 Solution

The ionic and covalent diabatic curves cross; coupling replaces the crossing by an avoided crossing of the adiabatic curves. Take the separated neutral atoms as the zero of valence energy. A covalent diabatic state has one electron in a lithium-centred $2s$ orbital and one in a hydrogen-centred $1s$ orbital, coupled to an overall singlet. An ionic diabatic state has an empty lithium valence orbital and a doubly occupied hydrogen $1s$ orbital.

At large internuclear separation $R$, the ionic limit lies above the neutral one by $I(\mathrm{Li})-A(\mathrm{H})$. At smaller $R$, its attractive Coulomb term $-e^2/(4\pi {ϵ}_{0}R)$ lowers it; short-range repulsion eventually raises both curves. A two-state Hamiltonian givesWhere the diabatic energies would coincide, nonzero coupling $W$ opens a gap $2|W|$. The lower adiabatic state changes continuously from predominantly ionic character near the bond minimum to predominantly neutral covalent character at dissociation. The upper state exchanges character in the opposite direction. Near the crossing both are mixtures, so a sketch assigning a fixed charge pattern to an entire adiabatic curve would miss the important physics.