Welcome to Chem Zipper.com......: 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]
(5) 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:

(1) Back bonding extent in boron trihalides decreases from BF₃ to BI₃ because increasing of size of p-orbital of halogen the strength of back bond decreases .Thus Extent of back Bonding:
[BF₃>BCl₃>BBr₃>BI₃]

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

[BF₃<BCl₃<BBr₃<BI₃]

Thus it is clear that BF3 is 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 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]

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)₃is 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)(1^st) N(SiH₃)₃,(2^nd) {(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 3^rd >1^st > 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? SiF_3, 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 sp²and (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 p_pi-p_pibond 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)]

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.