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 = rH + rCl).
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
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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