Bond Dissociation Energy Of - (X F, Cl, Br, I) Bonds In Halogen: Fill & Download for Free

In hydrogen halides the bond between hydrogen and halogen atoms is covalent but the bond has some ionic character. That means H—X bond is polar ( X = F, Cl, Br, I). Positive pole is on H atom and negative pole is on X atom. We know water is a highly polar solvent with negative charge density on O and positive charge density on H. When H-X is added to water, O atoms of water molecules attract H atoms of H-X and H atoms of water molecules attract X atoms of H-X. By this process H-X bonds are broken. Produced H+ ion are hydrated and form H3O+ ions and X- ions are also enveloped by the water molecules and the negative charge on X- are stabilized by decentralization. In this hydration process energy is released i. e. called hydration energy and that compensates the bond dissociation energy required for breaking of H-X bonds.Hope, this helps.

TLDR:HF is a weak acid, all other hydrohalic acids are strong! Just accept it!Corrosiveness and acidity are two different subjects. HF is very corrosive because fluor -as the most electronegative element -can form stable compounds with almost everything.Regarding the acidity of HF, I’m afraid it is not so easy to explain. I’m not claiming that this answer will solve all questions. (But at least I can try.)The question is puzzling chemists for a very long time. In 1956 Linus Pauling himself already devoted an article on the subject (Why is hydrofluoric acid a weak acid? J. Chem. Educ., 1956, 33 (1), p 16)I will start with some general definitions.Brønsted-Lowry acidity is a matter of equilibrium (in this case with water.)HX(aq)+ H[math]_2[/math]O(l) ⇌ H[math]_3[/math]O[math][/math][math]^+[/math] (aq) + X[math]^[/math][math]-[/math][math][/math](aq)This equilibrium is quantified via Kₐ, the acid dissociation constant. Chemists often use the negative logarithm of this constant.pK[math]_a[/math] = -log[math]_{10}[/math]K[math]_a[/math].The lower the pKₐ value, the stronger the acid.If we tabulate pKₐ of the hydrohalic acids:[math]\small{\begin{array}{cccc} \hline \text{HX} & \text{pK}_a \\\hline \text{HF} & 3.2 [/math][math][/math][math]\\ \text{HCl} & -5.9 ± 0.4 [/math][math][/math][math]\\ \text{HBr} & -8.8 ± 0.8 [/math][math][/math][math]\\ \text{HI} & -9.5 ± 1.0 [/math][math][/math][math]\\ \hline \end{array}}[/math]HF is the only weak acid all others are strong acids.The simplistic definition of a strong acid is ‘any acid that ionizes completely in solution’, but more formally it is an acid with a pKₐ value smaller than the hydronium ion. Or in other words, a strong acid has a negative pKₐ.Actually, a negative pKₐ means that the concentrations of HX in the above equilibrium are very small. Too small to measure, therefore the pKₐ value of HCl, HBr, and HI are not that reliable.I took the values from a 2016 article (J. Phys. Chem. A 2016, 120, pp 3663−3669) and it includes the potential errors. Please note that these values are logarithmic so the error of +/- 1 pKₐ units on HI actually implicates a factor of 10 on the Kₐ.Regardless of the values, it is clear that there is a large difference between HF and the other hydrohalic acids.If we compare this to another series: from methane CH₄ (or H-CH₃) to HF (i.e. moving left to right in the periodic table)[math]\small{\begin{array}{cccc} \hline \text{H-X} & \text{pK}_a & \chi \\\hline \text{H-CH}_3 & \sim 50 & 2.5\\ \text{H-NH}_2 & 35 & 3.0\\ \text{H-OH} & 14.0 & 3.5 [/math][math][/math][math]\\ \text{H-F} & 3.2 & 4.0\\ \hline \end{array}}[/math]The difference in acidity is nicely correlated to the difference in electronegativity of the different elements. A difference in electronegativity makes the bond more polar and this fits nicely with the trend in acidity: you indeed expected to have a greater tendency to go to the anionic form.But if we do the same when we down the table we should be able to make the same argument. Fluor is the most electronegative element and the bond is indeed more polar. But the difference in acidity is totally reversed.[math]\small{\begin{array}{cccc} \hline \text{HX} & \text{pK}_a & \chi \\\hline \text{HF} & 3.2 & 4.0 [/math][math][/math][math]\\ \text{HCl} & -5.9 & 3.0\\ \text{HBr} & -8.8 & 2.8 [/math][math][/math][math]\\ \text{HI} & -9.5 & 2.5 [/math][math][/math][math]\\ \hline \end{array}}[/math]This is a bit puzzling! Why would we use electronegativity to explain the difference in acidity in one case (moving horizontally) but completely disregard it if we go down the periodic table?Some explain it by the bond strength and claim that HF has the stronger bond in the series of HF to HI and therefore correlate acidity with bond strength.The reason for HF’s meager acidity is the strong bond between the hydrogen and the fluorine atom, resulting in only partial ionization in dilute solutions. (Why Hydrofluoric Acid Is a Weak Acid)The fact that the HF bond is stronger is correct.I tabulated them (data from Acc. Chem. Res., 2003, 36 (4), pp 255–263)[math]\small{\begin{array}{cccc} \hline \text{HX} & \text{pK}_a & \Delta H_{bond} \text{/kJ mol}^{-1} \\\hline \text{HF} & 3.2 & 571 [/math][math][/math][math]\\ \text{HCl} & -5.9 & 432\\ \text{HBr} & -8.8 & 367 [/math][math][/math][math]\\ \text{HI} & -9.5 & 299 [/math][math][/math][math]\\ \hline \end{array}}[/math]But the bond strenght explanation is again problematic and confusing. Especially if we would check the bond dissociation enthalpy in the series from CH₄ to HF. In that series the strongest bond gives the strongest acidl![math]\small{\begin{array}{cccc} \hline \text{HX} & \text{pK}_a & \Delta H_{bond} \text{/kJ mol}^{-1} \\\hline \text{H-CH}_3 & \sim 50 & 440 [/math][math][/math][math]\\ \text{H-NH}_2 & 35 & 451\\ \text{H-OH} & 14.0 & 498 [/math][math][/math][math]\\ \text{H-F} & 3.2 & 571 [/math][math][/math][math]\\ \hline \end{array}}[/math]Why would we use bond strength to explain the difference in acidity if we go down the periodic table but completely disregard if we move horizontally?It is almost like we can claim.[math]\boxed{\text{The main purpose [/math][math]of[/math][math] chemistry is to confuse students.}}[/math]In all seriousness. We are dealing with acidity in solution and therefore we can’t simplify this using one set of data.To start with the above data of the bond dissociation enthalpy is the energy required for homolytic bond cleavage into its two radicals. (We are interested in heterolytic bond cleavage or formations of ions!). In addition, this is thermodynamic data in the gas phase (not in water!).So it is clear that bond energy alone can not give you enough information.The full process can be described via a thought process using so-called Born–Haber cycles.Figure 1: Born-Haber Cycle of ion-formation of the halogen acids.Going from the neutral species in water we first need to remove it from water to bring it to the gas phase. Subsequently, cleave the bond (bond dissociation energy) and ionize (Ionization energy for hydrogen, Electron Affinities for X) As the last step, the ions are hydrated again.But we immediately come into serious problems using this thought process: we lack the data of all these steps. Actually, this is not that unexpected the data on the dehydration faces the same problem as a measurement of the concentration. The concentration of the neutral species is far too low in water, so thermodynamic data on the dehydration is not available. (For HF it is available, for the others, it might be not so reliable.)I tabulated the data coming from various sources. Electron affinity from Wikipedia, Bond energies from the above source, and dehydration energies from Ayotte’s article. (See below for the reference.)Table 1: Enthalpy changes for the Born Haber cycle of acidity of HX (see Figure 1 for codes)The total enthalpy difference includes values for Ionization Energy of hydrogen (1318.4 kJ/mol) and the hydration enthalpy of H+ (-1094 kJ/mol)As discussed previously the Bond energy (2) goes down towards Iodine. Also, the Electron Affinities (3) goes down towards Iodine, although F is not following the trend:The incoming electron is going to be closer to the nucleus in fluorine than in any other of these elements, so you would expect a high value of electron affinity. However, because fluorine is such a small atom, you are putting the new electron into a region of space already crowded with electrons and there is a significant amount of repulsion. This repulsion lessens the attraction the incoming electron feels and so lessens the electron affinity. A similar reversal of the expected trend happens between oxygen and sulfur in Group 16. The first electron affinity of oxygen (-142 kJ mol-1 ) is smaller than that of sulfur (-200 kJ mol-1) for exactly the same reason that fluorine’s is smaller than chlorine's.Why is Fluorine an Anomaly?Looking at the data one thing is striking: the difference in bond energy (2) is almost completely compensated by the difference the energy released by hydration (4) of the halogen ions.The total enthalpy change (ΔH total) seems to be negative for all halogens. Meaning it is an exothermic reaction for all of them, although overall the process is less exothermic for fluor.This lower exothermicity for the ionization of fluor in water is — based on the data -- mainly due to the differences in the first step: it cost more energy to dehydrate HF (1).Similar to ammonia and water, hydrogen fluoride is capable of forming strong hydrogen-bonds (with itself, but also with water). It costs a lot of energy to overcome these hydrogen bonds. This is e.g. also the reason for the higher boiling points of water, ammonia, and hydrogen fluoride.(scheme from Chemguide - intermolecular bonding - hydrogen bonds)Anyhow the process is exothermic for all of the hydrohalic acids. In order to explain the difference in acidity we need to dig further into the data:Based on the pKa values as quoted previously. We can calculate the total difference of Gibbs free energy. Using the formula ΔG° = 2.303 RT pKₐAnd the total entropy change. (based on ΔG° = ΔH° - TΔS°)Table 2: Gibbs Free energy and entropy changesBased on these values we can conclude that the true reason for the fact that HF is a weak base is mainly due to entropic reasons.The total ΔS° in the above table is the entropy for the full cycle but based on CRC handbook the standard entropy S° of aqueous F⁻ is –13.8 J/mol.K (compared to values of 56.5, 82.4, and 111.3 J/mol.K for Cl⁻, Br⁻ and I⁻)The fluor ion is small and has a large charge density. The solvation of fluor ions in water, therefore, produces a significant order in the surrounding water (and hydronium) molecules. Hence a negative entropy. This effect is much less in the larger halogen ions.This effect is demonstrated experimentally by Ayotte and coworkers in ‘Why is hydrofluoric acid a weak acid?’At lower temperature, the entropic part of the Gibbs free energy is indeed of lesser importance (ΔG° = ΔH° - TΔS°). By using infrared vibrational spectra at a very low temperature (40K) they could demonstrate that HF is indeed a much stronger acid at these lower temperatures.To be complete: There are alternatives explanations as well:There is good spectrocsopic evidence that hydrogen fluoride ionizes fairly completely in solution in water, but instead of producing free hydroxonium ions, H₃O+, and fluoride ions, there is such strong attraction between these that they form a strongly bound ion pair, H₃O+.F-Hydrogen halides as acidsAlthough other researchers have questioned that observation.I already tried to explain the difference between corrosivity and acidity in another answer so please read Kurt VdB's answer to What makes an acid so corrosive?

The iodide ion is a good nucleophile because it has a large atomic radius. As you go down the halogen group, the proton number increases and there is more shielding of the outer electron from the nucleus from more full inner electron shells. The increase in shielding is more significant than the increase in proton number, thus the outermost electron is further away from the nucleus at a higher energy level (it has a higher intrinsic energy) and the size of the electron cloud is bigger. This means that halogens further down the group are more polarisable, as the electrons of the anion (the electron cloud) are more easily distorted and drawn away from the anion, as they are further away from the positive nucleus so there is a lower electrostatic force of attraction between the outer free out electrons and the nucleus. Therefore, I- is a good nucleophile and a better nucleophile than F-, Cl-, and Br- as it is more able to donate a pair of its outer electrons to an electrophile, forming a dative covalent bond.The fact that the iodide ion is a good base/nucleophile as it is very capable of donating a lone pair of electrons to an electron deficient site also coincides with the fact that it is a good leaving group. This is because the stronger the acid that is formed from the leaving group, the better the leaving group. The I- ion is a great leaving group due to its large atomic radius, meaning that the x-I bond (where x is an atom bonded to the iodine in a specific compound) is weaker (has a lower bond enthalpy, as it is longer and there are less electrostatic forces of attraction between the two positive nuclei and the shared Pair of electrons between them), so it is more easily broken meaning that iodine is more capable of behaving as a leaving group and departing with a pair of electrons via heterolytic fission (forming the iodide anion). Consequently, the acid formed from this leaving group, HI, is a strong acid as the H-I bond is weak (again due to iodine's large atomic radius) so the acid dissociates more completely into its ions in solution (so is a more capable proton donator/electron acceptor) and the equilibrium HI <=> H+ + I- lies further to the right (as the formation of HI releases less energy, so the overall reaction is more endothermic and less energetically favourable). This means it is a stronger acid than HF, HCl, and HBr, hence iodine is a better leaving group.