On page 130 of Organic Chemistry by Clayden, it is stated that
Nucleophilic attack by the hydride ion, $\ce{H-}$, is an almost unknown reaction. This species, which is present in the salt sodium hydride, NaH, has such a high charge density that it only ever reacts as a base. The reason is that its filled 1s orbital is of an ideal size to interact with the hydrogen l atom’s contribution to the $\sigma ^*$orbital of an H–X bond (X can be any atom), but much too small to interact easily with carbon’s more diffuse 2p orbital contribution to the LUMO (π* of the C=O group.)"
We may draw a direct proportionality between molecular radius and charge density.
But according to radii values, Inorganic Chemistry: Principles of structure and Reactivity states,
it has a radius of 208 pm compared to 216 pm for the iodide ion
Calculating the charge density, as
$Q_\mathrm{density} = \frac{[\mathrm{Charge}]}{\mathrm{[Volume]}} = \frac{[1.6\times 10^{-19}]}{[4/3π \times 208^3]} $ (nearly equal to iodide which has low charge density.)
But in the next few lines it is written that the charge density is sufficient to interact with the σ* orbital of H in H-X molecule, but is too small to interact with 2p orbitals of carbon.
The two appear contradictory as on one hand size of hydride is big and soft, while on the other it cannot react with C as "as carbon has too diffused 2p orbitals."
Please explain the contradiction.
