Show which hybridized and unhybridized orbitals are involved in the formation of sigma and pi bonds in N3-.
AZIDE ION (e.g. N3-) is a very curious chemical species
As an ion, it is an electrically charged particle formed by Three Nitrogen Atoms.
So...WHICH IS HYBRIDIZATION STATUS OF THREE ATOMs?
Well, YOU HAVE CONSIDER THAT THREE ATOMs BY SAME CHEMICAL ELEMENTs MAY OBEY TO RESONANCE STRUCTUREs.
On the other hands, OCTET RULE STATE THAT "MOST STABLE AND MAIN STRUCTURE ASSURE EIGHT ELECTRONs AROUND EVERY ATOMs ENVIRON".
MAIN STRUCTURE FOR AZIDE ION ASSIGN
"sp or Linear Hybridization" for Central Nitrogen Atom whilst Outer Atoms play "sp2 or Trigonal Hybridization".
LEWIS DOT DIAGRAM FOLLOWS
where EVERY BAR DENOTES AN ELECTRON's COUPLE.
I hope this helps you.
N3 HybridizationSource(s): https://shrinke.im/bafqi
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Hybridization in N3-?
Show which hybridized and unhybridized orbitals are involved in the formation of sigma and pi bonds in N3-.Source(s): hybridization n3: https://tr.im/CIUNF
For the best answers, search on this site https://shorturl.im/avbwB
N3- no need of hybridisation. it is a monoatomic ion. No need of structure again. CO3- the hybridisation is sp2 on carbon. it is trigonal planar. NO3- too is trigonal planar. Hybridisation is sp2 again. In BF4- the hybridisation is sp3. And the shape is tetrahedral. Cheers
Formula to calculate hybridisation -
no. of atoms surrounding the central atom + 1 by 2 (grp no. of central atm - valency)
if the answer is 2....hybridisation is sp...if 3 sp2 if 4 sp3...!!