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# 化学代写|无机化学代考Inorganic Chemistry代写|CHEM2550 Electronegativity and bond enthalpy

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## 化学代写|无机化学代考Inorganic Chemistry代写|Electronegativity and bond enthalpy

The concept of electronegativity was introduced in Section $1.7 \mathrm{~d}$, where it was defined as the power of an atom of the element to attract electrons to itself when it is part of a compound. The greater the difference in electronegativity between two elements $\mathrm{A}$ and $\mathrm{B}$, the greater the ionic character of the $\mathrm{A}-\mathrm{B}$ bond.

Linus Pauling’s original formulation of electronegativity drew on concepts relating to the energetics of bond formation. For example, in the formation of $A B$ from the diatomic $\mathrm{A}{2}$ and $\mathrm{B}{2}$ molecules,
$$\mathrm{A}{2}(\mathrm{~g})+\mathrm{B}{2}(\mathrm{~g}) \rightarrow 2 \mathrm{AB}(\mathrm{g})$$
he argued that the excess energy, $\Delta E$, of the $\mathrm{A}-\mathrm{B}$ bond over the average energy of $\mathrm{A}-\mathrm{A}$ and $\mathrm{B}-\mathrm{B}$ bonds can be attributed to the presence of ionic contributions to the covalent bonding. He defined the difference in electronegativity as
$$\left|\chi_{\mathrm{P}}(\mathrm{A})-\chi_{\mathrm{P}}(\mathrm{B})\right|=0.102\left(\Delta E / \mathrm{kJ} \mathrm{mol}^{-1}\right)^{1 / 2}$$
where
$$\Delta E=B(\mathrm{~A}-\mathrm{B})-\frac{1}{2}[\mathrm{~B}(\mathrm{~A}-\mathrm{A})+\mathrm{B}(\mathrm{B}-\mathrm{B})]$$
with $B(\mathrm{~A}-\mathrm{B})$ the mean $\mathrm{A}-\mathrm{B}$ bond enthalpy. Thus, if the $\mathrm{A}-\mathrm{B}$ bond enthalpy is significantly greater than the average of the nonpolar $\mathrm{A}-\mathrm{A}$ and $\mathrm{B}-\mathrm{B}$ bonds, then it is presumed that there is a substantial ionic contribution to the wavefunction and hence a large difference in electronegativity between the two atoms. Pauling electronegativities increase with increasing oxidation number of the element, and the values in Table $1.7$ are for the most common oxidation state.

## 化学代写|无机化学代考Inorganic Chemistry代写|Oxidation states

The oxidation number, $\mathrm{N}{\mathrm{ox}},{ }^{3}$ is a parameter obtained by exaggerating the ionic character of a bond. It can be regarded as the charge that an atom would have if the more electronegative atom in a bond acquired the two electrons of the bond completely. The oxidation state is the physical state of the element corresponding to its oxidation number. Thus, an atom may be assigned an oxidation number and be in the corresponding oxidation state. ${ }^{4}$ The alkali metals are the most electropositive elements in the periodic table, so we can assume they will always be present as $\mathrm{M}^{+}$and are assigned an oxidation number of $+1$. Because oxygen’s electronegativity is exceeded only by that of $F$, we can regard it as $\mathrm{O}^{2-}$ in combination with any element other than $\mathrm{F}$, and hence it is ascribed an oxidation number of $-2$. Likewise, the exaggerated ionic structure of $\mathrm{NO}{3}^{-}$is $\mathrm{N}^{5}+\left(\mathrm{O}^{2-}\right)_{3}$, so the oxidation number of nitrogen in this compound is $+5$, which is denoted either $\mathrm{N}(\mathrm{V})$ or $\mathrm{N}(+5)$. These conventions may be used even if the oxidation number is negative, so oxygen has oxidation number $-2$, denoted $\mathrm{O}(-2)$ or more rarely $\mathrm{O}(-\mathrm{II})$, in most of its compounds.

In practice, oxidation numbers are assigned by applying a set of simple rules (Table 2.9). These rules reflect the consequences of electronegativity for the ‘exaggerated ionic’ structures of compounds and match the increase in the degree of oxidation that we would expect as the number of oxygen atoms in a compound increases (as in going from $\mathrm{NO}$ to $\mathrm{NO}_{3}^{-}$). This aspect of oxidation number is taken further in Chapter 5 . Many elements, for example nitrogen, the halogens, and the d-block elements, can exist in a variety of oxidation states (Table 2.9).

## 化学代写|无机化学代考Inorganic Chemistry代写|Electronegativity and bond enthalpy

Linus Pauling 最初的电负性表述借鉴了与键形成的能量学相关的概念。例如，在形成 $A B$ 从双原子 $\mathrm{A} 2$ 和 $\mathrm{B} 2$ 分子，
$$\mathrm{A} 2(\mathrm{~g})+\mathrm{B} 2(\mathrm{~g}) \rightarrow 2 \mathrm{AB}(\mathrm{g})$$

$$\left|\chi_{\mathrm{P}}(\mathrm{A})-\chi_{\mathrm{P}}(\mathrm{B})\right|=0.102\left(\Delta E / \mathrm{kJmol}^{-1}\right)^{1 / 2}$$

$$\Delta E=B(\mathrm{~A}-\mathrm{B})-\frac{1}{2}[\mathrm{~B}(\mathrm{~A}-\mathrm{A})+\mathrm{B}(\mathrm{B}-\mathrm{B})]$$

## 化学代写|无机化学代考Inorganic Chemistry代写|Oxidation states

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