Understand the reaction

The given reaction is: $2X + B_2H_6 \rightarrow [BH_2(X)_2]^+[BH_4]^-$

This is the formation of an amine–borane derived ionic compound from diborane.

Key idea about diborane reaction with amines

Diborane reacts with amines having at least one replaceable N–H hydrogen to form products of the type: $[BH_2(RNH_2)_2]^+[BH_4]^-,$ $[BH_2(R_2NH)_2]^+[BH_4]^-$ etc., after interaction with the Lewis basic nitrogen atom.

For the cation $[BH_2(X)_2]^+$ to form as written, the amine must be able to coordinate to boron and also participate in the proton/hydride transfer process that generates the ionic pair. This requires an amine with an N–H bond.

Check each option

Option A: $NH_3$

Ammonia has N–H bonds and a lone pair on nitrogen. It reacts with diborane to give the corresponding ionic adduct.

Therefore, A is correct.

Option B: $CH_3NH_2$

Methylamine is a primary amine. It has:

• a lone pair on N
• N–H bonds

So it can form the required ionic product.

Therefore, B is correct.

Option C: $(CH_3)_2NH$

Dimethylamine is a secondary amine. It has:

• a lone pair on N
• one N–H bond

Hence it can also give the ionic product.

Therefore, C is correct.

Option D: $(CH_3)_3N$

Trimethylamine is a tertiary amine. It has:

• a lone pair on N
• no N–H bond

Since it lacks N–H hydrogen, it cannot form the ionic species of the given type. Instead, tertiary amines generally form only neutral Lewis acid–base adducts with borane/diborane.

Therefore, D is not correct.

Final answer

The amines that can undergo the given reaction are: $NH_3,\ CH_3NH_2,\ (CH_3)_2NH$

So the correct options are: $\boxed{A,\ B,\ C}$

Comparison with stored correct answer

Stored correct answer: A, B, C

This matches the derived answer exactly.