Why is the ionic radius of hydride so large? The order of ionic radii for halides and hydride is apparently as follows:
$$\ce{F-} &lt; \ce{Cl-} &lt; \ce{Br-} &lt; \ce{H-} &lt; \ce{I-}$$
Why is the hydride ion so large, even larger than bromide which has filled 3d and 4p subshells? Some claim that it is because the ratio of positive to negative charge is 0.5, which is an unusually small ratio. But this explanation doesn't fully convince me.
Nat. Commun. 2014, 5 (1), No. 3515 also makes mention of the unusually large size:

The ionic radius of an $\ce{H-}$ ion is $\pu{134\pm 2 pm}$, which is slightly larger than the radius of $\ce{F-}$(II) ($\pu{128.5 pm}$) and almost the same as the radius of $\ce{O^2-}$(II) ($\pu{135 pm}$). This evaluation agrees with the generally accepted trend that the radius of an H− ion is nearly the same or slightly larger than that of an $\ce{F-}$ ion and increases with increasing electropositivity of the coordinating cation.
It is not.
Presenting the order of ionic radii like that is useless.
Ionic radius of a given ion is not constant and mainly rely on crystal structure (coordination environment, more precisely) and used experimental techniques (crystal structure determination and theoretical calculations).
As a consequence, you can only compare homogeneous data.
Proposing the ordering based on some cherry-picked values from random sources or their averaged values is pointless at best.
This very issue is discussed by Messer [1]:

The effective ionic radius of fluoride ion is given as 1.33 Å by Pauling (8) and 1.36 Å, by Zachariasen (9), for coordination number 6. The effective ionic radius of hydride ion varies from 1.27 to 1.52 Å (10), being much more highly sensitive to the particular cation present and to the particular assumptions used in deriving the radius than the fluoride value. For the alkali metal hydrides other than lithium hydride, with the same face-centered cubic structure as the corresponding fluorides, the effective hydride radius is 1.47 Å for $\ce{Na+},$ and 1.52–1.54 Å for $\ce{K+},$ $\ce{Rb+},$ and $\ce{Cs+}.$ For $\ce{LiH},$ $\ce{CaH},$ and $\ce{BaH},$ the hydride ion radius is 1.34–1.36 Å. The closest similarities between hydrides and fluorides should thus appear here.

You can find the order of ionic radii for group 1 hydrides and halides in a more recent work by Lang and Smith [2]:

$$\ce{Li+} &lt; \ce{H-} &lt; \ce{Na+} &lt; \ce{F-} &lt; \ce{K+} &lt; \ce{Cl-} &lt; \ce{Rb+} &lt; \ce{Br-} &lt; \ce{Cs+} &lt; \ce{I-}$$

Again, note that this doesn't only contradict your order, but it is no universal sequence and is only true for these compounds with soft-sphere model used for calculation.
References

Messer, C. E. Hydrides versus Fluorides: Structural Comparisons. Journal of Solid State Chemistry 1970, 2 (2), 144–155. DOI: 10.1016/0022-4596(70)90062-9.
Lang, P. F.; Smith, B. C. Ionic Radii for Group 1 and Group 2 Halide, Hydride, Fluoride, Oxide, Sulfide, Selenide and Telluride Crystals. Dalton Trans. 2010, 39 (33), 7786. DOI: 10.1039/c0dt00401d.