[thedy]

I have read some articles,but I don t understand one thing.In NO3- is one coordination bond between Oxygen and Nitrogen.
And now:Oxygen electron configuration is 1s2 2s2 2px2 2py1 2pz1.So,oxygen has two unpaired electrons,which can form bond.That means,that oxygen has not free orbital,which can make coordination bond,has it?How can lone pair from nitrogen get to free orbital,if oxygen has not free orbital?
Thanks

https://docs.google.com/drawings/d/1ZScRF9WEA6fGa_8roqwNAq41Ki0lKnonTUsacNGYxbc/edit

[Schrödinger]

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.

[thedy]

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..