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Chapter 8: Problem 11

Why are some bonds ionic and some covalent?

Short Answer

Expert verified
The type of bond formed between atoms depends on their relative electronegativities. When there is a large electronegativity difference, an ionic bond forms through the transfer of electrons, typically between metals and non-metals. If the electronegativity difference is small, a covalent bond forms through the sharing of electrons, usually between non-metal elements with similar electronegativities.

Step by step solution

01

Introduction to Bonds

Atoms can interact with each other in order to achieve a more stable state with a full outer electron shell, and this interaction can either involve the transfer of electrons (forming ionic bonds) or the sharing of electrons (forming covalent bonds). The type of bond that forms between atoms depends largely on their respective electronegativities.
02

Electronegativity

Electronegativity is the measure of the attraction an atom has for electrons in a chemical bond. In general, atoms with larger electronegativities tend to strongly attract electrons, while atoms with smaller electronegativities tend to have weaker attraction for electrons.
03

Ionic Bonds

Ionic bonds occur when there is a large electronegativity difference (\(\Delta EN\)) between the two bonding atoms. This causes one atom to "steal" electrons from the other, resulting in a transfer of electrons and the formation of charged species called ions. The positively charged ion (cation) and negatively charged ion (anion) are then attracted to each other, forming an ionic bond. Typical ionic bonds occur between metals (with lower electronegativities) and non-metals (with higher electronegativities).
04

Covalent Bonds

Covalent bonds occur when the electronegativity difference between the two bonding atoms is small, which means that both atoms have a similar attraction for electrons. In this case, sharing electrons between the atoms allows both to achieve a full outer electron shell, resulting in a more stable bond. Covalent bonds usually occur between non-metal elements with similar electronegativities.
05

Recap

To summarize, the type of bond formed between atoms depends on their relative electronegativities. If there is a large electronegativity difference between the atoms, an ionic bond will form through the transfer of electrons. If the electronegativity difference is small, a covalent bond will likely form through the sharing of electrons.

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Most popular questions from this chapter

Nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) has three possible Lewis structures: $$: \mathrm{N}=\mathrm{N}=\dot{\mathrm{O}}\longleftrightarrow: \mathrm{N} \equiv \mathrm{N}-\ddot{\mathrm{Q}} : \longleftrightarrow : \dot{\mathrm{N}}-\mathrm{N} \equiv \mathrm{O}$$ Given the following bond lengths, $\mathrm{N}-\mathrm{N} \qquad 167 \mathrm{pm} \quad \mathrm{N}=\mathrm{O} \quad 115 \mathrm{pm}$ $\mathrm{N}=\mathrm{N} \qquad 120 \mathrm{pm} \quad \mathrm{N}-\mathrm{O} \quad 147 \mathrm{pm}$ \(\mathrm{N} \equiv \mathrm{N} \quad 110 \mathrm{pm}\) rationalize the observations that the N-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?

Without using Fig. 8.3, predict the order of increasing electronegativity in each of the following groups of elements. a. \(\mathrm{Na}, \mathrm{K}, \mathrm{Rb} \quad\) c. $\mathrm{F}, \mathrm{Cl}, \mathrm{Br}$ b. \(\mathrm{B}, \mathrm{O}, \mathrm{Ga} \qquad\) d. $\mathrm{S}, \mathrm{O}, \mathrm{F}$

Use bond energies (Table \(8.5 ),\) values of electron affinities (Table 7.7\()\) , and the ionization energy of hydrogen \((1312 \mathrm{kJ} / \mathrm{mol})\) to estimate \(\Delta H\) for each of the following reactions. a. \(\mathrm{HF}(g) \rightarrow \mathrm{H}^{+}(g)+\mathrm{F}^{-}(g)\) b. \(\mathrm{HCl}(g) \rightarrow \mathrm{H}^{+}(g)+\mathrm{Cl}^{-}(g)\) c. \(\mathrm{HI}(g) \rightarrow \mathrm{H}^{+}(g)+\mathrm{I}^{-}(g)\) d. $\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{H}^{+}(g)+\mathrm{OH}^{-}(g)$

In general, the higher the charge on the ions in an ionic compound, the more favorable the lattice energy. Why do some stable ionic compounds have \(+1\) charged ions even though \(+4,+5,\) and \(+6\) charged ions would have a more favorable lattice energy?

The molecules $\mathrm{BF}_{3}, \mathrm{CF}_{4}, \mathrm{CO}_{2}, \mathrm{PF}_{5},\( and \)\mathrm{SF}_{6}$ are all nonpolar, even though they contain polar bonds. Why?
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