Two O-atoms of each tetrahedron are shared

  General formula is ${\left({\mathrm{Si}}_2{\mathrm{O}}_5\right)}_{\mathrm{n}}^{2\mathrm{n}-}$

  Co-ordination number of Si is 4

  Number of O-atoms per tetrahedra unit is $\frac{5}{2}$

  ${\mathrm{Ca}}_2{\mathrm{B}}_6{\mathrm{O}}_{11}5{\mathrm{H}}_2\mathrm{O}$

  ${\mathrm{CaB}}_4{\mathrm{O}}_7.{\mathrm{NaBO}}_{2}8{\mathrm{H}}_2\mathrm{O}$

  $2{\mathrm{Mg}}_3{\mathrm{B}}_8{\mathrm{O}}_{15}{\mathrm{MgCl}}_2$

  ${\mathrm{Ca}}_2{\mathrm{B}}_6{\mathrm{O}}_{11}.3{\mathrm{H}}_2\mathrm{O}$

  6

  4

  8

  7

  $\mathrm{CO}, {\mathrm{CO}}_2, {\mathrm{COCl}}_2 \mathrm{and} {\mathrm{HCONH}}_2$

  $\mathrm{CO}, {\mathrm{CO}}_2, H_{2}O, {\mathrm{COCl}}_2$

  ${\mathrm{CO}}_2, \mathrm{CO}, {\mathrm{H}}_2\mathrm{O}, {\mathrm{HCONH}}_2$

  $\mathrm{CO}, {\mathrm{CO}}_2, {\mathrm{H}}_2\mathrm{O}, {\mathrm{NH}}_2{\mathrm{CONH}}_2$

  ${\mathrm{CH}}_4$

  ${\mathrm{SiH}}_4$

  ${\mathrm{GeH}}_4$

  ${\mathrm{SnH}}_4$

  Each B and N atom is $s{p}^2$ hybridized

  Borazine satisfies the (4n + 2)$\pi$ electrons Huckel's rule

  Like organic benzene, borazine does not give addition product with HCI

  Borazine contains dative p$\pi$ - p$\pi$ bond

  It can not be titrated satisfactorily with NaOH, as a sharp endpoint is not obtain

  The acidity of its aqueous solution increases upon addition of ethylene glycol

  It is a weak electrolyte in water

  It has a three-dimensional structure due to hydrogen bonding

3

6

1

4

  ${\mathrm{PbO}}_2$

  ${\mathrm{SnO}}_2$

  ${\mathrm{Pb}}_2{O}_3$

  ${\mathrm{Pb}}_3{O}_4$

  The compound (A) is ${\left({\mathrm{CH}}_3\right)}_2{\mathrm{SiCl}}_2$

  The compound D is SiC

  The compound (B) is ${\left({\mathrm{CH}}_3\right)}_2\mathrm{Si}{\left(\mathrm{OH}\right)}_2$

  All of the above

  20

  25

  30

  36

  ${\mathrm{B}}_{\mathrm{n}}{\mathrm{H}}_{\mathrm{n}+6}$ are known as arachnoboranes

  Indium hydride is a polymeric solid

  ${\mathrm{AlH}}_3$ is a monomeric planar molecule

  Nido-boranes have the general formula ${\mathrm{B}}_{\mathrm{n}}{\mathrm{H}}_{\mathrm{n}+6}$

  ${\mathrm{Pbl}}_2$

  ${\mathrm{Pbl}}_4$

  $\mathrm{Pb}{\left({\mathrm{CH}}_3\mathrm{COO}\right)}_4$

  ${\mathrm{PbCl}}_2$

  ${\mathrm{Tl}}^{3+}$

  ${\mathrm{Ga}}^{3+}$

  ${\ln }^{3+}$

  ${\mathrm{Al}}^{3+}$

  Ethers

  ${\mathrm{H}}_2\mathrm{O}$

  ${\mathrm{NH}}_3$

  All of the above

  All the three B-F bond lengths are equal and each of them is shorter than the sum of the covalent radii of boron and fluorine

  The bond energy of the B-F bond is very high, higher than for any other single bond

  The unusual shortness and strength of the B-F bond may be explained by $p\pi -p\pi$ interaction between boron and fluorine atoms

  The unusual shortness and strength of the bonds may be explained by a $p\pi -d\pi$ interaction between boron and fluorine atoms

  ${\mathrm{CH}}_3{\mathrm{SiCl}}_3 \mathrm{and} \mathrm{Si}{\left({\mathrm{CH}}_3\right)}_4$

  ${\left({\mathrm{CH}}_3\right)}_2{\mathrm{SiCl}}_2 \mathrm{and} {\left({\mathrm{CH}}_3\right)}_3\mathrm{SiCl}$

  ${\left({\mathrm{CH}}_3\right)}_2{\mathrm{SiCl}}_2 \mathrm{and} {\mathrm{CH}}_3{\mathrm{SiCl}}_3$

  ${\mathrm{SiC}}_4 \mathrm{and} {\left({\mathrm{CH}}_3\right)}_3\mathrm{SiCl}$

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

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

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

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

  Fe

  Li

  K

  Al

  $\mathrm{CO}, {\mathrm{CO}}_2, {\mathrm{H}}_2\mathrm{O}$

  $\mathrm{CO}, {\mathrm{SO}}_2, {\mathrm{H}}_2\mathrm{O}$

  ${\mathrm{H}}_2\mathrm{S}, \mathrm{CO}, {\mathrm{H}}_2\mathrm{O}$

  $\mathrm{HCOOH}, {\mathrm{H}}_2\mathrm{S}, \mathrm{CO}$

  Only water

  Only acids

  Only alkalis

  Both acids and alkalis

  ${\mathrm{B}}_2{\mathrm{H}}_6 \mathrm{and} \mathrm{Nal} \mathrm{only}$

  ${\mathrm{B}}_2{\mathrm{H}}_6, {\mathrm{H}}_2 \mathrm{and} \mathrm{Nal}$

  ${\mathrm{Bl}}_3 \mathrm{and} \mathrm{NaH}$

  ${\mathrm{NaBl}}_4 \mathrm{and} \mathrm{Hl}$

  4 and 5

  4 and 3

  5 and 4

  5 and 3

  ${\mathrm{H}}_3{\mathrm{BO}}_3 \mathrm{and} {\mathrm{NaBO}}_2$

  ${\mathrm{H}}_3{\mathrm{BO}}_3 \mathrm{and} {\mathrm{Na}}_2{B}_4{O}_7$

  ${\mathrm{B}}_2{\mathrm{O}}_3 \mathrm{and} {\mathrm{NaBO}}_2$

  $B_2{O}_3 \mathrm{and} {\mathrm{Na}}_2{B}_4{O}_7$

  N

  P

  Al

  Na

  ${\mathrm{BBr}}_3$ < ${\mathrm{BCl}}_3$ < ${\mathrm{BF}}_3$

  ${\mathrm{BCl}}_3$ < ${\mathrm{BF}}_3$ < ${\mathrm{BBr}}_3$

  ${\mathrm{BF}}_3$ < ${\mathrm{BCl}}_3$ < ${\mathrm{BBr}}_3$

  ${\mathrm{BBr}}_3$ < ${\mathrm{BF}}_3$ < ${\mathrm{BCl}}_3$

  N is more electronegative than B

  The atomic mass of N is higher than that of B

  The B-F bond gets a partial double bond character due to p-p overlap

  N has a lone pair of electrons while B does not have

Formation of Al(OH)₃

Dimerisation of AlCl₃

Hydrolysis of AlCl₃ give HCl gas

Absorption of moisture 

+4, +2

+2, +4

+4, +4

+2, +2

 ${\mathrm{NaBH}}_4 + {\mathrm{I}}_2 \to$

 ${\mathrm{BF}}_3 + \mathrm{NaH} \to$

 ${\mathrm{B}}_3{\mathrm{N}}_3{\mathrm{H}}_6 + {\mathrm{H}}_2\mathrm{O} \to$

 ${\mathrm{Mg}}_3{\mathrm{B}}_2 + \mathrm{dil} \mathrm{HCl}\to$