A strong oxidizing agent among the following is -

1.  ${\mathrm{Tl}}^{3+}$

2.  ${\mathrm{Ga}}^{3+}$

3.  ${\ln }^{3+}$

4.  ${\mathrm{Al}}^{3+}$

From the following, the compound(s) that react(s) with \(\mathrm{BF}_3\) is/are -

${\mathrm{Na}}_2{\mathrm{B}}_4{\mathrm{O}}_7\cdot 10{\mathrm{H}}_2\mathrm{O}<mspace\; width="1em"></mspace>\underset{∆}{\overset{{740}^{\circ }\mathrm{C}}{\to }}<mspace\; width="1em"></mspace>2{\mathrm{NaBO}}_2+\left(\mathrm{X}\right)+10{\mathrm{H}}_2\mathrm{O}$
$\left(\mathrm{X}\right)+{\mathrm{CaF}}_2+{\mathrm{H}}_2{\mathrm{SO}}_4⟶\mathrm{Y}+{\mathrm{CaSO}}_4+{\mathrm{H}}_2\mathrm{O}$
$\left(\mathrm{Y}\right)+{\mathrm{NaBH}}_4⟶\left(\mathrm{Z}\right)+{\mathrm{NaBF}}_4$

Select the incorrect statement regarding compounds (X), (Y) and (Z).

1.  Compound (X) is ${\mathrm{B}}_2{\mathrm{O}}_3$ and compound (Z) contains two 3C - 2e bonds

2.  Compound (X) is ${\mathrm{B}}_2{\mathrm{O}}_3$ and compound (Y) contains one 3C - 2e bonds 

3.  Compound (Y) is Lewis acid and stable due to $p\pi -p\pi$ back bonding

4.  Compound (Z) reacts with excess of ${\mathrm{NH}}_3$ at high temperature gives inorganic graphite 

Hydrolysis of ${\mathrm{BCl}}_3$ gives X which on treatment with sodium carbonate produces Y. X and Y, respectively, are 

1.  ${\mathrm{H}}_3{\mathrm{BO}}_3<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{NaBO}}_2$

2.  ${\mathrm{H}}_3{\mathrm{BO}}_3<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{Na}}_2{B}_4{O}_7$

3.  ${\mathrm{B}}_2{\mathrm{O}}_3<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{NaBO}}_2$

4.  $B_2{O}_3<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{Na}}_2{B}_4{O}_7$

Ionization enthalpy (${∆}_{\mathrm{i}}{\mathrm{H}}_1\mathrm{kJ}<mspace\; width="1em"></mspace>{\mathrm{mol}}^{-1}$) for the elements of Group 13 follows the order.

A compound X, of boron, reacts with ${\mathrm{NH}}_3$ on heating to give another compound Y which is called inorganic benzene. The compound X can be prepared by treating ${\mathrm{BF}}_3$ with Lithium aluminium hydride. The compounds X and Y are represented by the formulas -

Which does not show inert pair effect

(1) Al

(2) Sn

(3) Pb

(4) Thallium

${\mathrm{BCl}}_3$ 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 $\mathrm{p\pi }-\mathrm{d\pi }$ 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

Carbon dioxide is a gas but silica is a solid because:

1. Carbon dioxide is composed of discrete covalent ${\mathrm{CO}}_2$ molecules whereas silica has a continuous tetrahedral structure.

2. ${\mathrm{CO}}_2$ molecules are lighter than ${\mathrm{SiO}}_2$ molecules.

3. ${\mathrm{CO}}_2$ is more acidic than ${\mathrm{SiO}}_2$.

4. The melting point of silica is very high.

${\mathrm{Na}}_2{\mathrm{B}}_4{\mathrm{O}}_7\cdot 10{\mathrm{H}}_2\mathrm{O}<mspace\; width="1em"></mspace>\underset{∆}{\overset{{740}^{\circ }\mathrm{C}}{\to }}<mspace\; width="1em"></mspace>2{\mathrm{NaBO}}_2+\left(\mathrm{X}\right)+10{\mathrm{H}}_2\mathrm{O}$
$\left(\mathrm{X}\right)+{\mathrm{CaF}}_2+{\mathrm{H}}_2{\mathrm{SO}}_4⟶\mathrm{Y}+{\mathrm{CaSO}}_4+{\mathrm{H}}_2\mathrm{O}$
$\left(\mathrm{Y}\right)+{\mathrm{NaBH}}_4⟶\left(\mathrm{Z}\right)+{\mathrm{NaBF}}_4$

Select the incorrect statement regarding compounds (X), (Y) and (Z).

1.  Compound (X) is ${\mathrm{B}}_2{\mathrm{O}}_3$ and compound (Z) contains two 3C - 2e bonds

2.  Compound (X) is ${\mathrm{B}}_2{\mathrm{O}}_3$ and compound (Y) contains one 3C - 2e bonds 

3.  Compound (Y) is Lewis acid and stable due to $p\pi -p\pi$ back bonding

4.  Compound (Z) reacts with excess of ${\mathrm{NH}}_3$ at high temperature gives inorganic graphite