (2+ and 2-) ensures overall neutrality of the compound. Ionic compounds are usually crystalline solids at room temperature, and they tend to have high melting and boiling points due to the strength of the ionic bonds.

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Most popular questions from this chapter

Some chemists believe that satisfaction of the octet rule should be the top criterion for choosing the dominant Lewis structure of a molecule or ion. Other chemists believe that achieving the best formal charges should be the top criterion. Consider the dihydrogen phosphate ion, \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-},\) in which the \(\mathrm{H}\) atoms are bonded to \(\mathrm{O}\) atoms. (a) What would be the predicted dominant Lewis structure if satisfying the octet rule is the top criterion? (b) What would it be if achieving the best formal charges is the top criterion? (c) Is there another Lewis structure you can draw that satisfies neither of these criteria?

A major challenge in implementing the "hydrogen economy" is finding a safe, lightweight, and compact way of storing hydrogen for use as a fuel. The hydrides of light metals are attractive for hydrogen storage because they can store a high weight percentage of hydrogen in a small volume. For example, \(\mathrm{NaAlH}_{4}\) can release \(5.6 \%\) of its mass as \(\mathrm{H}_{2}\) upon decomposing to \(\mathrm{NaH}(s), \mathrm{Al}(s),\) and \(\mathrm{H}_{2}(\mathrm{~g}),\) NaAlH \(_{4}\) possesses both covalent bonds, which hold polyatomic anions together, and ionic bonds. (a) Write a balanced equation for the decomposition of \(\mathrm{NaAlH}_{4}\) (b) Which element in \(\mathrm{NaAlH}_{4}\) is the most electronegative? Which one is the least electronegative? (c) Based on electronegativity differences, what do you think is the identity of the polyatomic anion? Draw a Lewis structure for this ion.

Draw Lewis structures for the following: (a) \(\mathrm{SiH}_{4},\) (b) CO, (c) \(\mathrm{SF}_{2}\), (d) \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{H}\) is bonded to \(\mathrm{O})\), (e) \(\mathrm{ClO}_{2}^{-},\) (f) \(\mathrm{NH}_{2} \mathrm{OH}\).

Based on data in Table 8.2 , estimate (within \(30 \mathrm{~kJ} / \mathrm{mol}\) ) the lattice energy for (a) LiBr, (b) CsBr, (c) \(\mathrm{CaCl}_{2}\)

(a) Does the lattice energy of an ionic solid increase or decrease (i) as the charges of the ions increase, (ii) as the sizes of the ions increase? (b) Arrange the following substances not listed in Table 8.2 according to their expected lattice energies, listing them from lowest lattice energy to the highest: \(\mathrm{MgS}, \mathrm{KI}\), \(\mathrm{GaN}, \mathrm{L} \mathrm{iBr}\)

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