Aluminium becomes passive in nitric acid because it-

1. Is a noble metal

2. Forms a thin film of oxide

3. Has a positive reduction potential

4.  None of the above

Hybridisation of boron in diborane is:

1.  $\mathrm{sp}$

2.  ${\mathrm{sp}}^2$

3.  ${\mathrm{sp}}^3$

4.  ${\mathrm{sp}}^3{\mathrm{d}}^2$

The blue coloured mineral 'Lapis Lazuli' used as semi precious stone is:

1.  sodium alumino silicate

2.  zinc cobaltate

3.  prussian blue

4.  basic copper carbonate

Red liquor is:

1.  ${\left({\mathrm{CH}}_3\mathrm{COO}\right)}_3 \mathrm{Al}$

2.  $\mathrm{Al}{\left(\mathrm{OH}\right)}_3$

3.  ${\mathrm{Al}}_2{\left({\mathrm{CO}}_3\right)}_3$

4.  ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$

${\mathrm{BCl}}_3$ does not exist as dimer but ${\mathrm{BH}}_3$ exist as dimer $\left({\mathrm{B}}_2{\mathrm{H}}_6\right)$ because:

1.chlorine is more electronegative than hydrogen

2.there is $\mathrm{p\pi }-\mathrm{p\pi }$ back bonding in ${\mathrm{BCl}}_3$ but ${\mathrm{BH}}_3$ does not contain such multiple bonding

3.large sized chlorine atoms do not fit in between the small boron atoms whereas small sized hydrogen atoms get fitted between boron atoms

4.none of the above

Which of the following imparts green colour to flame?

1.  $\mathrm{B}{\left(\mathrm{OMe}\right)}_3$

2.  $\mathrm{Na}\left(\mathrm{OMe}\right)$

3.  $\mathrm{Al}{\left({\mathrm{OBr}}_2\right)}_3$

4.  $\mathrm{Sn}{\left(\mathrm{OH}\right)}_2$

Borax bead test is responded by:

1.  Divalent metals

2.  Heavy metals

3.  Light metals

4.  Metal which forms coloured metaborates

Diborane does not undergo cleavage reaction with:

1.  trimethyl amine

2.  ammonia

3.  $\mathrm{CO}$

4.  ${\mathrm{CO}}_2$

Boric acid is not used:

1. As an antiseptic

2. As a flux in soldering

3. In making optical glasses

4. In making enamels and pottery glazes

Heating an aqueous solution of aluminium chloride to dryness will give:

1.  ${\mathrm{AlCl}}_3$

2.  ${\mathrm{Al}}_2{\mathrm{Cl}}_6$

3.  ${\mathrm{Al}}_2{\mathrm{O}}_3$

4.  $\mathrm{Al}\left(\mathrm{OH}\right){\mathrm{Cl}}_2$