Which of the following is a Lewis acid?

First, recall the definition of a Lewis acid and a Lewis base. According to G. N. Lewis,

$$\text{Lewis acid} = \text{a species that can accept an electron pair},$$

$$\text{Lewis base} = \text{a species that can donate an electron pair}.$$

Now, we examine every option one by one, asking whether the central atom in each molecule has an empty orbital that can accept an electron pair.

Option A is $$\text{PH}_3.$$ We note that phosphorus in $$\text{PH}_3$$ has the electronic configuration $$[Ne]\,3s^23p^3.$$ In the molecule, the three $$3p$$ electrons form σ-bonds with three hydrogens, and the remaining lone pair occupies the fourth position of a tetrahedron. All valence orbitals are filled; phosphorus does not possess an empty orbital of sufficiently low energy to accept another pair easily. Consequently, $$\text{PH}_3$$ behaves as a Lewis base (because of its lone pair) rather than a Lewis acid.

Option B is $$\text{NF}_3.$$ Nitrogen has the electronic configuration $$1s^22s^22p^3.$$ Just like in ammonia, nitrogen in $$\text{NF}_3$$ holds one lone pair. Again, this lone pair makes $$\text{NF}_3$$ a potential Lewis base, while no vacant orbital is available on nitrogen to accept an electron pair. Therefore, $$\text{NF}_3$$ is not a Lewis acid.

Option C is $$\text{NaH}.$$ Sodium hydride is best described as an ionic solid composed of $$\text{Na}^+$$ and $$\text{H}^-$$ ions. The hydride ion $$\text{H}^-$$ is a powerful electron-pair donor, hence unquestionably a Lewis base. The $$\text{Na}^+$$ ion, while electron-deficient, usually acts as a spectator cation and does not function as a conventional Lewis acid in simple molecular interactions. Thus, $$\text{NaH}$$ does not qualify as a Lewis acid.

Option D is $$\text{B(CH}_3)_3.$$ Boron in this molecule has the electronic configuration $$1s^22s^22p^1.$$ In $$\text{B(CH}_3)_3$$, boron forms three σ-bonds with the three $$\text{CH}_3$$ groups. After forming these bonds, boron possesses only six valence electrons:

$$\text{Total electrons around B}=3\times2\;(\text{from three σ-bonds})=6<8.$$

This electron deficiency means boron has an empty $$2p$$ orbital. Because that orbital can accept an electron pair from some donor, $$\text{B(CH}_3)_3$$ satisfies the definition of a Lewis acid. Indeed, trialkyl- and trialkoxy-boranes are classic textbook examples of Lewis acids, forming adducts such as $$\text{B(CH}_3)_3\!:\!NH_3$$ when a Lewis base like ammonia donates its lone pair into boron’s vacant orbital.

Comparing all four choices, only $$\text{B(CH}_3)_3$$ clearly possesses a vacant orbital capable of accepting an electron pair. Hence it alone behaves as a Lewis acid among the given compounds.

Hence, the correct answer is Option D.