The hydride that is NOT electron deficient is:

First, let us recall the meaning of the term “electron-deficient hydride”. A hydride is called electron deficient when the central atom does not attain an octet of electrons in the usual two-electron covalent bonds and therefore needs multi-centre (3-centre-2-electron) bonding or some other delocalisation to hold the molecule together.

We will now examine each option one by one and count the electrons present around the central atom in the simple Lewis picture.

For a single ordinary covalent bond, the shared pair contains $$2$$ electrons. To satisfy the octet rule, an atom ideally needs $$8$$ electrons in its valence shell.

Option A : $$SiH_4$$

Silicon belongs to group 14, so its valence electron count is $$4$$. In $$SiH_4$$ the silicon forms four Si-H single bonds. Each bond contributes $$2$$ electrons to the valence shell of silicon. Hence the total number of electrons around silicon is

$$4 \text{ bonds}\times 2 \text{ electrons per bond}=8 \text{ electrons}.$$

The octet is exactly satisfied, so $$SiH_4$$ is not electron deficient.

Option B : $$B_2H_6$$

Boron is in group 13 with $$3$$ valence electrons. Even if each boron made three normal B-H bonds it would receive only $$6$$ shared electrons, not $$8$$. The actual structure of $$B_2H_6$$ contains two three-centre two-electron (3c-2e) B-H-B bridges to compensate for the deficiency. The presence of 3c-2e bonds is a hallmark of electron-deficient compounds, so $$B_2H_6$$ is electron deficient.

Option C : $$GaH_3$$

Gallium too is a group 13 element with $$3$$ valence electrons. In $$GaH_3$$ the gallium would share $$3\times 2 = 6$$ electrons, falling short of the octet. Like the other group-13 trihydrides, it is electron deficient (polymeric with bridge bonds in the solid state).

Option D : $$AlH_3$$

Aluminium is also a group 13 element. In monomeric $$AlH_3$$, aluminium would again receive only $$6$$ electrons from the three Al-H bonds. To remedy this, $$AlH_3$$ forms polymeric chains with 3c-2e Al-H-Al bridges; therefore it is electron deficient.

From this detailed comparison we see that the only hydride in which the central atom achieves a complete octet through ordinary two-electron bonds is $$SiH_4$$.

Hence, the correct answer is Option A.