For a serendipity element, the shape function of a middle pt on a edge can be formed using the tensor product in each direction, e.g. point 5 on Quad8$$\phi_5(\xi,\eta)=\frac{1}{2}(1-\xi^2)(1-\eta)$$but the shape function on corner node (point 1) is$$\phi_1(\xi,\eta)=\frac{1}{4}(1-\xi)(1-\eta)-\frac{1}{4}(1-\xi^2)(1-\eta)-\frac{1}{4}(1-\eta^2)(1-\xi)$$Why the shape function of the corner point from a serendipity element (Q8)  can not be built by the tensor products like this$$\phi_1(\xi,\eta)=\frac{1}{4}\xi\eta(1-\xi)(1-\eta)$$Does this mean I just pick up the wrong basis?
Shape functions of serendipity element as tensor products
finite element