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:

BACK BONDING-THEORY:

1. Definition of Back Bond:

(1) Back bonding is a type of weaker π bond that is formed by sideways overlapping of filled orbital with empty orbital present on adjacent bonded atoms in a molecule.

(2) It is also considered an intermolecular Lewis acid-base interaction as it is a π bond.

(3) Back bonding is found to be effective and considerable in the following type of overlapping.

   [ (i) 2p-2p (ii) 2p-3p (iii) 2p-3d ]

(4) the extent of overlapping order is

    [2p-2p> 2p-3d >2p-3p]

(5) dx2-y2 and dx2 Orbital does not participate in back Bonding.

ILLUSTRATIVE EXAMPLE (1): Which of the following options is/are true about back Bonding?

(A) Sigma-dative bond 

(B) π- dative 

(C) Intermolecular Lewis acid-base interaction 

(D) Intramolecular Lewis acid-base interaction

SOLUTION: Options B and D are responsible for back Bonding and options A and C are responsible for Coordinate Bonding.

2. CONDITIONS FOR BACK BONDING:

(1) Both of the atoms bonded with back Bonding must be present in the 2nd-2nd or 2nd-3rd period.

(2) One of the atoms has lone pair and another has a vacant Orbital and the direction of back Bonding depends upon the vacant Orbital.

(3) The donor atom must have localized donatable electron pair. In general, these are later half-second period P - block elements (F, O, N and C).

(4) The acceptor atom must have low energy empty orbital which generally are np or nd orbitals. Small and similar-sized orbitals favour overlap.

3. EFFECTS OF BACK BONDING:

(1) It always leads to an increase in bond order between the participating atoms.

(2) It always leads to an increase in bond strength between participating atoms.

(3) It always leads to a decrease in bond length between participating atoms.

LEWIS ACID CHARACTER OF BORON HALIDES: 

(1) Back bonding extent in boron trihalides decreases from BF3 to BI3 because by increasing the size of the p-orbital of halogen the strength of the back bond decreases. Thus Extent of back Bonding:

[BF3>BCl3>BBr3>BI3]

(2) Lewis acid character of Boron Halides is inversely proportional to the extent of back bonding because on decreasing back bonding tendency to accept lone pair from the base increases thus the order of Lewis acid character is : 

   [BF3<BCl3<BBr3<BI3]

Thus it is clear that BF3 is the weakest Lewis acid due to stronger 2pπ-2pπ back bonding (stronger partial double bond character) in BF3 (lone pair orbital of fluorine into vacant orbital of boron) and consequently behaves as less electron deficient. The back bonding gradually decreases (From BF3 to BI3) and becomes weakest in BI3. So that BI3 become a strong Lewis acid.

(3) The Nucleophilicity order is inversely proportional to the Lewis acid character thus the   Nucleophilicity order is: (reaction with nucleophile/water

[BI3>BBr3>BCl3>BF3]

4. d-ORBITAL RESONANCE:

It is a phenomenon in which electrons of ms and np get delocalized to vacant nd orbital because this availability of vacant d orbital to expect back bond get reduced.

In those molecules species where the d orbital’s resonance exists back, Bonding is decreased.

ILLUSTRATIVE EXAMPLE (2): N(CH3)3 is pyramidal while (SiH3)3N is trigonal planer why?

SOLUTION:  N(CH3)3 has sp3 hybridization & pyramidal shape at N, but in (SiH3)3N again there is 2pπ—3dπ back bonding between lone pair orbital of nitrogen into vacant orbital of silicon. Hence trisilyl amines are sp2, planer & is less basic than trimethyl amine.

ILLUSTRATIVE EXAMPLE (3): (1st) N(SiH3)3,(2nd) {(Me)3Si}N---Si(Me)3 and(3rd) HN(SiH3)2

Q (1): Which is greater x or y?

Q (2): Which is greater x or z?

Q (3): Which has a greater extent of back Bonding?

SOLUTION:

Ans: (1) ‘y’ is greater than ‘x’ because of the steric repulsion of the -CH3 group.

Ans: (2)’z’ will be greater because one lone pair going two places.

Ans; (3) the extent of back bonding is 3rd >1st > 2nd 

ILLUSTRATIVE EXAMPLE (4): Give the correct order of (B-H) bond length of the following compounds. (1) B(OH)3  (2) B(OMe)3 and(3) B(Me)2OH

SOLUTION:

Extent Back Bonding is   3>1>2 and bond length order is  y>x>z  

ILLUSTRATIVE EXAMPLE (5): Arrange the silicon halides into decreasing order of Lewis acids Character?  SiF3, SiCl3, SiBr3, SiI3

SOLUTION:  In the case of silicone halides inductive effect dominate over back bonding hence Lewis acid character is decided by the inductive effect.

Hence order of lewis acid character SiF3 > SiCl3 > SiBr3 > SiI3

ILLUSTRATIVE EXAMPLE (6): Compare the acidic strength of silanol (SiH3OH) and methanol (CH3OH).

SOLUTION:  H3C-OH is less acidic than H3Si-OH due to stabilization of negative charge in H3Si-O- ion by 2pπ—3dπ back bonding

ILLUSTRATIVE EXAMPLE (7): Choose the correct statement about the structure of H3BO3 is/are. Statements are as: 

(1) Angle OBO =120   

(2) Angle  HOB>109 

(3) Hybridization  of atom O close to sp2 and   

(4) Molecule is non-planer and non-polar

SOLUTION: Statement (4) is wrong because the molecule is planer and polar

ILLUSTRATIVE EXAMPLE (8): Arrange in increasing order of X-O-X bond angle.

 (A) OMe2 

(B) O(SiH3)2 

(C) O(SiPh3)3

SOLUTION:  A>B> C

ILLUSTRATIVE EXAMPLE (9): Arrange increasing order of bond angle (X-O-X) in the following compounds:

(A) OMe2 

(B) H2O, 

(C) OF2, 

(D) OCl2.

SOLUTION: B<A<C<D

ILLUSTRATIVE EXAMPLE (10): Arrange increasing order of bond angle (X-N-X) in the following compounds:

(A) NH3, 

(B) NF3, 

(C) NCl3

(D) CCl2?

SOLUTION: B<A<C<D

ILLUSTRATIVE EXAMPLE (11): Correct statement about the structure of H3CNCS and H3SiNCS is/are?

(A) CNC bond angle in H3CNCS is >120 and hybridization is closed to sp2 

(B) Si-N-C bond angle is 180 in H3CNCS

(C) Both have Back Bonding

(D) Skeleton Si-N-C-S is linear but molecules are non-planer.

SOLUTION:  (A, B, D)

Due to back Bonding between nitrogen and silicon atom bond length decreases and shape become linear.

Hence options A, B, and D are correct.

ILLUSTRATIVE EXAMPLE (12): Correct statement about B3O6-3 and B3N3H6?

(A) Both are planer and non planer

(B) Both have aromatic character

(3) Both have a ppi-ppi bonds formed by the pairing of unpaired electrons

(4) Electrophilic reaction occurs at B3N3H6

SOLUTION:

SOLUTION:( A, B, D) In Boraxine ion boron and oxygen atom alternatively combined to form the six-member ring and also each boron atom is linked with extra oxygen atoms. Both boron and oxygen atoms have sp2 hybridization (by Back bonding and all oxygen atom involved in back bonding) and planer structure due to the fact ring become aromatic but due to sp3 hybridisation of the oxygen atom molecule become planer.

In the Borazine molecule, nitrogen is more electro-negative than boron. Nitrogen acquires a partial negative charge and boron acquires a partial positive charge and back bonding takes place between boron and nitrogen.

 Even though Borazine and Benzene have the same stricture their chemical properties are different.

(1) Organic benzene is C6H6 while Inorganic benzene is Borazine having chemical formula B3N3H6

 (2) Borazine is more reactive than Benzene with respect to addition reactions.

(3) Aromaticity of borazine is less than benzene  hence it is less reactive  toward Electrophilic  substitution reactions 

Hence options A B and D are correct.

5. BACK BONDING: CONCLUSIONS:

(1) Due to back Bonding, bond length always decreases.

(2) If the empty atomic orbital of the central atom of the molecule participates in back bonding then its hybridization does not change and its Lewis acid Character decreases.

(3) If the filled orbital of the central atom of a molecule participates in back Bonding then its hybridization may change and its Lewis basic Character may also change for example N(SiH3)3, however in some molecules, hybridization may not change.

(4)Due to back Bonding, bond angle either increases or remains the same but never decreases.

(5)In most molecules steric factor enhance (increase) the extent of back Bonding, for example, N(SiH3)3, OCl2, NCl3, O(SiH3)2 (disilyl ether) however in some cases steric Factors decreases extent of back Bonding for example O3BMe3, NSi(Me3)(N3).

(6) When the skeleton is planer then the steric Factor decreases the extent of back bonding.

(7) In the 2p-2p type of back Bonding, back Bonding dominates over the inductive effect while in 2p-3d and 2p-3p inductive effect dominates over back Bonding.

(8) Me3NO has a greater dipole moment than Me3PO as there is 2pπ—3dπ back donation from Oxygen into vacant d-orbitals of phosphorus (just like in CO).

(9) Me3C-OH is less acidic than Me3Si-OH due to stabilization of negative charge in Me3Si-O- ion by 2pπ—3dπ back bonding.

(10) Me2O forms an adduct with BF3 but (SiH3)2O does not react with BF3 due to the weakening of the basic character of Disilyl ether by back bonding.

(11) BH3 does not exist (it exists only as a dimer or higher boranes) but BX3 exist, (X=halogen). It can be attributed to the absence of the possibility of back bonding in BH3.

(12) BF3 is only partially hydrolysed into [BF3(OH)]- whereas BCl3 & BBr3 are completely hydrolysed into B(OH)3 or H3BO3 and HCl/HBr

(13) B-F bond length increases when BF3(130 pm) reacts with F- to form (BF4)- [143 pm]. It is due to the absence of Back-bonding in (BF4)- hence B-F bond has a completely single bond character.

(14) Si-O and P-O bonds are much stronger than expected to partial double character owing to the possibility of back-bonding.

(15) Bond angle of NF3(102 degrees) is lesser than in NH3 (107) as per VSEPR theory which suggests that in the case of less electronegative terminal atoms like H, Bond pairs would be closer to the more electronegative central atom, N and hence bonds open up due to repulsion between bond pairs electron density in the vicinity. But the bond angle of PF3 (100) is greater than PH3, it is due to the possibility of back bonding in PF3 between lone pair of fluorine and vacant d-orbital of phosphorous (2pπ—3dπ) henceforth P-F bond acquires partial double character and we know well that multiple bonds cause more repulsion so the bond angle is greater.

(16) SiCl4 has an abnormally low boiling point than CCl4,

(17) Due to the possibility of Back-bonding with metal (similar to carbonyl complexes), Ph3P or  R3P or PF3 behave as strong ligands in complexes.

NOTE- 3pπ—3pπ Back bonding in AlCl3 is not as effective hence it easily forms dimer in vapor phase or non-polar solvent.

6. BACK BONDING IN METAL CARBONYL:

(1) The carbon atom in carbon monoxide has a single pair of electrons that can be used to form a sigma bond with a metal. Because carbon monoxide has a low-lying orbital, it can accept electrons back from the metal, strengthening the bond between the metal and the carbon monoxide ligand even more.  Back bonding refers to the process of "accepting electrons back from the metal." It is critical to understand that because metal is more negatively charged, M-C back-bonding is stronger and C-O bonding is weaker than in CO.

(2)Back bonding is mostly observed in CO ligands which is a sigma donor as well as a pi- acceptor. [The typical example given for synergy in chemistry is the synergic bonding seen in transition metal carbonyl complexes. CO has much less dipole moment (0.11D) than expected due to back-donation from lone pair orbital of Oxygen into vacant orbital of carbon. (Similar behaviour from nitric acid, NO)

(3)Back bonding is also common in the organometallic chemistry of transition metals which have multi-atomic ligands such as carbon monoxide, ethylene or the nitrosonium cation e.g, Ni(CO)4 and Zeise’s salt, K[PtCl3(C2H4)]

BRIDGE BONDING-MULTI-CENTERED BOND:

STRUCTURE OF FULLERENCES :

STRUCTURE OF GRAPHITE:

Related Questions:

(1) Why trimethylamine amine ( N(CH3)3) is tetrahedral while trisilyl amine (N(SiH3)3) planner.?

(2) Why trimethyl amine {(CH3)3 N:} is pyramidal while trisilyl amine {(SiH3)3N:} is trigonal planer?

(3) Which is/are the correct statement/s about structure of methyl isothiocyanate (H3CNCS) and Silyl isothiocyanate(SiH3NCO)?

(4) What is the Si–N–C bond angle in Silyl isothiocyanate and methyl isothiocyanate (H3CNCS)?

(5) What are the order of extent back bonding, Lewis acid character and nucleophilicity of (BF3, BCl3, BBr3, BI3)boron trihalides?

(6) Arrange the silicon halides into decreasing order of Lewis acids Character? SiF4, SiCl4, SiBr4, SiI4

(7) Chloroform is more acidic than fluoroform why?

(8) Trisilyl amine, N(SiH3)3 is planar whereas trimethyl amines N(CH3)3 is pyramidal. Explain why?.

(9) Silianol (SiH3OH) is more acidic than methanol (CH3OH) why?

(10) Arrange the silicon halides into decreasing order of Lewis acids Character? SiF4, SiCl4, SiBr4, SiI4

(11) Why BF3 do not exist as dimer?. Explain.

(12) Why B-F bond length in BF3 is shorter (130 pm) than B-F bond Iength in BF4- (143 pm)?. Explain.

(13) Which is more basic N(CH3)3 or N(SiH3)3 , explain?

(14) Why trimethyl amine {(CH3)3 N:} is pyramidal while trisilyl amine {(SiH3)3N:} is trigonal planer?

(15) What are the order of extent back bonding, Lewis acid character and nucleophilicity of (BF3, BCl3, BBr3, BI3)boron trihalides?

(16) Why trimethylamine amine ( N(CH3)3) is tetrahedral while trisilyl amine (N(SiH3)3) planner.?

(17) Which of the following is correct order of (B-O) bond length of following compounds? (1) B(OH)3 (2) B(OMe)3 and (3) B(Me)2OH

(18) What are the structure of (1) N(SiH3)3 (2) {(Me)3Si}2N-Si(Me)3 (3) HN(SiH3)2 and give the Answers of following questions?

(19) What is structure (CF3)2-Al-O-Al-(CF3)2 and also find the number of pi bond or Back bond?

(20) How to compared bond angle (C-N-C) and (Si-N-Si) in trimethyl amine and trisilyl amine respectively?

(21) How to determine order of bond angle of Me-O-Me, H3Si-O-SiH3 and (ph)3Si-O-Si(Ph) ?

(22) Why is BCl3 a strongerLewis acid than BF3 ?