Why does (CH3)3P act as a stronger base than (CH3)3N towards B2H6?

Electronic Effects:
- The phosphorus atom in (CH3)3P is larger than the nitrogen atom in (CH3)3N, leading to poorer orbital overlap between the lone pair on the phosphorus atom and the vacant orbital on the boron atom in B2H6.
- This results in a weaker P-B bond in (CH3)3P compared to the N-B bond in (CH3)3N, making (CH3)3P a stronger base.

Steric Effects:
- The bulky methyl groups in (CH3)3P create steric hindrance around the phosphorus atom, preventing the boron atom in B2H6 from approaching as closely as it would with (CH3)3N.
- This reduces the effectiveness of the B-P bond formation in (CH3)3P, making it a stronger base compared to (CH3)3N.

Overall Stability:
- The combination of electronic and steric effects in (CH3)3P results in a stronger base character towards B2H6 as compared to (CH3)3N.
- The weaker P-B bond and the steric hindrance around the phosphorus atom make it more favorable for (CH3)3P to accept a proton from B2H6, making it a stronger base.
By considering both electronic and steric effects, we can conclude that (CH3)3P acts as a stronger base than (CH3)3N towards B2H6.