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## Hot answers tagged electronegativity

30

Until the (recent) redefinition of the IUPAC, the concept of oxidation states was not as well defined as one would expect. I have discussed the issues of the old version and outlined the new definition in more detail in an answer to Electronegativity Considerations in Assigning Oxidation States. When you apply the official pre-2016 definition (via the ...

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The following table contains some relevant data, the $\mathrm{p}K_\mathrm{a}$'s of the various haloforms along with the Pauling electronegativity of the corresponding halogen. \begin{array}{|c|c|c|} \hline \text{Haloform} & \mathrm{p}K_\mathrm{a} & \text{Electronegativity}\\ \hline \ce{CHF3} & 25\mathrm{-}28 & 3.98\\ \hline \ce{CHCl3} & ...

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Yes, this is a subtle thing. Using the Pauling electronegativities, one would expect CsF to have the larger electronegativity difference (3.2). So in principal, it should be "more ionic." Unfortunately, an ionic bond requires separating charge, so $\ce{Cs+F-}$. The problem is that $\ce{Cs+}$ is much larger than $\ce{K+}$ and so the dipole moment for a ...

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In the IUPAC Recommendations 2016 the definition of oxidation state underwent a significant and comprehensive change. It does now resemble the version wich was proposed be Hans-Peter Loock and is quoted in the earlier version of this answer, parts of which are included below. The electronegativity battle scheme is most helpful for all kinds of compounds ...

18

First of all as @chipbuster says $\ce{HF}$ in diluted solutions in water is nearly completely dissociated and therefore shouldn't be called weak. Wikipedia describes this nicely and cites several sources for this claim. It was rather difficult to prove (spectroscopic methods were used), because hydronium ions created in dissociation are mostly bound to ...

17

so wouldn't it be that you would have an even more positive carbon and 2 partially negative oxygens Yes, your analysis is correct to this point. A chemist would say that the bonds in $\ce{CO2}$ are polar (or polarized) and therefor each $\ce{C=O}$ bond has a bond dipole moment. However the molecule itself is linear and the two bond dipole moments are ...

16

Why do Krypton and Xenon have high electronegativity? As you point out, krypton ($\ce{Kr}$) and xenon ($\ce{Xe}$) are members of the Noble gas family. They are generally unreactive (noble) because all of their occupied orbitals are filled with electrons - they really don't want to gain or lose an electron. However, in the 1960's researchers found that $\... 16 The ratio of the speed of an electron traveling in the first Bohr orbit to the speed of light is given by the handy equation $$\mathrm{V_{rel}=\frac{[Z]}{[137]}}$$ where Z is the atomic number of the element under consideration and 137 is the speed of light in atomic units, also known as the fine structure constant. Consequently a 1s electron in the ... 14 As mentioned above there is no hard line between an ionic and a covalent bond, they are just two ends of a spectrum where a bond can have a certain degree of ionic or covalent character. At the lower end of this spectrum, bonds with small electronegativity differences between the atoms have very little ionic character because the charge distribution in the ... 14 You have to think about the whole process. When a metal loses electrons to make a metal ion the following happens: The metallic bonds holding the metal atoms together are broken. The metal atom loses the electrons. The resulting metal ion is hydrated. In your analysis you are only focusing on step 2. The enthalpy and entropy of the entire process factor ... 14 According to Wikipedia, bond dipole moment depends on: Distance between atoms and Overall charge difference, not just electronegativity difference. Resonance tells us that there is some amount of charge separation in$\ce{C=O}$bonds because of the$\ce{C+-O-}$contributor. This difference in charge, in addition to the electronegativity difference, is ... 13 Based on the the answer to this question: Why electronegativity difference greater than 1.7 are ionic? Here is a very nice graph of percent ionic character as function of electronegativity difference for some common binary compounds. Here ionic character can be thought of as the degree of charge separation across the covalent bond. Compounds with 50% or ... 13 Electronegativity is the power of an atom to attract bonding pairs of electrons to itself. It clearly depends on the nuclear charge: the larger it is, the more strongly the nucleus attracts electrons and the closer to the nucleus they stay, so the larger is the electronegativity. Besides, electronegativity also depends on the number of other electrons ... 11 Are the inductive effect and hyperconjugation both different ways of looking at the same phenomenon I think so, both inductive and resonance (that's what hyperconjugation is after all) effects move electrons around. I tend to separate the two effects as follows: inductive effects are associated with electron movement through sigma bonds due to ... 11 In understanding molecular polarity you need to take the whole structure into account. Your reasoning is correct as far as the parts of the molecule are concerned. The individual bonds are polar. But, a molecule can only be polar if it has a net dipole moment (that is, the charges don't balance out in direction across the whole molecule). So CO is polar as ... 11 Carbonates The quote from your text: Carbonates of alkaline earth metals are insoluble in water and can be precipitated by addition of a sodium or ammonium carbonate solution to a solution of a soluble salt of these metals. The solubility of carbonates in water decreases as the atomic number of the metal ion increases. All the carbonates ... 11 The inductive effect can be quantitatively measured by the Hammett equation $$\sigma(\ce{X}) = \mathrm{p}K_\mathrm{a}(\ce{H}) - \mathrm{p}K_\mathrm{a}(\ce{X})$$ where$\mathrm{p}K_\mathrm{a}(\ce{X})$refers to the$\mathrm{p}K_\mathrm{a}$of$\ce{X}$-substituted benzoic acid and$\mathrm{p}K_\mathrm{a}(\ce{H})$to the$\mathrm{p}K_\mathrm{a}$of ... 10 Yes absolutely, electronegativities are hardly static values when you start combining elements and forming ions or molecules. Electronegativity is simply a measure of an ability of an atom to attract electrons in a covalent framework. So keeping in mind that like charges repel (electrons repel electrons) ... it follows that elements in a high oxidation ... 10 Linus Pauling proposed an empirical relationship which relates the percent ionic character in a bond to the electronegativity difference$\Delta \chi$. Percent ionic character$= (1-e^{-(\Delta \chi/2)^2} )\times 100$But I'd like to correct the definition of percent ionic character in your question using dipole moment$\mu$(not Observed value of ionic ... 10 As the$\mathrm{s}$characteristic of an atom increases, so too does its electronegativity. Proof of this can be seen when comparing the$\mathrm{p}K_\mathrm{a}$'s of 2-carbon hydrocarbons. The$\mathrm{p}K_\mathrm{a}$'s of$\ce{CH3 -CH3}$,$\ce{CH2 =CH2}$, and$\ce{CH ≡CH}$are$51$,$44$, and$25$respectively. When considering phenyl group for what it is,... 10 Hydrofluoric acid is the least acidic hydrogen halide because of fluorine's electronegativity. Because of the fluoride ion's small size, it cannot disperse the negative charge over a larger space and will have an extremely high affinity for an electrophile (like$\ce{H+}$), and because of this it will remain mostly as$\ce{HF}$. At high concentrations, ... 10 Methyl groups are great at stabilising carbenium ions via an inductive effect — which should actually be considered a resonance effect — known as hyperconjugation. This effect, which is actually due to neighbouring$\ce{C-H}$bonds, not indiscriminately due to neighbouring carbon atoms, is electronic in nature. The cationic carbon has an empty p orbital, ... 9 I think your lecturer would have been more correct had he said that the carbon-sulfur bond reacts as if the sulfur is slightly$\delta^{-}$and the carbon slightly$\delta^{+}$. Sulphur is a larger atom so it has more, loosely held electrons than carbon. This means that a sulfur atom is more polarizable than a carbon atom. Although carbon and sulfur ... 9 The formula for the net dipole moment$\vec{\mu_{net}}$of an overall neutral system of$n$charged point particles is given by: $$\vec{\mu_{net}} = \sum\limits_{i}^{n} q_i\vec{r_i}$$ where$q_i$is the charge of the$i$th point particle and$\vec{r_i}$is the position vector for said particle; each individual dipole moment points from a negative charge ... 9 The answer lies in electronegativity. Sodium has a significantly lower electronegativity ($\approx 1.0$) than hydrogen ($\approx 2.1$) meaning that hydrogen pulls electron density towards itself, away from sodium to generate a sodium cation and a hydride anion. For a compound to be a Brønsted acid, you need to draw electron density away from the hydrogen, i.... 9 This can be argued on the basis of Bent's rule; concisely stated Atomic s character concentrates in orbitals directed toward electropositive substituents What follows below is a crude explanation. Before that, I'll note that we concern ourselves with the hybridisation of the orbitals at the central atom. Since s orbitals are lower in energy than p ... 9 Experimentally, according to The Relative Rates of Formation of Carbanions by Haloforms J. Am. Chem. Soc., 1957, 79 (6), pp 1406–1412 the rates of formation of the corresponding anions from the following haloforms was as follows: $$\begin{array}{|c|c|}\hline \text{Haloform}&\text{Rate}\\\hline \ce{CDCl2F}&0.89\\\hline \ce{CDBrClF}&21\\\hline \... 9 Electronegativity cannot be calculated from the number of subatomic particles in an atom. However, using Ionization Energy and Electron Affinity, one can calculate the Electronegativity of an atom using the Mulliken Electronegativity Equation. Ionization Energy : The amount of energy required to remove the electron that is least attracted to the nucleus in ... 8 Yes, you are on the right track, let's look at the situation in more detail. In the haloacid equilibrium$$\ce{HX <=> H+ + X-}$$anything that stabilizes HX will push the equilibrium to the left and make the$\mathrm{p}K_\mathrm{a}\$ more positive. Anything that stabilizes the proton and the halogen anion will push the equilibrium to the right and make ...

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