Dear Dr. @Mohammed Kooti

My thanks and appreciation for your valuable contributions.

Best regards.

hybridization of N is 1s2 2s2 2p3. three unpaired electron of p-orbital forms three bond and fourth one bond is formed by lone pair of electron that is why it shows 3+ and 5+ oxidation state with high electronegative atom than Nitrogen

Interesting discussion! My statements:

1) Oxidation state is the concept to be applied here. Find out the EN of the elements and assigne positive or negative oxidation states accordingly.

2) This has nothing to do with the nature of the bond. So please do not use hybridisation or related oversimplified concepts for binding.

3) NF4+: here F is definitely -1, so N is +5 (which is the highest oxidation state, reasonably feasible for N).

4) NF5: This would be an elusive species but has not been synthesised so far. The reason is probably quite simple: The N atom is too small to host five F-Atoms. These F-Atoms (or F- ions) have three lone-pairs, thus a strong repulsive force drives them away from each other, making NF5 sterically impossible. If the central atom is bigger, e.g. as P, PF5 is no problem at all.

5) The NF5 problem is an interesting start to understand  the chemistry of the elements of the 2nd period, mainly C and N on a sophisticated level: C and N are too small to host more than 4 ligands (atoms around the central atom). The space around them is so limited that they even agree with lower numbers of ligands such as 2 or 3. Consequently, they form multiple bonds in case of these lower coordination numbers. Of course, as for any element, four (or more) single bonds are more stable for C or N than multiple bonds, but what If they cannot? Then the SECOND BEST option is multiple bonds. Electronegative ligands carriying free electron pairs (lone pairs) are quite repulsive to each other. When combined with these small elements the coordination number tends to be small e.g. C(OH)4 is not very stable and will rapidly yield C(O)(OH)2 through cleavage of two water molecules. This is also called the Erlenmayer rule.

6) The chemistry of the heavier analogues Si and P is not limited by size and species such as Si(OH)4,  [Si(OH)5]-, Si(OH)6]2-, [SiF6]2-,  PF5, or [PF6]- are quite stable. Both Si and P can form multiple bonds as C and N. But remember, this is the SECOND BEST choice and usually Si and P have to be forced to accept lower coordination numbers than 4, 5, or 6.

Thank you very much gentlemen for great informations in your answers ..

Great thankfulness for you  ..

Best Regard

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Amines/Properties_of_Amines/Oxidation_States_of_Nitrogen

Good Morning dear Dr.Shaden :

This case noticed especially with flour and oxygen ,these ions have higher electromagnetically than nitrogen comparing to the other ions which have lower ability to make its oxidation state positive .

Dear Sir Ammar A. Oglat ,and Dear Dr. Israa Witwit ,

My thanks and appreciation for your valuable contributions.

Best regards.

Although Axel and other colleagues have given instructive answers to this question, I would like to add some points.

1) The oxdidatio state is different from coordination number.

2) N can have oxidation satate of five but can't have coordination number of 5.

3) The main point is that why we can have PF5 but not NF5?

In addition to the reason given by Axel, the other reason that N can't have 5-coordinate compound is that N does not have d orbitals to extend its coordination to 5. P on the hand can have 5 and 6 coordinate compounds because it is in the third period of the periodic Table.

Dear Dr. @Mohammed Kooti

My thanks and appreciation for your valuable contributions.

Best regards.

Great question. I follow the answers.

Nitrogen Penta Chloride is not possable but Phosphorous Penta Chloride is possable . Very simple since N does not avail d-orbital as phosphorous can .

PCl5 has a stable TBP structure with sp3d hybridisation .

Valence bond and hybridisation are oversimplyfied and should not be used!

Thus, Jaydips answer is wrong: All atoms have s, p, d, f, g.... orbitals. Thus N has also d orbitals as P.

Dear Prof. Klein

In my answer that I mentioned" nitrogen does not have d orbitals" I meant that it does not have d orbitals as it's valence orbitals to be used in bonding.

Dear Mohammad, sure N very probably does not use the d orbitals since they are energetically far away from the valence states.

But it has d orbitals as any atom.

This is the problem of these oversimplifying models: they start on wrong assumptions.

Thanks a lot Prof. Klein for your complementary answer and also many thanks for your visit to my country.