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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) BCl3 ,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 dimerize 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) Back bonding is a type of weaker π bond which is formed by sideways overlapping of filled orbital with empty orbital present on adjacent bonded atoms in a molecule.
(2) It is also considered as intermolecular Lewis acid-base interaction as it is a π bond.
(3) Back bonding is found to be effective and considerable in 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
(4)
dx2-y2 and dx2 Orbital’s 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.

CONDITIONS FOR BACK BONDING:
(1) Both of the atoms bonded with back Bonding are must be present in
2nd-2nd or 2nd-3rd period.
(2) One of the atoms has lone pair and another have vacant Orbital and direction of back Bonding depends upon 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 orbital’s favour overlap.

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: BF3 < BCl3 < BBr3 < BI3 is order of Lewis acidic character due to stronger 2pπ-2pπ back bonding in BF3 (lone pair orbital of fluorine into vacant orbital of boron) and gradually back bonding becomes weakest in BI3 (2pπ—5pπ) hence BF3 has stronger partial double bond character and consequently behaves as less electron deficient
Extent of back Bonding:
      BF3> BCl3>BBr3>BI3
Lewis acid Character:
      BF3<BCl3<BBr3<BI3
Reaction with nucleophile/water:9
      BI3>BBr3>BCl3>BF3

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 d orbital’s resonance exist of 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 is 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 have greater extent of back Bonding?
SOLUTION:

Ans: (1) ‘y’ is greater than ‘x’ because of steric repulsion of -CH3 group.
Ans: (2)’z’ will be greater because one lone pair going two places.
Ans; (3) the extant of back bonding is 3rd >1st > 2nd 

ILLUSTRATIVE EXAMPLE (4): Give the correct order of (B-H) bond length of 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 case of silicone halides inductive effect dominate over back bonding hence lewis acid character decided by inductive effect.
Hence order of lewis acid character   SiF3 >SiCl3 > SiBr3 > SiI3

ILLUSTRATIVE EXAMPLE (6): Compare the acidic strength of silianol (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 correct statement about 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
Ans: Statement (4) is wrong because molecule is planer and polar
ILLUSTRATIVE EXAMPLE (8): Arrange (A) OMe2 (B) O(SiH3)2 (C) O(SiPh3)3 in increasing order of X-O-X bond angle ?
SOLUTION:  A>B> C
ILLUSTRATIVE EXAMPLE (9): Arrange increasing order of bond angle (X-O-X) in (A) OMe2 (B) H2O, (C) OF2, and (D) OCl2?
SOLUTION: B<A<C<D
ILLUSTRATIVE EXAMPLE (10): Arrange increasing order of bond angle (X-N-X) in (A) NH3, (B)NF3, (C)  NCl3, (D) CCl2?
SOLUTION: B<A<C<D
ILLUSTRATIVE EXAMPLE (11): Correct statement about structure of H3CNCS,
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 molecule are non planer.
SOLUTION:  (A, B, D)
Due to back Bonding between nitrogen and silicon atom bond length decreases and shape become linear.
Hence option 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 ppi-ppi bond formed by 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 six member ring and also each boron atom 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 fact ring become aromatic but due to sp3 hybrisation of oxygen atom molecule become planer.
In Borazine molecule, nitrogen is more electro negative than the boron. Nitrogen acquires partial negative charge and boron acquires partial positive charge and back bonding take place between boron and nitrogen.
 Even though Borazine and Benzene have 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 Borazine undergoes addition reactions compared to benzene
(3) Aromaticity of borazine is less than benzene  hence it is less reactive  toward Eectrophilic  substitution reactions 
Hence options A B and D is correct

BACK BONDING: COLUSIONS:

(1) Due to back Bonding , bond length always  decrease .
(2) If empty atomic orbital of central atom of molecule participate in back bonding then its hybridization does not change and its Lewis acid Character decreases.
(3) If filled orbital of central atom of a molecule participate in back Bonding then it's hybridization may change and it's 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 increase or remain same but never decreases.
(5)In most molecules steric factor enhance (increases) the extent  of back Bonding for example N(SiH3)3 , OCl2 , NCl3 , O(SiH3)2 (disilylether) however in some cases steric Factors decreases extent of back Bonding for example O3BMe3 ,NSi(Me3)(N3).
(6) When skeleton is planer then steric Factor's decrease extent of back bonding.
(7) In 2p-2p type of back Bonding, back Bonding dominates over inductive effect while in 2p-3d and 2p-3p inductive effect dominates over back Bonding.

(8) Me3NO has greater dipole moment than Me3PO as there is 2pπ—3dπ back donation from Oxygen into vacant d-orbital’s 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 adduct with BF3 but (SiH3)2O do not react with BF3 due to weakening of basic character of Disilyl ether by back bonding.
(11) BH3 does not exist (it exist only as dimer or higher boranes) but BX3 exist, (X=halogen). It can be attributed to absence of 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]. Its due to absence of Back-bonding in (BF4)- hence B-F bond has completely single bond character.
(14) Si-O and P-O bonds are much stronger than expected to partial double character owing to possibility of back-bonding.
(15) Bond angle of NF3(102 degree) is lesser than in NH3 (107) as per VSEPR theory which suggests that in case of less electronegative terminal atoms like H, Bond pairs would be closer to more electronegative central atom, N and hence bonds open up due to repulsion between  bond pairs electron density in vicinity. But bond angle of PF3 (100) is greater than PH3, its due to 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 causes more repulsion so bond angle is greater.
(16) SiCl4 has abnormally low boiling point than CCl4,
(17) Due to possibility of Back-bonding with metal (similar to carbonyl complexes), Ph3P or  R3P or PF3 behave as strong ligand 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.
BACK BONDING IN METAL CARBONYL:
(A) The carbon atom in carbon monoxide has a lone pair of electrons that can be used to form a sigma bond with a metal. Because carbon monoxide has low-lying orbital’s, it can accept electrons back from the metal and further strengthen the bonding between the metal and the carbon monoxide ligand. This process of “accepting electrons back from the metal” is termed back bonding. Here  it’s important to understand that as Metal is more negatively charged; then M-C Back-bonding is stronger & C-O bond would have been weaker than in CO.
 (B) Back bonding is mostly observed in CO ligands which is a sigma donor as well as 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 behavior from nitric acid, NO)
(C) Back bonding is also common in 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)]

BENT'S RULE OF HYBRIDISATION:


CONCEPTUAL FACT: The more electronegative atom prefer to stay in that hybrid orbital having less S-character or more p- character and more electropositive atom prefer to that hybrid orbital which have more S-character or less p-character.
EXPLAINATION: The more electropositive atom or group will withdraw the bond pair more from central atom. it is easy when hybrid orbital is having less S-character and more P-character .
 S-orbital is closer to the nucleus so it electronegativity is more than p-orbital.
ILLUSTRATIVE EXAMPLE (1): PCl3F2 (sp3d Hybridization) and TBP (Trigonal bi pyramidal)
  Where angle < F-P-F =180 and <Cl-P-Cl =120 two P-Cl bond is axial while three P-Cl bond is equatorial                                  

(1) APICOPHIOLICITY IN TBP GEOMETRY:
(1) Trigonal bipiramidal geometry more electronegative atom prefers to stay in low electronegative PZdz2 orbital of sp3d hybridization. You can also say that more electronegative atom prefer that hybrid orbital having low s-character or no s-character.
(2) If electronegativity difference between central atom and surrounding atom is large then due to polarity some ionic character is developed and covalent character decreased.
(3)The poor covalence is not only due to electronegativity difference between bonding atom.  it is also generated due to poor overlapping, steric hindrance or mismatch of the overlapping orbitals.
(4) Position of lone pair: Bent rule is very important in predicting position of lone pair. lone pair is attracted by only one nucleus while bond pair is attracted by two nucleuses. Central atom hold lone pair cloud tightly if central atom is having more S-Character.
ILLUSTRATIVE EXAMPLE (2): SF4 (Sp3d) TBP and SF2Cl2 (Sp3d)
ILLUSTRATIVE EXAMPLE (3): XeF2 and XeO3F2
(2) ORBITAL ANALYSIS (CALCULATION OF (% S ) AND ( %P) CHARACTER:  
S% character is equatorial orbital at1 200c :
S% character for orbital at 900 C:



(3) EFFECT OF THE STRAINGTH OF COVALENCY: (Alternate Statement of bent)

RULE: The more electronegative atom not only prefer to stay in that orbital which having less  % S character (more p-character) but it also decreases % S-character and increases % P-character in its attached orbital from the central atom depending on circumstance.
On increasing % s character in hybrid orbital , the bond length  decreases while bond angle increases.
ILLUSTRATIVE EXAMPLE (3): : Explain C-H bond length of CH4 is longer than C-H bond length of Difloromethan (CH2F2) ?

EXCEPTIONS OF BENT’S RULE:
(1): Bent’s rule is applicable in those molecules where central atoms are same and they are also in same Hybridization. For example N-N bond length cannot be compared in N2H4 and N2O4 using Bent rule.
ILLUSTRATIVE EXAMPLE (4): Arrange PF3 ASH3, PH3, NH3, H2Se in decreasing order by Bent’s rule here we used DRAGO’S RUEE”
SOLUTION: NH3> NF3>PF3> PH3>AsH3>H2Se
(2): Bent’s rule violets in those molecules where steric factor’s plays dominating rule.
ILLUSTRATIVE EXAMPLE (5): Compared Bond angle among H2O OF2, OMe2, OCl2  
SOLUTION: 
ILLUSTRATIVE EXAMPLE (6):   IN CH2SF4 


Which of the following option is correct regarding ?

(1) 1800> 120                                                       (2) 1800> >120
(3) 1200> >900                                                         (4) 900 > >00

SOLUTION: S -character present in equatorial so more decrease in Bond angle in equatorial orbital than in axial. Because no S-character in present in axial orbital.
So 1200 > > 90 and 1800 > > 120 

ILLUSTRATIVE EXAMPLE (7):   Compare Dipole moment of following compound pairs

(1) Dipole moment of PCl2F3 is non zero while dipole moment of PCl3F2   is zero why?
(2) Dipole moment of P(CH3)2 (CF)3  is non zero while dipole moment of P(CH3)2(CF3)2 is zero why?

SOLUTION: According to bent rule more electronegative atom or group attached those orbital have minimum S- character there in Trigonal bipiramidal (TBP) Geometry we known that axial orbital hare no S- character so F and -CF3  are attached with axial positions. 
(1) :
(2)

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