I guess you pair up the unpaired electrons in the last 2 orbitals, px and pz and then use the newly created vacant orbital for the pair of e-s coming in from N. Either that, or hybridize the 4 orbitals (2s, 2px, 2py and 2pz) to form 4 sp3 orbitalsand do the same. 6 e-s go into the first 3 orbitals and the last one is left vacant for the coordinate bond.
It is also useful to consider the co-ordinate bond as a normal sigma bond with a +ve charge on the donor and a -ve charge on the acceptor. That way, the explanation is easier.
You said,,I guess you pair up the unpaired electrons in the last 2 orbitals, px and pz and then use the newly created vacant orbital for the pair of e
-s coming in from N."----Do you mean last two orbitals,so not px and pz,but py and pz,or not?
And thanks,but what about Hund's rules?If I hybridize orbitals to get 4 sp3 orbitals,I have to fill electrons exactly according Hund s rules.So I can t get vacant orbita,can I?And second problem is,that if I hybridize all p orbitals,I cannot use p(y) orbital to form pi bond..