A valence-bond configuration-interaction Slater-orbital wave function for LiH was constructed with all orbital exponents optimized. The orbitals used were Li $1s$, $2s$, $2p\ensuremath{\sigma}$, $3d\ensuremath{\sigma}$, and H $1s$, $2p\ensuremath{\sigma}$; and six configurations were included, all with undeformed inner shells ($1{s}^{2}$). The core deformation and the related correction to the nuclear quadrupole coupling constant were then calculated by the method of Sternheimer and Foley, i.e., by first determining the quadrupolarization of the $1s$ shell by the nuclear moment $Q$ and then calculating the interaction of the external molecular charges with the nuclear $Q$ shielded by the quadrupole moment induced in the $1s$ shell. Denoting by ${q}_{0}$ the value of the electric field gradient at the Li nucleus obtained from a wave function with no provision for core deformation and denoting by $\ensuremath{\Delta}q$ the correction to ${q}_{0}$ associated with the quadrupolarization of the core, we investigated the sensitivity of ${q}_{0}$ and of $\frac{\ensuremath{\Delta}q}{{q}_{0}}$ to changes in the wave function produced by varying the configuration mixture and altering the values of the orbital exponents. Although ${q}_{0}$ was found to be sensitive to these changes, the fractional Sternheimer correction, $\frac{\ensuremath{\Delta}q}{{q}_{0}}$, was insensitive and in all cases stayed between the limits of -0.22 and -0.24. The final values obtained for ${q}_{0}$ and $q$ are not highly accurate, because of the obvious limitations of the wave function, but the result that $\frac{\ensuremath{\Delta}q}{{q}_{0}}=\ensuremath{-}0.23$ is reliable enough to be useful in estimating Sternheimer corrections to the results of other workers who have used more elaborate wave functions. Combining the results of several calculations with implicit and explicit Sternheimer corrections, we estimate that $\frac{q}{2e}=(\ensuremath{-}0.0170\ifmmode\pm\else\textpm\fi{}0.0013){{a}_{0}}^{\ensuremath{-}3}$ and that $Q({\mathrm{Li}}^{7})=(\ensuremath{-}4.3\ifmmode\pm\else\textpm\fi{}0.3)\ifmmode\times\else\texttimes\fi{}{10}^{\ensuremath{-}26}$ ${\mathrm{cm}}^{2}$.