When metal X is treated with sodium hydroxide, a white precipitate A is obtained which is soluble in excess NaOH to give a soluble complex B. Compound A is soluble in dilute HCl to form compound C. When the compound A is heated strongly it gives compound D which is used to extract metal. a) Identify X, A, B, C, D supporting your answer(s) with appropriate chemical reactions. b) At which group and period does X fall?​

See explanation

Explanation:

If we look at the question closely, we will notice that the metal in question must be aluminum.

When aluminum is treated with sodium hydroxide, a precipitate, aluminium hydroxide is formed as follows;

Al(s) + 3NaOH(aq) ---> Al(OH)3(s) + 3Na(s)

In excess sodium hydroxide, the precipitate dissolves as follows;

Al(OH)3(s) + NaOH(aq) > [NaAlOH4]^-(aq)

The complex formed is sodium aluminum tetrahydroxo aluminate III.

The reaction of aluminum faith dilute hydrochloric acid occurs as follows to yield aluminum chloride;

2Al(s) + 6HCl(aq) > 2AlCl3(aq) + 3H2(g)

When aluminum metal is heated strongly, it yields aluminum oxide;

2Al(s) + 3O2(g) ---> Al2O3(s)

answer; economic order quantity;

c) should be the answer.