ALLOTROPES OF CARBON:

A characteristic property of the elements of carbon family is that these show allotropy. Example: carbon has two important allotropic forms i.e.  Crystalline and amorphous

(1) DIAMOND:

(1) Each carbon is linked to another atom and there is very closed packing in structure of Diamond.

(2) Density and hardness is very much greater for diamond because of closed packing in diamond due to sp3 hybrid and are tetrahedrally arranged around it. And C-C distance is 154 pm

(3) Diamond has sharp cutting edges that's why it is employed in cutting of glass.

(4) Diamond crystals are bad conductor of electricity because of absence of mobile electron.

(5) 1 carat of diamond = 200 mg.

(6) Diamond powder if consumed is fatal and causes death in minutes.

(2) GRAPHITE: 

(1) In Graphite Carbons are sp² hybridised out of the four valence electrons, three   involved in (sp²-sigma) covalent bonds form hexagonal layers and fourth unhybridised p– electron of each carbon forms an extended delocalized p-bonding with carbon atoms of adjacent layers

 (2) Each carbon is linked with 3 carbons and one carbon will be left and form a two dimensional shed like structure.

(3) Distance between two layers is very large so no regular bond is formed between two layers. The layers are attached with weak vander waal force of attraction.

(4) The carbon have unpaired electron so graphite is a good conductor of current.

(5) The C-C bond length within a layer is 141.5 pm while the inter layer distance is 335.4 pm shorter than that of Diamond (1.54 Å).

(6) Due to wide separation and weak interlayer bonds, graphite is sift , greasy and a lubricant character and low density.

(7) Graphite marks the paper black so it is called black lead or plumbago and so it is used in pencil lead.

(8) Composition of pencil lead is graphite plus clay .the percentage of lead in pencil is zero .

 (9) Graphite has high melting point so it is employed in manufacture of crucible.

(10) Graphite when heated with oxidizing agents like alkaline KMnO4 forms mellatic acid (Benzene hexa carboxylic acid).

(11) Graphite on oxidation with HNO₃ gives acid i.e. known as Graphite acid C₁₂H₆O₁₂

(3) FULLERENCES:

(1) A fascinating discovery was the synthesis of spherical carbon-cage molecules called fullerences. The discovery of fullerene was awarded the noble prize in chemistry (1996). Fullerenes   were first prepared by evaporation of graphite using laser.

(2) Fullerences are sooty material so formed consists of C₆₀ with small amount of C₇₀ and other fullerences containing an even number of carbon up to 350

(3) Fullerences have a smooth structure and unlike diamond and graphite, dissolved in organic solvent like toluene.

(4)  C₆₀ is the most stable fullerene. It has the shape of a football and called buckminsterfullerene

(5) C₆₀ consists of fused five and six membered carbon rings

(6) Six membered rings surrounded by alternatively by hexagons and pentagons of carbon.

(7) Five membered rings are surrounded by five hexagons carbon rings.

(8) There are 12 five –membered rings

(9) There are 20 Six –membered rings

(10) In fullerenes all the carbon sp2 hybridised each carbon formed three sigma bond and the fourth electron delocalized to formed pi bond .

(11) All the carbon atom are equivalent but all C-C bond are not equivalent.

(12)  In the structure C-C bonds of two different bond length occur at the fusion of two six membered rings the bond length is  C-C = 135.5 pm and at the fusion of  five and six membered rings C-C bond length is 146.7 pm.

(13) There are both single and double bonds

(14) The smallest fullerenes are C₂₀.

(15) Thermodynamically the most stable allotrope of carbon is considered to be graphite. This is due standard enthalpy of formation of graphite is taken zero .while enthalpy of formation of diamond and fullerenes are 1.90 KJ/Mole and 38.1 KJ/Mole respectively.      

ALKALI METALS - PHYSICAL PROPERTIES:

NAMING OF 1st GROUP: The group 1^st containing Li, Na, K, Rb, Cs & Fr , Francium is radioactive and has a very short life ( t_1/2=  21 minutes ), therefore very little is known about it. They are commonly called alkali metals because form hydroxide when react with water.

RELATIVE ABUNDANCE:  Na and K are 6^th and 7^th most abundance elements in earth crust respectively and both Na, K constitutes 4% of total earth crust.

                                        Na > K > Rb > Li > Cs

STRUCTURE OF ALKALI METALS: At normal temperature all the alkali metals are adopt BBC (Body Centred cubic) type lattice with coordination number (8) but at low temperature Li adopt HCP with coordination number (12).

ELECTRONIC CONFIGURATION:

The general electronic configuration of alkali metals may be represented by [noble gas]     ns¹ where n = 2 to 7

PHYSICAL PROPERTIES:

(1) SOFTNESS OF ALKALI METALS:

Except Francium (Fr) all the alkali metals are soft, malleable and metallic lusture when they are freshly cut due to oscillation of loosely binded electrons.  Down the group softness increases due to the decreasing of cohesive energy hence Li is the hardest element while Cs is the softest element in first group.

All the alkali elements are silvery white solid. The silvery luster of alkali metals is due to the presence of highly mobile electrons of the metallic lattice. There being only a single electron per atom,

These are highly malleable and ductile. the metallic bonding is not so strong. As the result, the metals are soft in nature. However, the softness increases with increase in atomic number due to continuous decrease in metallic bond strength on account of an increase in atomic size.

COHESIVE ENERGY: Cohesive energy is just reverse of atomization energy, magnitude is same but sign is different. The (energy) force by which atoms or ions are bind together in solid state called cohesive energy.

(2) ATOMIC AND IONIC RADII:

The atoms of alkali metals have the largest size in their respective periods. The atomic radii increase on moving down the group among the alkali metals.

                                  Li < Na < K < Rb < Cs < Fr

REASON: On moving down the group a new shell is progressively added. Although, the nuclear charge also increases down the group but the effect of addition of new shells is more predominant due to increasing screening effect of inner filled shell on the valence s-electrons. Hence the atomic size increases in a group.

IONIC RADIUS: Alkali metals change into positively charged ions by losing their valence electron. The size of cation is smaller than parent atom of alkali metals. However, within the group the ionic radii increase with increases in atomic number.

                                     Li⁺ < Na⁺ < K⁺ < Rb⁺ < Cs⁺ < Fr⁺

HYDRATED RADIUS: The alkali metal ions get extensively hydrated in aqueous solutions. Smaller the ion more is the extent or degree of hydration. Thus, the ionic radii in aqueous solution follow the order

                                     Li⁺ > Na⁺ > K⁺ > Rb⁺ > Cs⁺ > Fr⁺

The charge density on Li⁺ is higher in comparison to other alkali metals due to which it is extensively hydrated.

(3) IONIZATION ENERGY (ENTHALPY):

 The first ionization energy of the alkali metals are the lowest as compared to the elements in the other group. The ionization energy of alkali metals decreases down the group.

REASON: The size of alkali metals is largest in their respective period. So the outermost electron experiences less force of attraction from the nucleus and hence can be easily removed.

The value of ionization energy decreases down the group because the size of metal increases due to the addition of new shell along with increase in the magnitude of screening effect.

(4) OXIDATION STATE:

The alkali metals show +1 oxidation state. The alkali metals can easily loose their valence electron and change into uni-positive ions

REASON: Due to low ionization energy, the alkali metals can easily lose their valence electron and gain stable noble gas configuration. But the alkali metals cannot form M⁺²
ions as the magnitude of second ionization energy is very high.

(5) REDUCING PROPERTIES:

The alkali metals have low values of reduction potential and therefore have a strong tendency to lose electrons and act as good reducing agents. The reducing character increases from sodium to caesium. However lithium is the strongest reducing agent.

                               Li > Na > K > Rb > Cs > Fr

REASON: The alkali metals have low value of ionization energy which decreases down the group and so can easily lose their valence electron and thus act as good reducing agents.  The reducing character of any metal is best measured in terms of its electrode potential which among other things depends upon its

(1) Heat of vaporization

 (2) Ionization energy and

 (3) Heat of hydration.

Since Li⁺ ion has the smaller size, its heat of hydration has the highest value. Therefore, among the alkali metals Li has the highest negative electrode potential (E⁰ cell=3.05 volts) and hence is the strongest reducing agent.

(6) ELECTROPOSITIVE CHARACTER:

On account of their low ionization energies, these metals have a strong tendency to lose their valence electrons and thus change into positive ions. Consequently, alkali metals are strongly electropositive or metallic in character. As this tendency for losing electrons increases down the group, the electropositive character increases.

                                   Li < Na < K < Rb < Cs

(7) COLOUR: The compounds of alkali metals are typically white

(8) MAGNETIC BEHAVIOR: The compounds of alkali metals are diamagnetic. Superoxides of alkali metals are, however, paramagnetic.

(9) HYDRATION: Most of alkali metal salts dissolve in water. In solution alkali metal ions are hydrated. Since Li+ ion is smallest in size it is most heavily hydrated. Salts of lithium such as LiF, Li₂CO₃, and Li₃PO₄ are insoluble in water.

(10) MELTING AND BOILING POINTS:

The melting and boiling points of alkali metals are very low because the intermetallic bonds in them are quite weak. And this decreases with increase in atomic number with increases in atomic size. 

(11) DENSITY:

The densities of alkali metals are quite low as compared to other metals. Li, Na and K are even lighter than water. The density increases from Li to Cs.

REASON: Due to their large size, the atoms of alkali metals are less closely packed. Consequently have low density. On going down the group, both the atomic size and atomic mass increase but the increase in atomic mass compensates the bigger atomic size. As a result, the density of alkali metals increases from Li to Cs. Potassium is however lighter than sodium. It is probably due to an unusal increase in atomic size of potassium.

(12) NATURE OF BOND FORMATION:

All the alkali metals form ionic (electrovalent) compounds. The ionic character increases from Li to Cs because the alkali metals have low value of ionization energies which decreases down the group and hence tendency to give electron increases to form electropositive ion.

(13) CONDUCTIVITY:

The alkali metals are good conductors of heat and electricity. This is due to the presence of loosely held valence electrons which are free to move throughout the metal structure.

(14) IONIC MOBILITY:

Ionic mobility of ion is inversely proportional to size of hydrated ion

Size of the hydrated ion is = Li⁺(aq) > Na⁺(aq) > K⁺(aq) > Rb⁺(aq) > Cs⁺(aq)

Order of ionic mobility        = Li⁺(aq) < Na⁺(aq) < K⁺(aq) < Rb⁺(aq) < Cs⁺(aq)

(15) PHOTOELECTRIC EFFECT:

Alkali metals (except Li) exhibit photoelectric effect (A phenomenon of emission of electrons from the surface of metal when light falls on them). The ability to exhibit photoelectric effect is due to low value of ionization energy of alkali metals. Li does not emit photoelectrons due to high value of ionization energy.  Generally K, Rb, Cs used photoelectric cell (mainly Cs).

(16) FLAME COLOURATION:

The alkali metals and their salts impart a characteristic colour to flame

REASON: On heating an alkali metal or its salt (especially chlorides due to its more volatile nature in a flame), the electrons are excited easily to higher energy levels because of absorption of energy. When these electrons return to their ground states, they emit extra energy in form of radiations which fall in the visible region thereby imparting a characteristic colour to the flame.

 SUMMARY TABLE:

:ILLUSTRATIVE EXAMPLES:

ILLUSTRATIVE EXAMPLES (1): Why are Group 1 elements called alkali metals?

SOLUTION: The Group 1 elements are called alkali metals because they form water soluble hydroxides.

ILLUSTRATIVE EXAMPLE (2): What is the most reactive alkali metal and why?

SOLUTION: The most reactive alkali metal is cesium due to its lowest first ionization enthalpy and lowest electronegativity.

ILLUSTRATIVE EXAMPLE (3): The alkali metals have low densities. Explain. ?

SOLUTION: The alkali metals have low densities due to their large atomic sizes. In fact, Li, Na and K are even lighter than water.

ILLUSTRATIVE EXAMPLE (4): Write three general characteristics of the elements of s-block of the periodic table which distinguish them from the elements of the other blocks.

SOLUTION: The three general characteristics are:

            (1)       The compounds of the s-block elements are mainly ionic.

            (2)       The valency is equal to the group number.

            (3)       Due to low ionization energy, s-block elements are good reducing agents.

ILLUSTRATIVE EXAMPLE (5) which alkali metal is most abundant in earth’s crust?

SOLUTION: Sodium is the most abundant alkali metal in the earth’s crust.

ILLUSTRATIVE EXAMPLE (6): Why is the density of potassium less than sodium?

SOLUTION: This is due to abnormal increase in the atomic size of potassium.

ILLUSTRATIVE EXAMPLE (7): Why is lithium the strongest reducing agent in the periodic table?

SOLUTION: The E^o value (reduction potential) depends on the three factors i.e. sublimation, ionization and hydration enthalpies. With the small size of its ion lithium has the lightest hydration enthalpy which accounts for its high negative E^o value and its reducing power

ILLUSTRATIVE EXAMPLE (8): Name the metal which floats on water without any apparent reaction with it.

SOLUTION: Lithium

ILLUSTRATIVE EXAMPLE (9): Which is softer – Na or K and why?

SOLUTION: Potassium is softer than sodium due to weak metallic bonding because of the large size of K atoms. 

ILLUSTRATIVE EXAMPLE (10): What makes sodium highly reactive?

SOLUTION: Low ionization enthalpy, strongly electropositive nature, tendency to attain noble gas configuration by the loss of one valence electron makes sodium highly reactive.

ILLUSTRATIVE EXAMPLE (11): Alkali metals impart colour to Bunsen flame due to

            (A) The presence of one electron in their outermost orbital

            (B) Low ionization energies                      

            (C) Their softness

            (D) Their reducing nature

SOLUTION: (B)

ILLUSTRATIVE EXAMPLE (12): The metallic lustre exhibited by sodium is explained by

            (A) Diffusion of sodium ions         

            (B)Oscillation of loose electrons

            (C) Excitation of free protons

            (D) Existence of body–centered cubic lattice

SOLUTION:  (B)             

ALKALI METALS - CHEMICAL PROPERTIES:

                                                                    Continuous…..   

BRIDGE BONDING-MULTI-CENTERED BOND:

Formation of bridge bonds is properly explained by MOT. According to which these bonds are formed by filling electrons into molecular orbital’s which lies over three nuclei hence such bonds are called specified as three centre bonds.
Bond angle between bridge bonds is less than bond angle between terminal bonds.

Bridge bonds are longer than terminal bonds
Bond energy of 3C-2e bond is found to be higher than 2C-2e bond for same substitute. It may also be true for 4C-4e bond.
During formation of bridge bond empty atomic orbitals of central atom participate in hybridization.
ILLUSTRATIVE EXAMPLES (1):

ILLUSTRATIVE EXAMPLE (2): BCl₃ do not dimerised due to back bonding

ILLUSTRATIVE EXAMPLE (3): AlCl₃   will have also    very less back bonding due to crowding  

ILLUSTRATIVE EXAMPLE (4): Steric crowding will there in B(CH₃)₃

IMPORTANT NOTES:

(1) If there is no steric crowding and back bonding in a molecules then bridge bond formed and molecules dimerised and stabilized and dimerisation are more stable than back bonding.

(2) Most of the electron deficient compound attains stability by performing back bonding or they undergo dimerisation provided certain conditions are fulfilled

(3) BCl_{3 ,}BBr₃ BI₃ and B(Me)₃ although  they are electron deficient compound but do not undergo dimerisation  because  of steric factor in demmeric formed

TYPE OF BRIDGE BOND:
(1) 3C-2e Bond Or Banana Bond         (2) 4C-4e Bond

(1) 3C-2e BOND OR BANANA BOND:

ILLUSTRATIVE EXAMPLE (5): FORMATION OF B₂H₆:

(1) Formation of 3C-2e bond in B₂H₆ is best explain by MOT and total number of bond in B₂H₆ is 6 (3C-2e=2 and 3C-4e=4)

(2) Bridge bonds are longer than terminal bond because at bridge bonds electrons are delocalized at three centres

(3) Bond energy (441kj/mole) of B-H-B bond is greater than bond energy (381 K j/mole) of   B-H bond.

(4) Hybridization of B atom is sp3, so non planer, and non polar (U=0)

(5)  B₂H₆ Methylated up to B₂H₂ Me₄

(6) B₂H₆ is hypovalent molecule hence act as Lewis acid and undergoes two type of cleavage when react with Lewis base:

(A) UNSYMETRICAL CLEAVAGE: 

B₂H₆ Undergo unsymmetrical cleavage with small size strong Lewis base like NH₃ NH₂Me and NH (Me) ₂ etc.

(B) SYMETRICAL CLEAVAGE:

B₂H₆ undergoes symmetrical cleavage with large size weak Lewis base like PH₃, PF₃,Me₃N , OEt , OMe₃, pyridine , THF , Thiophene , SMe₂ ,Set₂

(2) 3C-4e BOND or 3C-4e BRIDGE BOND: 

AL₂Cl₆ Dimmerised by 3C-4e bond bridge bond:

Al₂Cl₆ is neither hypovalent nor hypovalent rather its octet is complete. We will used  MOT here  it cannot act as Lewis acid  due to crowding in spite   having vacant d orbital’s   however Alcl₃  act as Lewis acid .

Al2Cl6 contains six bond having two bridge bond(3c-4e) and four bond is (2C-2e)

Boron do not formed bridge bond because boron experience steric crowding.

REASON OF DIMERISATION:

(1) By formation of 3C-2e bond

(2) By formation of 3C-4e bond

(3) By pairing of unpaired electrons.

ILLUSTRATIVE EXAMPLE (6): Which of the following molecule is/are dimerized by co-ordination bond?

(A) AlCl₃                     (B) BeCl₂                    (C) ICl₃                         (D) All of these

ILLUSTRATIVE EXAMPLE (7): The geometry with respect to the central atom of the following molecules are:   N (SiH₃)₃ ; Me₃N ; (SiH₃)₃P

(A) Planar, pyramidal, planar

(B) Planar, pyramidal, pyramidal

(C) Pyramidal, pyramidal, pyramidal

(D) Pyramidal, planar, pyramidal

ILLUSTRATIVE EXAMPLE (8): Which one of the following statements is not true regarding diborane?

(A) It has two bridging hydrogens and four perpendicular to the rest.

(B)When methylated, the product is Me₄B₂H₂.

(C) The bridging hydrogens are in a plane perpendicular to the rest.

(D) All the B–H bond distances are equal.

ILLUSTRATIVE EXAMPLE (9): The structure of diborane (B₂H₆) contains

(A) Four (2C–2e–) bonds and two (2C–3e–) bonds

(B) Two (2C–2e–) bonds and two (3C–2e–) bonds

(C) Four (2C–2e–) bonds and four (3C– 2e–) bonds

(D) None of these

ILLUSTRATIVE EXAMPLE (10): The molecular shapes of diborane is shown:

Consider the following statements for diborane:

1. Boron is approximately sp3 hybridized

2. B-H-B angle is 180°

3. There are two terminal B-H bonds for each boron atom

4. There are only 12 bonding electrons available

Of these statements:

(A) 1, 3 and 4 are correct                                               (B) 1, 2 and 3 are correct

(C) 2, 3 and 4 are correct                                               (D) 1, 2 and 4 are correct

Assertion & Reason:

ILLUSTRATIVE EXAMPLE (11):

Statement-1 : BeH₂ undergoes polymerisation while BH3 undergoes dimerisation.

Statement-2 : After dimerization of BH3 molecules into B₂H₆, no vaccant orbital at B

atom isleft to carry on further polymerization. However, in case of BeH₂, after dimerization of BeH₂molecules into Be₂H₄ each Be atom still contain sone empty 'p' orbital which brings further polymerization. 

(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation forstatement-1.

(C) Statement-1 is true, statement-2 is false.

(D) Statement-1 is false, statement-2 is true.

ILLUSTRATIVE EXAMPLE (12):

Statement-1: The B–F bond length in BF3 is not identical with that in –BF₄

Statement-2: Back bonding is involved in –BF₄ but not in BF₃

(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation forstatement-1.

(C) Statement-1 is true, statement-2 is false.

(D) Statement-1 is false, statement-2 is true.

ILLUSTRATIVE EXAMPLE (13):

Statement-1: (CH₃)₃Si – OH is more acidic than (CH₃)₃C – OH.

Statement-2: (CH₃)₃ Si – OH has back bonding.

(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.

(C) Statement-1 is true, statement-2 is false.

(D) Statement-1 is false, statement-2 is true.

Answers Key: