Advertisements
Advertisements
Question
Aluminium dissolves in mineral acids and aqueous alkalies and thus shows amphoteric character. A piece of aluminium foil is treated with dilute hydrochloric acid or dilute sodium hydroxide solution in a test tube and on bringing a burning matchstick near the mouth of the test tube, a pop sound indicates the evolution of hydrogen gas. The same activity when performed with concentrated nitric acid, reaction doesn’t proceed. Explain the reason.
Advertisements
Solution
Al being amphoteric dissolves both in acids and alkalies evolving H2 gas which bums with a pop sound.
\[\ce{2Al + 6HCl -> 2AlCl3 + 3H2}\]
\[\ce{2Al + 2NaOH + 2H2O -> \underset{Sod meta-aluminate}{2NaAlO2} + 3H2}\]
With conic. HNO3, Al becomes passive and the reaction does not proceed. This passivity is due to the formation of a thin protective layer of its oxide (Al2O3) on the surface of the metal which prevents further action.
\[\ce{2Al + 6HNO3 -> Al2O3 + 6NO2 + 3H2O}\]
APPEARS IN
RELATED QUESTIONS
If B–Cl bond has a dipole moment, explain why BCl3 molecule has zero dipole moment.
How would you explain the lower atomic radius of Ga as compared to Al?
Write a balanced equation for B2H6 + NH3 → ?
The exhibition of highest co-ordination number depends on the availability of vacant orbitals in the central atom. Which of the following elements is not likely to act as central atom in \[\ce{MF^{3-}6}\]?
A compound X, of boron reacts with NH3 on heating to give another compound Y which is called inorganic benzene. The compound X can be prepared by treating BF3 with Lithium aluminium hydride. The compounds X and Y are represented by the formulas.
Which of the following statements are correct. Answer on the basis of Figure.

(i) The two birdged hydrogen atoms and the two boron atoms lie in one plane;
(ii) Out of six B – H bonds two bonds can be described in terms of 3 centre 2-electron bonds.
(iii) Out of six B – H bonds four B – H bonds can be described in terms of 3 centre 2 electron bonds;
(iv) The four-terminal B – H bonds are two centre-two electron regular bonds.
Explain why the following compounds behave as Lewis acids?
BCl3
Explain the following:
PbX2 is more stable than PbX4.
Explain the following:
Electron gain enthalpy of chlorine is more negative as compared to fluorine.
Identify the compounds A, X and Z in the following reactions:
\[\ce{A + 2HCl + 5H2O -> 2NaCl + X}\]
Identify the compounds A, X and Z in the following reactions:
\[\ce{X ->[Δ][370 K] HBO2 ->[Δ][> 370 K] Z}\]
Match the species given in Column I with properties given in Column II.
| Column I | Column II |
| (i) Diborane | (a) Used as a flux for soldering metals |
| (ii) Galluim | (b) Crystalline form of silica |
| (iii) Borax | (c) Banana bonds |
| (iv) Aluminosilicate | (d) Low melting, high boiling, useful for measuring high temperatures |
| (v) Quartz | (e) Used as catalyst in petrochemical industries |
Account for the following observations:
PbO2 is a stronger oxidising agent than SnO2
BCl3 exists as monomer whereas AlCl3 is dimerised through halogen bridging. Give reason. Explain the structure of the dimer of AlCl3 also.
Boron fluoride exists as BF3 but boron hydride doesn’t exist as BH3. Give reason. In which form does it exist? Explain its structure.
Boron compounds behave as Lewis acids because of their ______.
Which one of the following is the correct statement?
