Complete the following chemical equations : Z + 3 LiAlH 4 → X + 3 LiF + 3 AlF 3 X + 6 H 2 O → Y + 6 H 2 X + 3 O 2 → ∆ + B 2 O 3 + 3 H 2 O

Diborane is prepared by treating boron triflouride with lithium aluminium hydride in diethyl ether. Boranes are readily hydrolysed by water to give boric acid. Diboranes catches fire spontaneously upon exposure to air. 4 BF 3 ( Z ) + 3 LiAIH 4 → 2 B 2 H 6 ( X ) + 3 LiF + 3 AlF 3 B 2 H 6 g ( X ) + 6 H 2 O g → 2 B OH 3 aq ( Y ) + 6 H 2 g B 2 H 6 ( X ) + 3 O 2 → ∆ B 2 O 3 + 3 H 2 O Z = BF 3 , X = B 2 H 6 and Y = B OH 3 or H 3 BO 3

Match the species given in Column I with the properties mentioned in Column II. Column I Column II (i) BF 4 - (a) Oxidation state of the central atom is +4 (ii) AlCl 3 (b) Strong oxidising agent (iii) SnO (c) Lewis acid (iv) PbO 2 (d) Can be further oxidised (e) Tetrahedral shape

There are four atoms around the central boron atom, the fifth position will be occupied by lone pair of electrons. Hence, the molecular geometry will be tetrahedral. Hence, the hybridization and shape of BF 4 - is sp 3 and tetrahedral respectively. A Lewis acid is a chemical species that contains an empty orbital which is capable of accepting an electron pair from a Lewis base to form a Lewis adduct. Octet of Al is not complete, acts as Lewis acid. SnO is composed of tin and oxygen where tin has the oxidation state of + 2 . It is a reducing agent. The stable oxidation state of Pb is + 2 , due to inert pair effect. As Oxidising agent oxidises other species and itself undergoes reduction therefore, Pb in PbO 2 (i.e. + 4 oxidation state ) acts as a strong oxidising agent. Column I Column II (i) BF 4 - (e)Tetrahedral shape (ii) AlCl 3 (c) Lewis acid (iii) SnO (d) Can be further oxidised (iv) PbO 2 (a) Oxidation state of central atom is + 4 (b) Strong oxidising agent

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) Gallium (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 Note: The spelling of GALLIUM is wrong in text book.

A 4 -centre 2 -electron ( 4 c – 2 e ) bond is a type of chemical bond in which four atoms share two electrons in bonding, with a net bond order of 1 / 2 . This type of bonding differs from the usual covalent bond, which involves two atoms sharing two electrons ( 2 c – 2 e bonding). Gallium has a unique property of melting below room temperature. Borax is used as a flux for soldering metals. For welding of iron and steel, a mixture of borax and ammonium chloride is used as a flux. The mixture lowers the melting point of the unwanted iron oxide (scale), allowing it to run off. Aluminosilicate is used as catalyst in petrochemical industries. It is used in process like cracking and hydrocracking. Quartz is a crystalline form of silica. It is a firm mineral which is colourless. Column I Column II (i) Diborane (c) Banana bonds (ii) Galluim (d) Low melting, high boiling, useful for measuring high temperatures (iii) Borax (a) Used as a flux for soldering metals (iv) Aluminosilicate (e) Used as catalyst in petrochemical industries (v) Quartz (b) Crystalline form of silica

Match the species given in Column I with the hybridization given in Column II. Column I Column II (i) Boron in [ B ( OH ) 4 ] - (a) sp 2 (ii) Aluminium in [ Al ( H 2 O ) 6 ] 3 + (b) sp 3 (iii) Boron in B 2 H 6 (c) sp 3 d 2 (iv) Carbon in Buckminsterfullerene (v) Silicon in SiO 4 4 - (vi) Germanium in [ GeCl 6 ] 2 -

(i) B ( OH ) 3 due to its incomplete octet accepts electron pair (as OH - ) to give [ B ( OH ) 4 ] - . Boron in this ion involves one 2 s orbital and three 2 p orbitals. Thus, hybridisation of B in [ B ( OH ) 4 ] - is sp 3 . Al is in + 3 oxidation state. The orbitals of 3 s , 3 p and 3 d are hybridized (re-organised into orbitals of equal energy) and used to accept lone pairs from 6 water molecules. Boron has three valence electrons, so it is supposed to make three bonds in a molecule with hybridisation, sp 2 as only s and two p orbitals are used in hybridisation and last orbital is vacant. But in diborane, B 2 H 6 contains two electrons each, three centered bonds (banana bond). Each boron atom is in a link with four hydrogen atoms. This makes the tetrahedral geometry. Hence, each boron atom is sp 3 hybridised. According to VSEPR theory, each carbon atom forms three sigma bonds with its sp 2 hybrid orbitals and one pi-bond with the remaining p-orbital. Each carbon atom is a part of one pentagon and two hexagons and has two single bonds and one double bond. The silicate units have a tetrahedral structure formed as a result of sp 3 hybridisation. The silicon atom has its complete octet but each oxygen atom is still short of one electron to complete its octet. They can complete their octet by taking up 4 electrons from a metal, getting converted to an anion SiO 4 4 - . The electronic configuration of Ge is as follows- Ge ( ground state ) = [ Ar ] 3 d 10 4 s 2 4 p 2 4 d 0 Ge ( + 4 ) = [ Ar ] 3 d 10 4 s 0 4 p 0 4 d 0 Now, 6 Cl atoms donate 6 lone pairs of electrons respectively in one 4 s , three 4 p and two 4 d orbitals forming sp 3 d 2 hybrid orbitals. Hence, hybridisation of Ge in [ GeCl 6 ] 2 - is sp 3 d 2 . Column I Column II (i) Boron in [ B ( OH ) 4 ] - (b) sp 3 (ii) Aluminium in [ Al ( H 2 O ) 6 ] 3 + (c) sp 3 d 2 (iii) Boron in B 2 H 6 (b) sp 3 (iv) Carbon in Buckminsterfullerene (a) sp 2 (v) Silicon in SiO 4 4 - (b) sp 3 (vi) Germanium in [ GeCl 6 ] 2 - (c) sp 3 d 2

HARD

11th CBSE

IMPORTANT

Earn 100

Explain the following :

Why does the element silicon, not form a graphite like structure whereas carbon does.

Important Questions on The p-Block Elements

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 11 - The p-Block Elements>Short Answer Type>Q 34

Identify the compounds $A, X$ and $Z$ in the following reactions :

(i) $A + 2 HCl + 5 H_{2} O \to 2 NaCl + X$

$X \to_{370 K}^{∆} {HBO}_{2} \to_{> 370 K}^{∆} Z$

MEDIUM

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 11 - The p-Block Elements>Short Answer Type>Q 35

Complete the following chemical equations :

$Z + 3 {LiAlH}_{4} \to X + 3 LiF + 3 {AlF}_{3} X + 6 H_{2} O \to Y + 6 H_{2} X + 3 O_{2} \overset{∆}{\to} + B_{2} O_{3} + 3 H_{2} O$

EASY

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 11 - The p-Block Elements>Matching Type>Q 36

Match the species given in Column I with the properties mentioned in Column II.

Column I		Column II
(i)	${BF}_{4}^{-}$	(a)	Oxidation state of the central atom is +4
(ii)	${AlCl}_{3}$	(b)	Strong oxidising agent
(iii)	$SnO$	(c)	Lewis acid
(iv)	${PbO}_{2}$	(d)	Can be further oxidised
		(e)	Tetrahedral shape

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 11 - The p-Block Elements>Matching Type>Q 37

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)	Gallium	(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

Note: The spelling of GALLIUM is wrong in text book.

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 11 - The p-Block Elements>Matching Type>Q 38

Match the species given in Column I with the hybridization given in Column II.

Column I		Column II
(i)	Boron in $[B (OH)_{4}]^{-}$	(a)	${sp}^{2}$
(ii)	Aluminium in $[Al (H_{2} O)_{6}]^{3 +}$	(b)	${sp}^{3}$
(iii)	Boron in $B_{2} H_{6}$	(c)	${sp}^{3} d^{2}$
(iv)	Carbon in Buckminsterfullerene
(v)	Silicon in ${SiO}_{4}^{4 -}$
(vi)	Germanium in $[{GeCl}_{6}]^{2 -}$

MEDIUM

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 11 - The p-Block Elements>Long Answer Type>Q 41

Describe the general trends in the following property of the elements in Groups $13$ and $14$ .

Atomic size

MEDIUM

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 11 - The p-Block Elements>Long Answer Type>Q 41

Describe the general trends in the ionisation enthalpy of the elements in Groups $13$ and $14$ .

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 11 - The p-Block Elements>Long Answer Type>Q 41

Describe the general trends in the following property of the elements in Groups $13$ and $14$ .

Metallic character

Explain the following : Why does the element silicon, not form a graphite like structure whereas carbon does.

Important Questions on The p-Block Elements

Learn and Prepare for any exam you want !

Explain the following :

Why does the element silicon, not form a graphite like structure whereas carbon does.