Volume of anion and comparison

Question

Recently, I came across a book in which I was solving the exercises and found out that the solution to a question says that the fluoride ion, $\ce{F^-}$, is smaller in size than the hydride ion, $\ce{H-}$.

As far I know, the hydrogen has only one shell in contrast to fluorine. Addition of a electron to hydrogen adds a electron to the 1s orbital and thus to the first shell. Similarly in fluorine, the electron is added to the 2p orbital and therefore to the same shell.

Anions are a bit bigger than their parent atom, and there is some serious electron repulsion, but how can such a repulsion make the hydride ion bigger than the fluoride ion, given that the fluoride ion also faces some repulsion within itself. Are there some other factors - or is the fact which I reported is very wrong?

Answer 1

The quick answer to which ion, $\ce{H-}$ or $\ce{F-}$, is bigger is "it depends". The general reasoning behind the claimed fact the hydride anion is larger than the fluoride anion typically comes down to charge density. In $\ce{H-}$, there are two electrons for the one proton, giving it a charge ratio of 2, while in $\ce{F-}$, there are ten electrons and nine protons for a charge ratio of roughly 1.1. The significantly higher charge ratio of hydride leads to a much more diffuse orbital, and a potentially larger ionic radius.

However, there are a few issues that arise, including:

• How do we measure the ionic radius? Ions and atoms don't have a hard boundary; the probability of finding an electron a certain distance from the nucleus drops, but there's no firm boundary.
• Ionic radii depend on the nature of the counterion

One potential way to measure the ionic radius is to look at lattice parameters of solids, determined by X-ray crystallography. This does not give much information about what the ions look like in solution, but this is not of huge importance in this case as formal hydride is unlikely to exist in solution. More importantly, X-ray crystallography gives the distance between the two ions, but it does not give the ionic radii. However, we can directly compare the lattice parameters of two solids sharing the same cation with the idea that a larger lattice parameter would be indicative of a larger anion.

For the alkali metal binary salts, the fluoride ion does typically lead to a smaller lattice constant than hydride (e.g. 4.027 and 4.083 Å for lithium salts, respectively). However, for calcium hydride and fluoride, the latter gives a larger orthorhombic cell, implying that fluoride is a larger anion.¹ Indeed, for most of the analogous binary hydrides and fluorides in this reference, the fluorides produce larger lattice constants. Using different approaches to measuring ionic radii, including the soft sphere and hard sphere approaches, Lang and Smith succinctly declare that "the ions maintin the following order of size: $$\ce{Li+}<\ce{H-}<\ce{Na+}<\ce{F-}<\ce{K+}<\ce{Cl-}<\ce{Rb+}<\ce{Br-}<\ce{Cs+}<\ce{I-}$$ All the anions are larger than isoelectronic cations, but the Pauling order of ion size is not maintained."²

Therefore, while it may be true that the hydride anion appears larger than the fluoride anion in a few cases, this is not true in general.

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¹ J. Solid State Chem., 1970, 2 (2), 144

² Dalton Trans., 2010, 39, 7786