Four elements coded A, B, C and D form a series of substances e.g. AB, $$\text{B}_2$$, $$\text{CB}_3$$, $$\text{DB}_2$$ and $$\text{DB}_3$$. If the atomic number of these elements are not necessarily in the order are 13, 19, 26 and 35. Write down extra nuclear electronic structures of these elements. From this information and the formula of the compounds, allocate A, B, C and D with appropriate atomic numbers. Mention the nature of bonding in each of them.

PLACEHOLDER ANSWER - the answer is a set of allocations. The configurations are 13 (Al) 2, 8, 3; 19 (K) 2, 8, 8, 1; 26 (Fe) 2, 8, 14, 2; and 35 (Br) 2, 8, 18, 7. B forms the diatomic $$\text{B}_2$$, so B is Br (35); then AB is KBr (A = 19), $$\text{CB}_3$$ is $$\text{AlBr}_3$$ (C = 13), and $$\text{DB}_2$$, $$\text{DB}_3$$ are $$\text{FeBr}_2$$ and $$\text{FeBr}_3$$ (D = 26, showing valencies 2 and 3). KBr is ionic, $$\text{Br}_2$$ is non-polar covalent, $$\text{AlBr}_3$$ is largely covalent, $$\text{FeBr}_2$$ is mainly ionic and $$\text{FeBr}_3$$ has more covalent character.