The FFT output coefficients (for complex input of size N) are from 0 to N - 1 grouped as [LOW,MID,HI,HI,MID,LOW] frequency.
I would consider that the element at k has the same frequency as the element at N-k since for real data, FFT[N-k] = complex conjugate of FFT[k].
The order of scanning from LOW to HIGH frequency is
0,
1,
N-1,
2,
N-2
...
[N/2] - 1,
N - ([N/2] - 1) = [N/2]+1,
[N/2]
There are [N/2]+1 groups of frequency from index i = 0 to [N/2], each having the frequency = i * SamplingFrequency / N
So the frequency at bin FFT[k] is:
if k <= [N/2] then k * SamplingFrequency / N
if k >= [N/2] then (N-k) * SamplingFrequency / N