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$\ceHCl$ is a molecular compound because there is a covalent bond between $\ceH+$(a proton) and also $\ceCl-$ (a chloride ion)

$\ceNH4NO3$ is an ionic compound because there is an ionic bond in between $\ceNH4+$ (an ammonium ion) and also $\ceNO3-$ (a nitrate ion) even though every the atoms making up the molecule room non-metal.

Then, i beg your pardon of the adhering to is true?

$\ceH3C-COOH$ is one ionic compound since there is one ionic bond in between $\ceC2H3O-$ (an acetate ion) and $\ceH+$ (a proton)? OR

$\ceH3C-COOH$ is a molecular compound because there is a covalent bond in between $\ceC2H3O2-$ (an acetate ion) and also $\ceH+$ (a proton)?

How do you call if a bond in between the 2 ions space molecular or ionic? (regardless of whether they room polyatomic or monoatomic ions)


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Safdar Faisal
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In principle, when studying a structure favor HCl, CH3COOH, or any kind of other compound, the first thing to carry out is illustration the price of the atoms in a reasonable geometry ~ above a paper of paper. Then think about the outer electron class : put enough points around each symbol to gain a reasonable Lewis structure. Because that example, one suggest near $\ceH$ and six points around $\ceO$. Then shot to join atoms through their neighbors to make covalencies (or doublets). Once all atoms are joined come neighbors, count the variety of electrons roughly each atom, taking into account the the 2 electrons contained in a covalence between A and also B space at the periphery of both atoms A and B. Check all atoms. If an atom respects the octet law, don"t change it : its framework is covalent. If an atom A does no respect octet law, try to remove one electron from its external layer, and also send this electron to the B atom, which becomes negatively charged. The covalency disappears indigenous the an ar between A and B, and also the bond abdominal muscle becomes ionic v a positive charge on the A atom, which becomes $\ceA+$, and a an adverse charge on $\ceB$ which becomes $\ceB-$.

Using this method, you will view that $\ceHCl$ and also $\ceCH3COOH$ space both do of covalencies, since all their atoms room respecting the octet rule. You will likewise see the $\ceNaCl$ may very first be drawn in a covalent structure. Yet then the sodium atom does not fit the octet rule. It will if one electron is removed from the sodium atom, to be provided to the chlorine atom, producing then $\ceNa+$ and also $\ceCl-$ ions.

Hydroxides consists of two sorts of bonds. Hydroxides choose $\ceNaOH$ room made of covalent bonds between $\ceO$ and also $\ceH$ atoms, and also of ionic bonds between $\ceNa^+$ and $\ceO-$.

Oxygenated acids space a distinct case. Lock may acquire two sorts of bonds in between $\ceA$ and $\ceB$. For example sulfuric acid $\ceH2SO4$ have the right to be drawn with this method to give very first a covalent structure with two bonds $\ceH-O$, two bonds $\ceO-S$, and two twin bonds $\ceS=O$. This is ok for all atoms except the central Sulfur atom, i beg your pardon is surrounded by 6 doublets or 6 covalences. In ~ this point, some chemists room thinking the it is a case of hyper valence. Part think the the octet dominion may be derived by removed the 2nd bond indigenous each twin bond native the sulfur atom, and sending it come the external Oxygen atom. This provides a main sulfur atom fee $2+$ (respecting the octet rule) and also two oxygen atoms charged $1-$. Both theories deserve to be discussed.

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The only trouble with this reasoning is the truth that it does not occupational properly for atoms the the third column. Because that example, $\ceBF3$ have to be ionic making $\ceB^3+$ and also $\ceF-$ ions. Without doubt $\ceBF3$ remains covalent, and it is hard to justify. Maybe, on might admit that Boron atom prefers to stay covalent fairly than to loosened three electrons. 3 is probably too much.