Based on the English alphabetical order, three of the following four letter-clusters are alike in a certain way and thus form a group. Which letter-cluster does not belong to that group?
(Note: The odd letter-cluster is not based on the number of consonants/vowels or their position in the letter-cluster.)

The correct answer is KON.

To determine which letter-cluster does not belong to the group, let us analyze the alphabetical positions of each letter in the given options.

Let us write down the numerical positions of the letters in the English alphabet:

A = 1, B = 2, C = 3, D = 4, E = 5, F = 6, G = 7, H = 8, I = 9, J = 10, K = 11, L = 12, M = 13, N = 14, O = 15, P = 16, Q = 17, R = 18, S = 19, T = 20, U = 21, V = 22, W = 23, X = 24, Y = 25, Z = 26.

Now, let us analyze the pattern for each letter-cluster:

1. CGE

Position of C = 3
Position of G = 7
Position of E = 5

Pattern:
First to Second letter: $3+4=7$ (G)
Second to Third letter: $7-2=5$ (E)

2. MQO

Position of M = 13
Position of Q = 17
Position of O = 15

Pattern:
First to Second letter: $13+4=17$ (Q)
Second to Third letter: $17-2=15$ (O)

3. GKI

Position of G = 7
Position of K = 11
Position of I = 9

Pattern:
First to Second letter: $7+4=11$ (K)
Second to Third letter: $11-2=9$ (I)

4. KON

Position of K = 11
Position of O = 15
Position of N = 14

Pattern:
First to Second letter: $11+4=15$ (O)
Second to Third letter: $15-1=14$ (N)

Conclusion:
All letter-clusters except KON follow the pattern of adding 4 to the first letter's position to get the second letter, and then subtracting 2 from the second letter's position to get the third letter ($+4$, $-2$). KON follows a pattern of ($+4$, $-1$), which makes it the odd one out.