how the bond length decreases with the increse in the s character and how the s character increases ...book says that as s orbital is smaller than p orbital...but how..explain briefly with identical proofs.

Dear Aakriti,

Definition of Bond Length

“The average distance between the centre of the nuclei of the two bonded atoms is called bond length”.

It is expressed in terms of Angstrom (1 Å = 10^–10 m) or picometer (1 pm = 10^–12 m).

In an ionic compound, the bond length is the sum of their ionic radii (d = r₊ + r_–) and in a covalent compound, it is the sum of their covalent radii (e.g., for HCl, d = r_H + r_Cl).

Factors affecting bond length

(i) The bond length increases with increase in the size of the atoms. For example, bond length of H – X are in the order, HI > HBr > HCl > HF.

(ii) The bond length decreases with the multiplicity of the bond. Thus, bond length of carbon–carbon bonds are in the order, C ≡ C < C = C < C – C.

(iii) As an s-orbital is smaller in size, greater the s-character shorter is the hybrid orbital and hence shorter is the bond length.

For example, sp³ C – H > sp² C – H > sp C – H

(iv) Polar bond length is usually smaller than the theoretical non-polar bond length.

More is the s- character in hybridisation of carbon, more is its electronegativity and hence carbon will hold its electron pair more tightly and bond length decreases... as shorter bond is much hard to break than a longer bond and hence bond is more stronger

Cracking IIT just got more exciting,It s not just all about getting assistance from IITians, alongside Target Achievement and Rewards play an important role. ASKIITIANS has it all for you, wherein you get assistance only from IITians for your preparation and win by answering queries in the discussion forums. Reward points 5 + 15 for all those who upload their pic and download the ASKIITIANS Toolbar, just a simple  to download the toolbar….

So start the brain storming…. become a leader with Elite Expert League ASKIITIANS

Thanks

Aman Bansal

Askiitian Expert

CONSIDER SP , SP² AND ETC 

IN THE FIRST CASE THERE ARE 1 S ORBITAL AND 1 P ORBITAL IN TOTAL 2 ORBITAL THEREFORE  %S CHARACTER = (1/2)*100 

THEREFORE 50% S AND 50% P

 SIMILARLY IN THE SECOND CASE 1 ORBITAL OF S AND 2 ORBITAL OF P TOTAL NO. OF ORBITAL = 3 THEREFORE %S CHARACTER=(1/3)*100

  AND AS %S CHARACTER INCRESES ELECTRONEGITIVITY INCRESES THEREFORE BOND LENGTH DECREASES

 PLZZZZZZZZZZZ APPROVE