Name: predicting ionic stability with lattice energy
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The Born-Haber Cycle

La formation de liaisons ioniques nécessite un transfert d'électrons d'un métal à un atome non métallique un processus souvent endothermique. Cependant, lorsque le sodium élémentaire et le chlore réagissent pour former un cristal solide de chlorure de sodium, c'est un processus hautement exothermique. Mais d'où vient l'énergie Selon la loi de Coulomb, les cations et les anions sont attirés les uns aux autres par de fortes forces électrostatiques dans un réseau solide, ou réseau.La structure de réseau résultante est stabilisée en diminuant l'énergie potentielle, qui est libérée sous forme de chaleur, une réaction exothermique. L'énergie totale associée à la formation ou à la dégradation d'un réseau cristallin en ses constituants gazeux est appelée énergie du réseau. Dans un composé ionique solide, un grand nombre de particules chargées interagissent les unes avec les autres, ce qui rend difficile la détermination expérimentale de la valeur exacte de l'énergie du réseau.Cependant, il peut être calculé en utilisant la loi de Hess dans une série hypothétique d'étapes appelées le cycle Born Haber, qui représente la formation d'un composé ionique à partir de ses éléments constitutifs. Par exemple, le cycle de Born-Haber pour la formation de chlorure de sodium considère deux voies alternatives, l'une directe et l'autre indirecte. La voie directe représente l'enthalpie standard de formation de NaCl à partir du sodium élémentaire et du chlore.L'itinéraire indirect comporte cinq étapes. Dans la première étape, le sodium à l'état solide est transformé en sa forme gazeuse. Ensuite, les molécules de chlore diatomiques se dissocient en atomes de chlore gazeux.Les troisième et quatrième étapes représentent le transfert d'électrons pour former des ions. Un électron est retiré du sodium gazeux pour former un cation sodium. L'électron est ensuite absorbé par le chlore gazeux pour former un anion chlorure.Dans l'étape finale, l'attraction électrostatique entre les ions gazeux conduit à la formation de la structure en maillage. La loi de Hess stipule que le changement dans l'enthalpie globale d'un processus par étapes est la somme des changements d'enthalpie de chaque étape. Cela signifie que la valeur de l'enthalpie de la voie directe est égale à la somme des enthalpies des cinq étapes.En résolvant l'équation pour l'énergie du réseau, une grande valeur négative est déterminée, ce qui signifie que la réaction est exothermique

le cycle de born-haber

<h4>Énergie réticulaire </h4>
 
 Un composé ionique est stable en raison de l'attraction électrostatique entre ses ions positifs et négatifs. L'énergie réticulaire d'un composé est une mesure de la force de cette attraction. L'énergie réticulaire (&DeltaHréticulaire) d'un composé ionique est défini comme l'énergie nécessaire pour séparer une mole du solide en ions gazeux le composant.  
 
 Ici, la convention de signe est utilisée lorsque le solide ionique est séparé en ions, ce qui signifie que les énergies réticulaires seront endothermiques (valeurs positives). Une autre façon est d'utiliser une convention équivalente, mais de signe opposée, pour laquelle l'énergie réticulaire est exothermique (valeurs négatives) et décrite comme l'énergie libérée lorsque les ions se combinent pour former un réseau. Par conséquent, assurez-vous de confirmer la définition utilisée lorsque vous regardez les énergies réticulaires d'une autre référence.  
 
 Dans les deux cas, une plus grande valeur algébrique pour l'énergie réticulaire indique un composé ionique plus stable. Pour le chlorure de sodium, &DeltaHréticulaire = 769 kJ. Ainsi, il faut 769 kJ pour séparer une mole de NaCl solide en ions Na+ et Cl – gazeux. Lorsqu'une mole de chaque ions Na+ et Cl– gazeux forme du NaCl solide, 769 kJ de chaleur sont libérés. 
 
 <h4>Détermination de l'énergie réticulaire d'un composé ionique </h4>
 
 Il n'est pas possible de mesurer directement les énergies réticulaires. Cependant, l'énergie réticulaire peut être calculée à l'aide d'un cycle thermochimique. Le cycle de Born-Haber est une application de la loi de Hess qui décompose la formation d'un solide ionique en une série d'étapes individuelles : 
 
 <table style="margin-left: auto; margin-right: auto;">
 <tbody style="border: 1px solid black;">
 <tr>
 <td style="color: black; border: 1px solid black; text-align: center;">Enthalpie de sublimation de Cs (s)</td>
 <td colspan="2" style="color: black; border: 1px solid black; text-align: center;">Cs (s) &arr; Cs (g)</td>
 <td style="color: black; border: 1px solid black; text-align: center;">ΔH = ΔHs° = 76,5 kJ/mol </td>
 </tr>
 <tr>
 <td style="color: black; border: 1px solid black; text-align: center;">La moitié de l'énergie de liaison de F2</td>
 <td colspan="2" style="color: black; border: 1px solid black; text-align: center;">½ F2 (g) &arr; F (g)</td>
 <td style="color: black; border: 1px solid black; text-align: center;">&DeltaH = frac12; D = 79,4 kJ/mol </td>
 </tr>
 <tr>
 <td style="color: black; border: 1px solid black; text-align: center;">Énergie d'ionisation de Cs (g)</td>
 <td colspan="2" style="color: black; border: 1px solid black; text-align: center;">Cs (g)  &arr; Cs+ (g) + e−</td>
 <td style="color: black; border: 1px solid black; text-align: center;">ΔH = Ei = 375,7 kJ/mol </td>
 </tr>
 <tr>
 <td style="color: black; border: 1px solid black; text-align: center;">Affinité électronique de F </td>
 <td colspan="2" style="color: black; border: 1px solid black; text-align: center;">F (g) + e− &arr; F− (g) </td>
 <td style="color: black; border: 1px solid black; text-align: center;">ΔH = Ae = −328,2 kJ/mol </td>
 </tr>
 <tr>
 <td style="color: black; border: 1px solid black; text-align: center;">Valeur négative de l'énergie réticulaire de CsF (s)</td>
 <td colspan="2" style="color: black; border: 1px solid black; text-align: center;">Cs+ (g) + F− (g) &arr; CsF (s)</td>
 <td style="color: black; border: 1px solid black; text-align: center;">&DeltaH = −Delta;Hréticulaire = ?</td>
 </tr>
 <tr>
 <td style="color: black; border: 1px solid black; text-align: center;">Enthalpie de formation de CsF (s), ajouter les étapes 1 à 5 </td>
 <td colspan="2" style="color: black; border: 1px solid black; text-align: center;">ΔH = ΔHf° = ΔHs°+ &frac12 D + Ei + (Ae) + (−ΔHréticulaire) 
 Cs (s) + ½ F2 (g) &arr; CsF (s)</td>
 <td style="color: black; border: 1px solid black; text-align: center;">Δem>H = −553,5 kJ/mol </td>
 </tr>
 </tbody>
 </table>
 
 <ol>
 <li>Considérez les éléments dans leur état le plus courant, Cs (s) et F2 (g). </li>
 <li>Le ΔHs° représente la conversion du césium solide en gaz (sublimation), puis l'énergie d'ionisation transforme les atomes de césium gazeux en cations.  </li>
 <li>Dans l'étape suivante, l'énergie nécessaire pour rompre la liaison F–F afin de produire des atomes de fluor doit être prise en compte.  </li>
 <li>La conversion d'une mole d'atomes de fluor en ions fluorure est un processus exothermique, de sorte que cette étape fournit de l'énergie (l'affinité électronique)  </li>
 <li>Maintenant, une mole de cations Cs et une mole d'anions F sont présentes. Ces ions se combinent pour produire du fluorure de césium solide. La variation d'enthalpie dans cette étape est la valeur négative de l'énergie réticulaire, donc c'est aussi une quantité exothermique.  </li>
 <li>L'énergie totale impliquée dans cette conversion est égale à l'enthalpie de formation, ΔHf°, du composé à partir de ses éléments, qui est déterminée expérimentalement. Dans ce cas, la transformation globale est exothermique. </li>
 </ol>
 
 Les énergies réticulaires calculées pour les composés ioniques sont généralement beaucoup plus élevées que les énergies de dissociation des liaisons mesurées pour les liaisons covalentes. Alors que les énergies réticulaires se situent généralement dans la plage de 600 – 4000 kJ/mol (certaines encore plus élevées), les énergies de dissociation des liaisons covalentes se situent généralement entre 150 – 400 kJ/mol pour les liaisons simples. Gardez à l'esprit, cependant, que ces valeurs ne sont pas directement comparables. Pour les composés ioniques, les énergies réticulaires sont associées à de nombreuses interactions, car les cations et les anions se rassemblent dans un réseau étendu. Pour les liaisons covalentes, l'énergie de dissociation de liaison est liée à l'interaction de seulement deux atomes. 
 
 <h4>Énergie réticulaire en fonction du rayon ionique et de la charge  </h4>
 
 L'énergie réticulaire d'un cristal ionique augmente rapidement à mesure que les charges des ions augmentent, et la taille des ions diminue. Lorsque tous les autres paramètres sont maintenus constants, doubler la charge à la fois du cation et de l'anion quadruplera l'énergie réticulaire. Par exemple, l'énergie réticulaire de LiF (Z+ et Z– = 1) est de 1023 kJ/mol, alors que celle de MgO (Z+ et Z– = 2) est de 3900 kJ/mol (Ro = la distance interionique définie comme la somme des rayons des ions positifs et négatifs, est presque la même — environ 200 pm pour les deux composés). 
 
 Différentes distances interatomiques produisent des énergies réticulaires différentes. Par exemple, comparez l'énergie réticulaire de MgF2 (2957 kJ/mol) à celle de MgI2 (2327 kJ/mol), ce qui démontre l'effet sur l'énergie réticulaire de la taille de l'ion F– plus petite par rapport à I–. 
 
 <h4>Autres applications du cycle de Born-Haber </h4>
 
 Le cycle de Born-Haber peut également être utilisé pour calculer l'une des autres grandeurs dans l'équation de l'énergie réticulaire, à condition que le reste soit connu. Par exemple, si l'enthalpie de sublimation  ΔHs°,  l'énergie d'ionisation (Ei), l'enthalpie de dissociation de liaison (D), l'énergie réticulaire &DeltaHréticulaire et l'enthalpie standard de formation Delta;Hf°  en question sont connues, le cycle de Born-Haber peut être utilisé pour déterminer l'affinité électronique d'un atome. 
 
 Ce texte est adapté de <a href="https://openstax.org/books/chemistry-2e/pages/7-5-strengths-of-ionic-and-covalent-bonds">Openstax, Chimie 2e, Section 7.5 : Forces des liaisons ioniques et covalentes.</a>

Le cycle de Born-Haber

Énergie du réseau

Un composé ionique est stable en raison de l’attraction électrostatique entre ses ions positifs et négatifs. L’énergie du réseau d’un ...

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predicting ionic stability with lattice energy

Lattice energy helps explain why ionic compounds are stable. It measures the strength of attraction between positive and negative ions in a crystal. In one common definition, lattice energy is the energy required to separate one mole of an ionic solid into gaseous ions, so it has a positive value. Another definition uses the opposite sign and describes the energy released when gaseous ions form a lattice. When checking data, it is important to confirm which convention is being used. In either case, a larger lattice energy magnitude means a more stable ionic compound.

Sodium chloride gives a simple example. Its lattice energy is 769 kJ/mol, which means 769 kJ is needed to pull one mole of solid NaCl apart into Na+ and Cl- ions in the gas phase. The same amount of heat is released when those ions come together to form solid NaCl. This value shows how strongly the ions attract each other in the crystal.

Lattice energy cannot be measured directly. Instead, it is found with a thermochemical cycle. The Born-Haber cycle uses Hess’s law to break the formation of an ionic solid into separate steps. For cesium fluoride, these steps include sublimation of Cs(s) to Cs(g), half of the F2 bond energy to make F(g), ionization of Cs(g) to Cs+(g) plus an electron, electron affinity of F(g) to make F-(g), and formation of solid CsF from the gaseous ions. The enthalpy change for forming the lattice is the negative of the lattice energy.

For CsF, the overall reaction Cs(s) + 1/2 F2(g) -&gt; CsF(s) has a standard enthalpy of formation of -553.5 kJ/mol. The cycle combines the sublimation enthalpy of cesium, 76.5 kJ/mol, one-half of the F2 bond energy, 79.4 kJ/mol, the ionization energy of cesium, 375.7 kJ/mol, and the electron affinity of fluorine, -328.2 kJ/mol, with the lattice term. Because the total is exothermic, energy is released overall when the solid forms from the elements in their most common states, Cs(s) and F2(g).

Lattice energies for ionic compounds are usually much larger than bond dissociation energies for covalent bonds. Ionic lattice energies often fall between 600 and 4000 kJ/mol, and some are even higher. Single covalent bond dissociation energies are usually about 150 to 400 kJ/mol. These values are not directly comparable because lattice energy reflects many attractions in an extended crystal, while a covalent bond energy describes the interaction of just two atoms.

Ion charge and ion size strongly affect lattice energy. When all other factors stay the same, larger charges give much larger lattice energies, and smaller ions also increase lattice energy. Doubling the charge on both ions can quadruple the lattice energy. For example, LiF has a lattice energy of 1023 kJ/mol, while MgO is 3900 kJ/mol, even though their interionic distances are nearly the same at about 200 pm. Size also matters in compounds such as MgF2 and MgI2, where the smaller F- ion leads to a higher lattice energy than the larger I- ion.

The Born-Haber cycle can also solve for other unknowns in the lattice energy equation. If the other terms are known, the cycle can be used to find a missing quantity such as electron affinity. This makes the method useful for linking measured enthalpy values to the energy changes inside ionic compounds.

Predicting Ionic Stability with Lattice Energy

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the born-haber cycle

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Learn how lattice energy, ion charge, ion size, and the Born-Haber cycle help explain ionic stability and calculate values for compounds like CsF.

Predicting Ionic Stability with Lattice Energy | JoVE K12

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An ionic compound is stable because of the electrostatic attraction between its positive and negative ions. The lattice energy of a compound is a measure ...

Covalent Bonding and Lewis Structures

Compared to ionic bonds, which results from the transfer of electrons between metallic and nonmetallic atoms, covalent bonds result from the mutual ...

Electronegativity

Whether a bond is nonpolar or polar covalent is determined by a property of the bonding atoms called electronegativity.&#xA0;

Electronegativity values of ...

Bond Polarity, Dipole Moment, and Percent Ionic Character

Bond Polarity

The absolute value of the difference in electronegativity (&Delta;EN) of two bonded atoms provides a rough measure of the polarity to be ...

Lewis Structures of Molecular Compounds and Polyatomic Ions

To draw Lewis structures for complicated molecules and molecular ions, it is helpful to follow a step-by-step procedure as outlined:


	Determine the ...

Resonance

The Lewis structure of a nitrite anion (NO2&minus;) may actually be drawn in two different ways, distinguished by the locations of the N-O and N=O ...

Formal Charges

In some cases, there are seemingly more than one valid Lewis structures for molecules and polyatomic ions. The concept of formal charges can be used to ...

Exceptions to the Octet Rule

Many covalent molecules have central atoms that do not have eight electrons in their Lewis structures. These molecules fall into three categories:


	...

Analyze and interpret data to explain the mechanisms and properties of the two main types of intramolecular (ionic and covalent) bonds.

Analyze the structure of atoms and isotopes.

Predicting Ionic Stability with Lattice Energy

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