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Chapter 8: Problem 18

Most atoms in nature are found bonded together to form compounds instead of existing as individual atoms. Why is this true?

Short Answer

Expert verified
Most atoms in nature are found bonded together to form compounds because bonding allows them to achieve a stable electron configuration, resulting in lower energy states and greater stability. Atoms with incomplete valence electron shells participate in chemical bonding, either through ionic bonds (electron transfer) or covalent bonds (electron sharing), to attain a stable configuration similar to noble gases. This process lowers the overall energy of the atoms and contributes to the stability of the formed compound.

Step by step solution

01

Understanding Atom Stability

Atoms seek to achieve a stable state, where their overall energy is minimal. This stability can be achieved by having a filled outer electron shell. Elements with a full outer electron shell, such as noble gases, are naturally stable and do not participate in chemical bonding because they already have a full valence electron configuration.
02

Role of Valence Electrons

Valence electrons are the electrons in the outermost energy level of an atom. These are the electrons that participate in chemical bonding. Atoms with incomplete valence electron shells tend to give away, accept, or share electrons to achieve a stable configuration similar to that of a noble gas.
03

Types of Chemical Bonds

There are two main types of chemical bonds: ionic and covalent. 1. Ionic Bonds: Occurs when electrons are transferred from one atom to another, resulting in the formation of oppositely charged ions. These ions are then attracted to each other by electrostatic forces, forming an ionic compound. 2. Covalent Bonds: Occurs when two or more atoms share outer shell electrons to achieve a full valence electron shell. The shared electrons contribute to the stability of each atom involved in the bond.
04

Formation of Stable Compounds

Most atoms in nature bond together to form compounds in order to achieve a stable electron configuration. This bonding lowers the overall energy of the atoms, making them more stable than they would be as individual atoms. Additionally, bonded atoms often have a lower potential energy compared to when they were separated, which also contributes to the overall stability of the compound.
05

Conclusion

In summary, most atoms in nature are found bonded together to form compounds because this process allows them to achieve a stable electron configuration, which results in lower energy states and greater stability. This is achieved through the formation of ionic or covalent bonds, as the atoms either transfer or share their outer shell electrons.

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Most popular questions from this chapter

Use the formal charge arguments to rationalize why \(\mathrm{BF}_{3}\) would not follow the octet rule.

The standard enthalpies of formation for $\mathrm{S}(g), \mathrm{F}(g), \mathrm{SF}_{4}(g),\( and \)\mathrm{SF}_{6}(g)\( are \)+278.8,+79.0,-775,$ and \(-1209 \mathrm{kJ} / \mathrm{mol}\) respectively. a. Use these data to estimate the energy of an \(\mathrm{S}-\) F bond. b. Compare your calculated value to the value given in Table \(8.5 .\) What conclusions can you draw? c. Why are the \(\Delta H_{f}^{\circ}\) values for \(\mathrm{S}(g)\) and \(\mathrm{F}(g)\) not equal to zero, since sulfur and fluorine are elements?

The following electrostatic potential diagrams represent \(\mathrm{CH}_{4}\) , \(\mathrm{NH}_{3},\) or \(\mathrm{H}_{2} \mathrm{O}\) . Label each and explain your choices.

A polyatomic ion is composed of \(\mathrm{C}, \mathrm{N},\) and an unknown element \(\mathrm{X}\) . The skeletal Lewis structure of this polyatomic ion is \([\mathrm{X}-\mathrm{C}-\mathrm{N}]^{-} .\) The ion \(\mathrm{X}^{2-}\) has an electron configuration of \([\text { Ar }] 4 s^{2} 3 d^{10} 4 p^{6} .\) What is element \(X ?\) Knowing the identity of \(X,\) complete the Lewis structure of the polyatomic ion, including all important resonance structures.

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?
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