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Important Applications of Backbonding
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Based on the concepts of backbonding.

Tamanna Chaudhary is teaching live on Unacademy Plus

Tamanna Chaudhary
नमस्ते 🙏 The Newton's apple. 🍎

Unacademy user
7/10...Thank you sir...
In example N(SiH3)3 will have 2p-3d
decreasing order of bond length not increasing order ... the order is wrong
As long as you know what '>' and '<' means while writing an order, there shouldn't be any problem at all.
The order given in the lesson is absolutely right as the signs are used correctly.
thank you
I didn't understand how flourine haven't any empty orbital but chlorine have.....Mam pls give me some hint to understand clearly..
Sure Shruti. Here we are talking about back-bonding for which d-orbitals are needed. Chlorine ki configuration mein 3d aata hai, so d orbitals are involved. But in Flourine these is no involvement if d orbital as it's configuration is 1S2 2S2 2P5
Shruti Sandhya
a year ago
Lekin ,mam chlorine ki configuration - 1s2,2s2,2p6,3s2,3p5 hoti hai then how 3d orbitals are involved...??
Lekin n=3 to valid hai na Cl ke liye, isiliye 3d orbitals hongi Cl ke pass and they will be empty. F ke liye n=2 hai, and we know that for n=2, d orbital is not present. So, F will lack d orbitals.
Shruti Sandhya
a year ago
ok ,mam . thanks
Rashmi rana
a year ago
thanks mem
Rashmi rana
a year ago
I am also face this problem but after seen this reply I understand betterly thank you so much mem
thank u so much mam
basic to complex knowledge, a very good explanation
  1. Name: Tamanna Chaudhary Ex-Hans Raj (DU) Student. - AIIMS and NEET Aspirant. Topic of Discussion. Applications of Backbonding

  2. Finding the more stable molecule with the help of Backbonding CONCEPT. I. A molecule with backbonding is more stable than a molecule without backbonding 2. When dealing with two molecules in which Backbonding is present in both, the molecule with stronger backbonding will be more stable.

  3. CL ackbondine fresent PT-3p which one mare stable? SiH STH Tho no empty orbital Back banding abent Back bondinq

  4. To find out which one is the better acid. 1. When given to compare the acidic strength of two acids, make their Remove a proton from it) Weaker the conjugate bases, more stronger is the acid. Using this concept look for the backbonding in the acids, the one with more stronger BB will be less acidic. conjugate bases 2. which one of th louloutng ts better acid? CH OR CHF 3 Co ugote Base ce CL No Be, thus weaker

  5. To explain the exceptions in hybaidisation Concept- Due to BB, sometimes the molecule hybridises to lower H. CHy SiH 88 ocaurs Se i^bidisatien

  6. To find the molecule longer bond length. Concept Stronger the back bonding, shorter is the bond length. Arrange tn order a tnereasina Bond length 9 6 >B Neither them BB absenb