Metal A has electronic configuration of 2, 8, 1 and metal B has 2, 8, 8, 2 which is more reactive. Identify these metals and give their reactions with dil HCl.

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Solution

_{According to the given electronic configuration:}

_{Metal A is Sodium (Na) - 2,8,1}

Metal B is Calcium (Ca) - 2,8,8,2.

Metal A (Sodium) is more reactive than metal B (Calcium).

Reaction with dilute HCl:

Sodium metal reacts vigorously with dilute hydrochloric acid to form sodium chlorides and hydrogen.

2Na_(a) + 2HCl_(aq) → 2NaCl_(aq) + H₂_(g)

Calcium reacts less vigorously to form calcium chloride and hydrogen.

Ca + 2HCl → CaCl₂ + H₂

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A

B

A

Metal A has an electronic configuration of 2,8,1 and metal B has 2,8,8,2 which is a more reactive metal.

(a) Identify A and B and give their reactions with dil.HCl and dil. $H_{2} S O_{4}$ .

(b) Give the effect of heat on their:

(i) oxides, (ii) hydroxides,

(iii) carbonates, (iv) nitrates.

A

2, 8, 1

2, 8, 2

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