Study the sequence of reactions of diborane $$\text{B}_2\text{H}_6$$ given below and identify the products K, L, M and N. The shown reagents are taken in excess.

$$\text{K} = \text{BNH}_6$$, $$\text{L} = (\text{BN})_x$$, $$\text{M} = \text{B}_2\text{H}_5\text{Cl}$$, $$\text{N} = \text{B}_3\text{N}_3\text{H}_3\text{Cl}_3$$

$$\text{K} = [\text{B}_2\text{H}_6 \cdot 2\text{NH}_3]$$, $$\text{L} = \text{B}_3\text{N}_3\text{H}_6$$, $$\text{M} = \text{B}_2\text{H}_5\text{Cl}$$, $$\text{N} = \text{B}_3\text{N}_3\text{Cl}_6$$

$$\text{K} = \text{BNH}_6$$, $$\text{L} = (\text{BN})_x$$, $$\text{M} = \text{BCl}_3$$, $$\text{N} = \text{B}_3\text{N}_3\text{Cl}_6$$

$$\text{K} = [\text{B}_2\text{H}_6 \cdot 2\text{NH}_3]$$, $$\text{L} = \text{B}_3\text{N}_3\text{H}_6$$, $$\text{M} = \text{BCl}_3$$, $$\text{N} = \text{B}_3\text{N}_3\text{H}_3\text{Cl}_3$$