
Be and Al exhibit a diagonal relationship. Which of the following statements about them is/are not true?
(i) Both react with HCl to liberate ${{H}_{2}}$.
(ii) They are made passive by $HN{{O}_{3}}$.
(iii) Their carbides give acetylene on treatment with water.
(iv) Their oxides are amphoteric.
(A) (iii) and (iv)
(B) (i) and (iii)
(C) (i) only
(D) (iii) only
Answer
511.8k+ views
Hint: Both Be and Al react with HCl to give their hydroxides as one of the products. As they show diagonal relationship, their oxides show the same nature. Here, metal becomes passive means it becomes unreactive.
Complete step by step answer:
Let’s check all the given options whether they are true or not.
i) Be is an s-block element and as it is a metal it will react with HCl to produce hydrogen gas. The reaction can be given as below.
\[Be+2HCl\to BeC{{l}_{2}}+{{H}_{2}}\uparrow\]
- Aluminum even being a p-block element, shows properties of metals and also liberates hydrogen gas on treatment with HCl. The reaction is,
\[2Al+6HCl\to 2AlC{{l}_{3}}+3{{H}_{2}}\]
Thus, sentence (i) is true.
ii) When Be and Al are dipped in $HN{{O}_{3}}$, the upper surface of the metal reacts with nitric acid and the upper layers get converted into their oxides. So, in beryllium, BeO gets formed and in Aluminum, $A{{l}_{2}}{{O}_{3}}$ gets formed on the surface. So, this layer then protects the inner layer of the metal to get further oxidized. Thus, it is true that Be and Al turn passive if we add them into nitric acid.
iii) Carbides of both beryllium and Aluminum give methane gas and corresponding hydroxides upon its reaction with water. The reactions can be given as,
\[A{{l}_{4}}{{C}_{3}}+12{{H}_{2}}O\to 3C{{H}_{4}}+4Al{{(OH)}_{3}}\]
\[B{{e}_{2}}C+4{{H}_{2}}O\to C{{H}_{4}}+2Be{{(OH)}_{2}}\]
Thus, we can say that acetylene does not get formed upon reaction of Al or Be with water. Thus, sentence (iii) is not true.
iv) BeO being a metal, its oxide is not basic but it is amphoteric means it can act as both acid a
So, the correct answer is “Option D”.
Note: Remember that Be is a metal but its oxide BeO is not basic as other oxides of metal, actually it is amphoteric in nature. Aluminum is a metal which is situated in the p-block of the modern periodic table and it exhibits most of the properties of metal.
Complete step by step answer:
Let’s check all the given options whether they are true or not.
i) Be is an s-block element and as it is a metal it will react with HCl to produce hydrogen gas. The reaction can be given as below.
\[Be+2HCl\to BeC{{l}_{2}}+{{H}_{2}}\uparrow\]
- Aluminum even being a p-block element, shows properties of metals and also liberates hydrogen gas on treatment with HCl. The reaction is,
\[2Al+6HCl\to 2AlC{{l}_{3}}+3{{H}_{2}}\]
Thus, sentence (i) is true.
ii) When Be and Al are dipped in $HN{{O}_{3}}$, the upper surface of the metal reacts with nitric acid and the upper layers get converted into their oxides. So, in beryllium, BeO gets formed and in Aluminum, $A{{l}_{2}}{{O}_{3}}$ gets formed on the surface. So, this layer then protects the inner layer of the metal to get further oxidized. Thus, it is true that Be and Al turn passive if we add them into nitric acid.
iii) Carbides of both beryllium and Aluminum give methane gas and corresponding hydroxides upon its reaction with water. The reactions can be given as,
\[A{{l}_{4}}{{C}_{3}}+12{{H}_{2}}O\to 3C{{H}_{4}}+4Al{{(OH)}_{3}}\]
\[B{{e}_{2}}C+4{{H}_{2}}O\to C{{H}_{4}}+2Be{{(OH)}_{2}}\]
Thus, we can say that acetylene does not get formed upon reaction of Al or Be with water. Thus, sentence (iii) is not true.
iv) BeO being a metal, its oxide is not basic but it is amphoteric means it can act as both acid a
So, the correct answer is “Option D”.
Note: Remember that Be is a metal but its oxide BeO is not basic as other oxides of metal, actually it is amphoteric in nature. Aluminum is a metal which is situated in the p-block of the modern periodic table and it exhibits most of the properties of metal.
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