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1

In principle, when studying a structure like HCl, CH3COOH, or any other compound, the first thing to do is drawing the symbol of the atoms in a reasonable geometry on a sheet of paper. Then consider the outer electron layers : put enough points around each symbol to get a reasonable Lewis structure. For example, one point near $\ce{H}$ and six points around $...


5

I think, what actually OP meant to ask is: Are there any known organic compounds with carbon involving in an ionic bond? The reason for my suggestion is the phrase "to make an ionic bond and to exhibit ionic properties" in OP's question, even though it did not describe the involvement of carbon. Otherwise, any ionic carbonate (e.g., $\ce{Na2CO3}$)...


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On the surface, the answer is no, because that's not what we would call a hydrogen bond. On a deeper level, the answer is... probably still no for NaH or CsH, because they are too ionic for this. Other hydrides, however, actually do have a thing which looks mighty like an "inverted" hydrogen bond (though we don't normally call it that, either). ...


0

Well, you can say that they are kind of similar...the only major difference is that Schottky defects are only for ionic compounds, and the number of missing cations and anions must be equal. Vacancy and interstitial defects can be shown by non-ionic solids. Ionic solids must always maintain electrical neutrality. Rather than simple vacancy or interstitial ...


2

Without electrolysis, there would be just an extremely short spike of a temporary initial current, until capacitors formed by electrodes charge. There is no significant accumulation of ions, as even a tiniest such charge separation would cause strong electrostatic forces acting against it.


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TeF4 is polymeric and low-melting. Not ionic. Covalent. In general, the elements that are metalloids (on the "border" of metal and non-metal) like Te and Sb, tend to behave chemically like nonmetals (looking at their compounds). But physically as metals (shiny, malleable, conductive). So that's another good clue that these elements would be ...


3

Your analogy isn't correct. As @Ivan pointed out in the comments, oxidation numbers apply to covalent compounds as well as they do to ionic ones. You calculate oxidation numbers assuming that the more electronegative atom takes up both the electrons in a bond. Formal charge, on the other hand, is determined assuming no difference in electronegativity. Take ...


3

The oxidation number is the charge that one atom would have if all the covalent bonds of the molecule (or of the ion) were transformed into ionic bonds, with a negative charge on the most electronegative atom of the bond, and a positive charge on the least electronegative atom.


1

Suppose $x$ and $y$ are the amount of $\ce{BaCl2}$ and $\ce{NaCl}$ in $\pu{mol}$, respectively in original mixture. Thus the mass of $\ce{BaCl2}$ in the mixture is $208.232x$ and that of $\ce{NaCl}$ is $58.4425y$, both in $\pu{g}$. Thus, $$208.232x + 58.4425y = 0.1036 \tag1$$ Now, the the amount of $\ce{Cl-}$ ions in original mixture is $2x+y$ in $\pu{mol}$. ...


3

Don't get your hopes up. Wikipedia identifies magnesium fluoride as insoluble in ethanol. Calcium fluoride fares no differently in acetone. While these data are hardly exhaustive, they suggest strongly that Group 2 fluorides are not a good way to get fluoride ions into solution. Of course, we would not expect Group 2 ionic compounds with their doubly ...


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