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Determine which of the atoms in each pair possess a partial positive charge and which a partial negative, the \(\mathrm{O}-\mathrm{F}\) bond, (b) the \(\mathrm{O}-\mathrm{N}\) bond, (c) the \(\mathrm{O}-\mathrm{S}\) bond. Electronegativity values for these elements can be found from a table of electronegativities.

Short Answer

Expert verified
In the O-F bond, fluorine (F) has a partial negative charge and oxygen (O) has a partial positive charge. In the O-N bond, oxygen (O) has a partial negative charge and nitrogen (N) has a partial positive charge. In the O-S bond, oxygen (O) has a partial negative charge and sulfur (S) has a partial positive charge.

Step by step solution

01

Finding the Electronegativity Values

Check a table of electronegativity values to find the electronegativity values of oxygen (O), fluorine (F), nitrogen (N), and sulfur (S). Using the Pauling scale, we have: - Oxygen: \(3.44\) - Fluorine: \(3.98\) - Nitrogen: \(3.04\) - Sulfur: \(2.58\)
02

Comparing Electronegativity Values for the O-F Bond

Compare the electronegativity values of oxygen and fluorine: - Oxygen (O): \(3.44\) - Fluorine (F): \( 3.98\) Since fluorine has a higher electronegativity value than oxygen, fluorine will have a partial negative charge, and oxygen will have a partial positive charge.
03

Comparing Electronegativity Values for the O-N Bond

Compare the electronegativity values of oxygen and nitrogen: - Oxygen (O): \(3.44\) - Nitrogen (N): \(3.04\) Since oxygen has a higher electronegativity value than nitrogen, oxygen will have a partial negative charge, and nitrogen will have a partial positive charge.
04

Comparing Electronegativity Values for the O-S Bond

Compare the electronegativity values of oxygen and sulfur: - Oxygen (O): \(3.44\) - Sulfur (S): \(2.58\) Since oxygen has a higher electronegativity value than sulfur, oxygen will have a partial negative charge, and sulfur will have a partial positive charge.
05

Conclusion

Based on the electronegativity values and comparisons, we can conclude the following: (a) In the O-F bond: - Oxygen (O): partial positive charge - Fluorine (F): partial negative charge (b) In the O-N bond: - Oxygen (O): partial negative charge - Nitrogen (N): partial positive charge (c) In the O-S bond: - Oxygen (O): partial negative charge - Sulfur (S): partial positive charge

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Key Concepts

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

Partial Charge
When atoms form chemical bonds, they share electrons to some extent, but this sharing is not always equal. The concept of partial charge emerges when two atoms with different electronegativities form a bond. Electronegativity is a measure of an atom's tendency to attract and hold onto electrons. In a bond between two atoms, the more electronegative atom will pull the shared electrons closer to itself, gaining a slightly negative character, while the less electronegative atom will become slightly positive as the shared electrons are pulled away from it.

This results in the formation of partial charges, denoted by \( \delta+ \) for the slightly positive charge and \( \delta- \) for the slightly negative charge. The imbalance of electron distribution in a chemical bond is crucial as it affects the physical properties of the molecules, like boiling and melting points, solubility, and reactivity. For instance, water's polarity, as well as its excellent solvent properties, is a result of the oxygen atom's partial negative charge and the hydrogen atoms' partial positive charges.
Chemical Bonding
Chemical bonding is the physical process that enables atoms to be held together in molecules or compounds. There are several types of chemical bonds, including ionic, covalent, and metallic bonds. Covalent bonds are formed when two atoms share a pair of electrons; this is typical between two nonmetals. For instance, in a water molecule (\(H_2O\)), oxygen and hydrogen share electrons to form covalent bonds.

Importantly, the way electrons are shared during chemical bonding can have profound effects on the resulting compound. In a covalent bond, if electrons are shared equally between the atoms, the bond is considered to be nonpolar. However, if electrons are shared unequally due to differing electronegativity values, the bond becomes polarized, leading to the creation of partial charges as seen with the oxygen and hydrogen in water.
Pauling Scale
The Pauling scale is a widely used method of quantifying the electronegativity of elements. Developed by Linus Pauling, this scale assigns a numerical value, typically between 0.7 and 4.0, to the relative ability of an atom to attract electrons when forming a bond. At one end of the scale is fluorine with a value of 3.98, making it the most electronegative element, while elements like cesium and francium lie at the other end with the lowest values.

The Pauling scale helps predict the type and polarity of bonds that can form between different atoms, allowing us to understand the behavior of molecules. For example, in the exercises provided, oxygen with an electronegativity of 3.44 is less electronegative than fluorine but more so than nitrogen and sulfur. This difference in electronegativity within chemical bonds determines the direction and magnitude of the partial charges on the atoms involved.

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Most popular questions from this chapter

Write the Born-Haber cycle for the formation of crystalline sodium fluoride \(\left(\mathrm{Na}^{+} \mathrm{F}^{-}\right)\), starting with solid \(\mathrm{Na}\) and gaseous F. Then, using the thermochemical data supplied below, determine its heat of formation: (1) \(\mathrm{Na}(\mathrm{s}) \quad \rightarrow \mathrm{Na}(\mathrm{g}) \quad \Delta \mathrm{H}=+26.0 \mathrm{Kcal}:\) sublimation (2) \(\mathrm{F}_{2}(\mathrm{~g}) \quad \rightarrow 2 \mathrm{~F}(\mathrm{~g})\) \(\Delta \mathrm{H}=+36.6 \mathrm{~K}\) cal : dissociation (3) \(\mathrm{Na}(\mathrm{g}) \rightarrow \mathrm{Na}^{+}(\mathrm{g})+\mathrm{e}^{-} \quad \Delta \mathrm{H}=+120.0 \mathrm{~K}\) cal : ionization (4) \(\mathrm{F}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{F}^{-}(\mathrm{g})\) \(\Delta \mathrm{H}=-83.5 \mathrm{Kcal}:\) electron addition (5) \(\mathrm{Na}^{+}(\mathrm{g})+\mathrm{F}^{-}(\mathrm{g}) \rightarrow \mathrm{Na}^{+}, \mathrm{F}^{-}(\mathrm{s}) \Delta \mathrm{H}=-216.7 \mathrm{Kcal}\) : lattice formation.

Distinguish a metallic bond from an ionic bond and from a covalent bond.

An ionic bond is established between positive ion \(\mathrm{A}\) and negative ion \(B\). How would one expect the strength of the bond to be affected by each of the following changes: (a) Doubling the charge on \(\mathrm{A}\), (b) Simultaneously doubling the charge on \(\mathrm{A}\) and \(\mathrm{B}\), (c) Doubling the radius of \(\mathrm{B}\), and Simultaneously doubling the radius of \(\mathrm{A}\) and \(\mathrm{B}\) ?

A chemist has one mole of \(X\) atoms. He finds that when half of the \(\mathrm{X}\) atoms transfer one electron to the other half, \(409 \mathrm{~kJ}\) must be added. If all of the resulting \(\mathrm{X}^{-}\) ions are subsequently converted to \(\mathrm{X}+\) ions, an additional \(733 \mathrm{~kJ}\) must be added. Find the ionization potential, and the electron affinity of \(\mathrm{X}\).

Assuming the ionization potential of sodium is \(5.1 \mathrm{eV}\) and the electron affinity of chlorine is \(3.6 \mathrm{eV}\), calculate the amount of energy required to transfer one electron from an isolated sodium (Na) atom to an isolated chlorine (C1) atom.

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