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Why are some bonds ionic and some covalent?

Short Answer

Expert verified
Ionic bonds form when there is a large electronegativity difference between two elements (typically greater than 1.7), causing electron transfer and resulting in charged particles (ions). Covalent bonds form when the electronegativity difference is smaller (typically less than 1.7), leading to the sharing of electrons between atoms. The type of bond in a compound depends on the difference in electronegativity of the elements involved.

Step by step solution

01

Introduction to Ionic and Covalent Bonds

Ionic bonds are formed when electrons are transferred from one atom to another, resulting in charged particles (ions). This typically happens between a metal and a non-metal. Covalent bonds, on the other hand, occur when atoms share electrons, and this usually happens between two non-metals. The type of bond formed depends on the differences in electronegativity between the two elements.
02

Electronegativity

Electronegativity is the measure of the ability of an atom to attract electrons towards itself. The higher the electronegativity value, the stronger the atom attracts electrons. The electronegativity difference between the two elements involved in a bond determines whether it's an ionic or covalent bond.
03

Ionic Bonds

When there is a large electronegativity difference between the two elements (typically greater than 1.7), the electrons are transferred from the less electronegative element (usually a metal) to the more electronegative element (usually a non-metal), resulting in an ionic bond. For example, when sodium (a metal with low electronegativity) reacts with chlorine (a non-metal with high electronegativity), sodium loses an electron to become a positively charged ion (Na+), and chlorine gains the electron to become a negatively charged ion (Cl-). The electrostatic attraction between these oppositely charged ions forms an ionic bond, resulting in sodium chloride (NaCl).
04

Covalent Bonds

When the electronegativity difference between the two elements is smaller (typically less than 1.7), the electrons are shared between the atoms forming a covalent bond. The shared electrons are attracted to both nuclei, which holds the atoms together. For example, consider the formation of a bond between two hydrogen atoms. The electronegativity difference between the two hydrogen atoms is zero, so they share their electrons equally, forming a covalent bond.
05

Polar and Nonpolar Covalent Bonds

Depending on the electronegativity difference, covalent bonds can be classified into nonpolar covalent bonds (less than 0.5) and polar covalent bonds (0.5 to 1.7). In nonpolar covalent bonds, the electron pair is shared equally between the two atoms, while in polar covalent bonds, the electron pair is shared unequally, resulting in a dipole. In summary, the type of bond (ionic or covalent) in a compound depends on the difference in electronegativity of the elements involved, with ionic bonds forming when the difference is large and covalent bonds forming when the difference is smaller.

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Most popular questions from this chapter

Write the formula for each of the following compounds: a. chromium(VI) oxide b. disulfur dichloride c. nickel(II) fluoride d. potassium hydrogen phosphate e. aluminum nitride f. ammonia g. manganese(IV) sulfide h. sodium dichromate i. ammonium sulfite J. carbon tetraiodide

Which of the following ions have noble gas electron configurations? a. \(\mathrm{Fe}^{2+}, \mathrm{Fe}^{3+}, \mathrm{Sc}^{3+}, \mathrm{Co}^{3+}\) b. \(\mathrm{Tl}^{+}, \mathrm{Te}^{2-}, \mathrm{Cr}^{3+}\) c. \(\mathrm{Pu}^{4+}, \mathrm{Ce}^{4+}, \mathrm{Ti}^{4+}\) d. \(\mathrm{Ba}^{2+}, \mathrm{Pt}^{2+}, \mathrm{Mn}^{2+}\)

Nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) has three possible Lewis structures: $$\therefore N=N=O^{\cdot} \leftrightarrow: N \equiv N-\vec{O}: \longleftrightarrow: N-N \equiv 0$$ Given the following bond lengths, $$\begin{aligned} &\mathrm{N}-\mathrm{N} \quad 167 \mathrm{pm} \quad \mathrm{N}=\mathrm{O} \quad 115 \mathrm{pm}\\\ &\mathrm{N}=\mathrm{N} \quad 120 \mathrm{pm} \quad \mathrm{N}-\mathrm{O} \quad 147 \mathrm{pm}\\\ &\mathrm{N} \equiv \mathrm{N} \quad 110 \mathrm{pm} \end{aligned}$$ rationalize the observations that the \(\mathrm{N}-\mathrm{N}\) bond length in \(\mathrm{N}_{2} \mathrm{O}\) is \(112 \mathrm{pm}\) and that the \(\mathrm{N}-\mathrm{O}\) bond length is \(119 \mathrm{pm}\). Assign formal charges to the resonance structures for \(\mathrm{N}_{2} \mathrm{O}\). Can you eliminate any of the resonance structures on the basis of formal charges? Is this consistent with observation?

For each of the following atomic numbers, use the periodic table to write the formula (including the charge) for the simple ion that the element is most likely to form in ionic compounds. a. 13 b. 34 c. 56 d. 7 e. 87 f. 35

Order the following species with respect to carbon-oxygen bond length (longest to shortest). $$\mathrm{CO}, \quad \mathrm{CO}_{2}, \quad \mathrm{CO}_{3}^{2-}, \quad \mathrm{CH}_{3} \mathrm{OH}$$ What is the order from the weakest to the strongest carbonoxygen bond? \(\left(\mathrm{CH}_{3} \mathrm{OH} \text { exists as } \mathrm{H}_{3} \mathrm{C}-\mathrm{OH} .\right)\)

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