2006-12-28 00:39:06 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
N2 has 4 bonding molecular orbital and 1 anti bonding molecular orbital.
So bond order in N2 is 3.
NO has 4 bonding molecular orbital and 1.5 anti bonding molecular orbital (1 sigma* 2S, 0.5 pi* 2P)
So bond order in NO is 2.5.
2006-12-28 00:53:46 · answer #1 · answered by Ash 2 · 0⤊ 0⤋
Nitrogen = N2 (14 electrons). Its molecular orbital configuration is:
sigma 1s2, antisigma 1s2, sigma 2s2, antisigma 2s2, pi 2py2, pi 2pz2, sigma 2p2
NO = 15 electrons (use same orbital list as for N2(-1)
same orbitals as N2 except you end with antipi 2pz1
Bond order = (bonding electrons - antibonding electrons)/2
For N2 = (10-4)/2 = 3
For NO = (10-5)/2 = 2.5
2006-12-28 00:48:46 · answer #2 · answered by The Old Professor 5 · 0⤊ 0⤋
N = 1s2 2s2 2p3 so N has in basic terms 3 valancies to end its octet N makes 3 bonds.... O=N-O N makes a double bond with 1st oxygen and a million valency remains in N so its make a unmarried bond with different oxygen
2016-12-18 20:28:44 · answer #3 · answered by ? 4 · 0⤊ 0⤋
remember that you determine bond order by subtracting the number anti-bonded electrons from the number of bonded electrons... so once you fill in the MOs, you can just count and subtract.
oops forgot... divide by 2!
2006-12-28 04:03:16 · answer #4 · answered by Anonymous · 0⤊ 0⤋