 Solution

The intended chemically matched surface approximation is ${\gamma }_{SL}\simeq 0$. It gives $S\simeq 23-21.3=1.7\mathrm{mN}{\mathrm{m}}^{-1}>0$, so increasing the wetted area lowers free energy. The initially steep drop spreads and flattens toward complete wetting:The initial $90^{\circ }$ angle describes a nonequilibrium shape and must not be inserted into equilibrium Young's equation to fix ${\gamma }_{SL}$. More precisely, complete wetting requires ${\gamma }_{SL}\le 1.7\mathrm{mN}{\mathrm{m}}^{-1}$ in this surface-energy model; chemical similarity alone is not a measurement of that interfacial energy.

Water on such a low-energy, oil-like surface is expected to bead into a comparatively rounded, high-angle drop because the substrate does not compensate its cohesive interactions. Its numerical contact angle needs the water-solid interfacial energy, which is not supplied.