Oxidation state of hydrogen in the LaNi5H intermetallic

This question is from an introductory chemistry course:

What is the oxidation state of $\ce{H}$ in $\ce{NaH}$, $\ce{HCl}$, and $\ce{LaNi5H}$?

I've looked in the textbook to no avail.

Google doesn't help me much either. It does tell me that $\ce{LaNi5H}$ is an "intermetallic" compound.

The best I've found is this:

Allen electronegativities also yield zero for the oxidation state of all such $\eta^2$ hydrogen atoms.

What is the justification? Preferably, is there an "intro to chem" level explanation for this?

At an intro chem level, the oxidation state of all elements in an alloy or intermetallic is zero.

At a deeper level of analysis, fractional, integer, and/or decimal oxidation states may be more appropriate, depending on the system in question.

Per Wikipedia:

An intermetallic, also called an intermetallic compound, intermetallic alloy, ordered intermetallic alloy, and a long-range-ordered alloy, is a solid-state compound exhibiting metallic bonding, defined stoichiometry and ordered crystal structure. Many intermetallic compounds are often simply called alloys.

Schulze in 1967, defined intermetallic compounds as solid phases containing two or more metallic elements, with optionally one or more non-metallic elements, whose crystal structure differs from that of the other constituents.

So, in a high-level description, an intermetallic containing hydrogen such as the $\ce{LaNi5H}$ of the question can be considered essentially to be an alloy with hydrogen as a constituent. Based on this, assigning a $0$ oxidation state to all elements is warranted as a first approximation:

For intermetallic compounds, the ultimate choice of the oxidation state zero at all atoms is best if needed in redox chemistry. (source | doi)

A more detailed investigation might turn up evidence supporting assignment of nonzero fractional, integer, and/or decimal oxidation states to various atoms:

Subtler estimates and round offs are required for compounds with electrons delocalized over non-equivalent atoms, as expressed by several resonance formulas with weights in arbitrarily long decimal numbers. Without round offs of bond orders in Lewis formulas, decimal values of oxidation states would be obtained for certain bonding connectivities. [...] [U]nambiguous and reasonable fractions of small integers are obtained for oxidation states in compounds such as dithiolate and catecholate (see Section 11) or in (car)boranes such as $\ce{B6H10}$ (see Section 6) and $\ce{B10C2H12}$ (p1041), or when vicinal oxidation states are indistinguishably mixed, such as in $\ce{YBaFe2O5}$ (p1058). Reasonable fractional oxidation states appear also in ions where the charge is distributed over several equivalent atoms such as $\ce{C7H7+}$, $\ce{B6H6^2−}$ (p1040), $\ce{I3−}$, and $\ce{N3−}$ (p1037). (source | doi)

Both nickel and lanthanum have electronegativities lower than hydrogen’s $2.2$. Thus, even if the compound is intermetallic, we can treat it as a hydride and give hydrogen an oxidation state of $\mathrm{-I}$. Oxidation states lower than $\mathrm{-I}$ are highly improbable.

Determining the oxidation states of lanthanum and nickel would be a lot harder, but summed up they must be $\mathrm{+I}$. It would need a structure to determine the rest though.