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3. Part 2. Show your calculation or explanation for (b) and (c). (a) (3pt) The Molecular Orbital energy diagram of NO (nitrog
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Answer #1

* NO = 11 electrons from 2nd shell.

*2 sigma bonding M.O. and antibond. M.O. both filled(4 e) and 3sigma bonding Pz (2e) and both pi bonding 4Px and 4Py having equal energy (4 electrons) and 5 pi Px antibonding has 1 electron.

*B.O. of NO = 8-3/2 = 2.5

*B.O. of NO+ = 8-2/2 = 3.0

*B.O. of NO- = 8-4/2 = 2.0

* B.O. is inversely prop. to B. Length

*Shortest B. Length for NO+ (Highest B. O.)

2 4 mol 5 2 No 2. NO= 13 e-

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3. Part 2. Show your calculation or explanation for (b) and (c). (a) (3pt) The Molecular Orbital energy diagram of NO (nitrogen monoxide) is given in the right figure. Write the electron configur...
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