when we talk about stability of halogen anions we say, $\ce{I-} \gt \ce{Br-} \gt \ce{Cl-} \gt \ce{F-}$, and the reason is that "the negative charge (the electrons) have more space to move about in the larger halogen". But when we talk about stability of a negative charge on an orbital, why do we say that the negative charge on the $sp$ hybridized orbital (which is the smaller one) is more stable than $sp^2$ and then $sp^3$ as the least stable where as the $sp^3$ orbital is the biggest orbital among all of the three and can allow the electrons to spread out more.
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$\begingroup$ The primary stability factor is energy. Also, what do you mean by stability, particularly what do you mean saying I- is more stable than F- ? It is easy to oxidize I-, but try it (chemically) with F-. $\endgroup$– PoutnikCommented Oct 5, 2020 at 8:27
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$\begingroup$ I meant that the iodine anion is more stable than the flourine anion, as there is much more space for the electron density to spread out in iodine anion. @Poutnik $\endgroup$– FloatossCommented Oct 5, 2020 at 8:41
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2$\begingroup$ No, that is considered reason of stability, not what you mean by stability itself. Do you mean hydrated or naked ion ? Do you mean acido-basic stability or general chemical stability ? You have put too little context in the question, what usually leads to unnecessary clarification ping-pong. $\endgroup$– PoutnikCommented Oct 5, 2020 at 8:44
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$\begingroup$ oh that was a question on checking which compound was more acidic $\endgroup$– FloatossCommented Oct 5, 2020 at 8:57
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1$\begingroup$ and that has not been included in the question, because....... ? For the basicity of hydrated halogenides part, or acidity of hydrated hydrogen halogenides, it is much more complex, as the environment interaction and thermodynamics has to be considered. See chem.libretexts.org/Bookshelves/Inorganic_Chemistry/… $\endgroup$– PoutnikCommented Oct 5, 2020 at 9:18
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When we talk about hybrid orbitals and stability of charge on them, the electronegativity factor plays a more prominent role. The sp hybridised orbital is more electronegative than the others, owing to Bent’s rule which states - Greater the s character of a hybrid orbital, greater is its electronegativity.
Therefore, species having the negative charge on an sp hybridised atom tend to be less basic.
(You might be interested in knowing that, an sp hybridised carbon atom is actually more electronegative than a nitrogen atom.)
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2$\begingroup$ Looking at the other comments, I also get suggestions that there may be a slight confusion between basicity and nucleophilicity. Basicity is a characteristic property, and that depends only on how stable the negative charge on an atom is. Nucleophilicity is, however, dependent on a few other things besides that. It’s not a characteristic property, it’s more like a kinetic property. They do go hand in hand sometimes, but not always. $\endgroup$ Commented Oct 5, 2020 at 9:59
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2$\begingroup$ In the context of my previous comment, for example - the iodide ion is a weaker base than the fluoride ion. And that won’t change no matter what environment (solvent) you want to test their basicities in. But, the iodide ion happens to be a stronger nucleophile in a polar protic solvent, since the fluoride ion is so heavily solvated (or hydrated) that it loses it’s ability to act as a nucleophile to a large extent. The iodide ion, however, is not solvated effectively because of its huge size and a lesser charge density. $\endgroup$ Commented Oct 5, 2020 at 10:03
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1$\begingroup$ The s orbitals are closer to the nucleus than p orbitals, which makes the electron pair in an s orbital more strongly bond. That’s what makes the trend so. $\endgroup$ Commented Oct 5, 2020 at 15:13
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1$\begingroup$ And I totally understand your point regarding the size of the orbitals. But here’s the thing - electronegativity factor and size factor are giving conflicting suggestions. It just happens to be so, that the electronegative factor is more prominent. I’m not saying size factor is incorrect. It’s just that, considering electronegativity gives a better (in this case the totally opposite) suggestion. $\endgroup$ Commented Oct 5, 2020 at 15:16
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1$\begingroup$ If I may add this, it’s just one of those things where we can’t quite understand where we went wrong...one thing being more fit than other doesn’t imply the first thing is unreasonable. Because again, I see no fault in your reasoning. $\endgroup$ Commented Oct 5, 2020 at 15:22