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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)]

5 comments:

  1. Nice work...it is like my chemistry sir's notes...it is more helpful and it makes it easier to understand concepts well...��

    ReplyDelete
  2. thank you for appreciation .......

    ReplyDelete
  3. Not intermolecular but intramolecular Lewis acid-base interaction

    ReplyDelete
  4. just awesome explanation,it helped me in neet preparation

    ReplyDelete

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