Fields don't have bases, so I think you mean to ask about a standard basis for C^{2} over a field F, or more generally for n-tuples of a field F over F.

Answer: It depends on the scalar field F because the dimension could change thereby changing the number of elements in any basis.

Consider the vector space C over R. Then {1, i} is the standard basis, and the dimension is two. So {1, i} is a linearly independent spanning set for the space. In other words, 1 and i are not R-multiples of each other, and every complex number in C can be written as a linear combination of 1 and i with coefficients from R.

Consider the vector space C over C. Then {1} is the standard basis, and the dimension is one. Every complex number can be written as a C-multiple of 1, so {1} is a spanning set for the space.

Consider C^{2} over R. Then {(1,0), (0,1), (i,0), (0,i)} is the standard basis, and the dimension is four. Every ordered pair of complex numbers can be written as a linear combination of these four elements, (a + bi, c + di) = a(1,0) + c(0,1) + b(i,0) + d(0,i), and so we have a spanning set. It should be clear by inspection that the set is linearly independent with respect to the scalars R.

Consider C^{2} over C. Then {(1,0), (0,1)} is the standard basis, and the dimension is two. Any pair (a + bi, c + di) = (a + bi)(1,0) + (c + di)(0,1) and we have spanning. For independence assume (a + bi)(1,0) + (c + di)(0,1) = (0,0). Then a + bi = 0 and c + di = 0, which can only happen when a = b = c = d = 0; hence, the set is linearly independent as well. Thus, {(1,0), (0,1)} is a basis for the space.