(N/A) For carbon hydrides of type $C_{n}H_{2n+2},$ the following hydrides are possible:
$n=1 \Rightarrow CH_{4}$
$n=2 \Rightarrow C_{2}H_{6}$
$n=3 \Rightarrow C_{3}H_{8}$
For a hydride to act as a Lewis acid,it must be electron-deficient (electron acceptor). For it to act as a Lewis base,it must be electron-rich (electron donor).
Taking $C_{2}H_{6}$ as an example,the total number of valence electrons is $14$ and the total number of covalent bonds is $7$. Hence,the bonds are regular $2e^{-}-2$ centered bonds.
Thus,$C_{2}H_{6}$ has sufficient electrons to be represented by a conventional Lewis structure. It is an electron-precise hydride,where all atoms have complete octets (or duplets for hydrogen). Therefore,it can neither donate nor accept electrons to act as a Lewis acid or Lewis base.