Question

“Be and Al are made passive by $HN{O_3}$”. Answer whether the above statement is true or false. If true enter 1, else enter 0.

Answer 1

Hint: Passivation is the process of formation of a protective layer on the surface which protects the bulk of the metal from further oxidation. Beryllium and Aluminum show resemblance as both are diagonally opposite elements in the periodic table. Hence, they show certain properties which are similar to each other. One such similarity is seen when both react with nitric acid.

Complete step by step answer:
“Be and Alare made passive by $HN{O_3}$”.
First of all it is very important to note that concentrated nitric acid $\left( {HN{O_3}} \right)$ is a very strong oxidizing agent. Nitric acid attacks metal which form its corresponding oxides.
Beryllium and Aluminum which are diagonally present in the periodic table show similar property when attacked by nitric acid.
Both Be and Al are rendered positive by reaction with nitric acid due to formation of inert, insoluble and impervious oxide layer on their surface.
$3Be + 8HN{O_3} \to 3Be{\left( {N{O_3}} \right)_2} + 2NO + 4{H_2}O$
$2Al + 6HN{O_3} \to A{l_2}{O_3} + 6N{O_2} + 3{H_2}O$
This passivation protects the bulk of the metal from further oxidation. Thus, the statement “Be and Al are made passive by $HN{O_3}$” is true.

Note:
It is very interesting to note that such a diagonal relationship occurs because crossing and descending the periodic table have opposing effects. A diagonal relationship is said to exist between certain pairs of diagonally adjacent elements in the second and third period (first 20 elements) of the periodic table. Students might also think of other metals which show similar passivation. Well, yes metals like chromium, nickel, cobalt also react with nitric acid in similar fashion.