CO Analogs

Question 1

Name 4 CO Analogs

Answer 1

• Isocyanide
• Dinitrogen
• Nitrogen monoxide
  -Thiocarbonyl

Question 2

Compare isocyanide to carbonyl

Answer 2

• Electronical very similar ( both have triple bond and lone pair )
• Back donation the same: from MRNC and MCO
• IR stretching frequencies act the same: ν lower when ligand accept π electrons
• RNC bridging and terminal (like CO)
• RNC stronger σ donor than CO

Question 3

What is v

Answer 3

Wave number - how much the molecule stretches

Question 4

Explain IR stretching frequencies act the same: ν lower when ligand accept π electrons for isocyanide to carbonyl

Answer 4

• When a ligand accepts more π-electrons from the metal, its bond order decreases, leading to a lower IR stretching frequency
• Both CO and RNC ligands show this behavior: when they are bonded to electron-rich metals, the π backbonding effect weakens the C–O or C–N bond, reducing their IR stretching frequencies

Question 5

Compare dinitrogen and carbonyl

Answer 5

• CO and N2 isoelectronic ( same electron properties )
• M−NN bond is similar to that of M−CO ( bc of isoelectric same bonding nature due to lone pairs etc)
• NN occurs as bridging and terminal (like CO)
• σ donation: NNM like COM
• π back donation: MNN like MCO
• N≡N has weaker σ donor than CO ( N = higher electronegativity and is NP)
• N≡N has weaker π acceptor than CO
• MNN complexes less stable than MCO

Question 6

Explain the π back donation: MNN like MCO for dinitrogen and carbonyl

Answer 6

• Both ligands accept π electron density from the metal into their empty π rbitals
• This back-donation weakens the internal bond (N≡N or C≡O) and lowers the IR stretching frequency

Question 7

Explain cynanide

Answer 7

• CN− is a strong σ donor
• CN− is a weak π acceptor (weaker than CO, NO, RNC)
• CN− forms linear bridges: M−CN−M
• M−CN complexes usually anionic
• M−CN π bonding essential for stability of its complexes

Question 8

Proof of MNN back donation

Answer 8

• Stronger M−N bond and weaker N≡N bond
• Confirmed crystallographically: N≡N distance in M-NN longer than N≡N distance in free N2
• N≡N in M-NN still short enough to indicate multiple bond character

Question 9

Thiocarbonyl

Answer 9

• Free CS does not exist like CO under normal conditions
• CS exists in complexes (stabilized)

Stretching frequencies:

ν of CS = 1274 cm-1

ν of CS in MCS = 1270 - 1360 cm-1 (terminal CS)

Question 10

Nitrogen monoxide

Answer 10

• Linear, terminal MNO
• Bonding in M-CO: Metal has empty σ orbital and a pair of filled dπ orbitals (4 electrons)

Question 11

Bonding in M-NO

Answer 11

• Metal has empty σ orbital, dπ orbitals contains only 3 electrons
• The 4th electron is supplied by NO (NO has one more electron than CO, and it is in the π* orbital

Question 12

Metal Carbonyl Chemistry

Answer 12

T- he metal carbonyl bond is the bond between the metal and the carbon. This bond shows both the sigma and pi characters

• A sigma bond generally is made when the carbonyl carbon present in the complex gives an electrons’ lone pairs to the unoccupied orbital of the given metal. A pi bond usually made by the donation or contribution of an electrons’ pair from the occupied metal

Question 13

Metal carbonyls show the two types of bonding

Answer 13

• The formation of sigma bond from carbonyl to metal that is (M C) by the donation of electrons’ lone pair of carbon which is present in carbonyl group into unfilled orbital of the given metal atom. This is one of the dative overlaps.
• The formation of pi bond from metal to carbonyl that is (M C) by the donation of electrons from complete metal d orbitals into unfilled π* antibonding molecular orbitals of the carbonyl group 12
• The back bonding produces a synergic effect and it makes stronger the bond between carbonyl and the metal

Question 14

Bond strength

Answer 14

• Structural evidence: M-CO and M-NO equal strength
• Substitution reactions: CO displaced easier than NO

B. Bent, terminal MNO

• A type of M-NO bond not found with CO
• The bent angle is between 120 and 140°

Question 15

Bridging NO

Answer 15

• Less common than CO
• Capable of double and triple bridges
• Bridging wavenumber lower than terminal wavenumber

Question 16

Group 5B and 6B briefly

Answer 16

• P, As, Sb, Bi (trivalent)
• S, Se (divalent)

Question 17

Group 5B and 6B fully

Answer 17

• Strong electron donors (Lewis bases)
• Have dπ orbitals available for back acceptance
• Order of decreasing accepting ability (π acidity):
  CO ≈ PF3 > PCl3 > P(OR)3 > PR3
• PF3 is a better acceptor than PPh3….Why?
• Electronegativity?
  - More electronegative groups enhance back donation to 3d of P (PF3 > PCl3 > PR3)
• PR3 and the other ligands replace CO in a complex.
• All CO’s in a complex seldom replaced

[Rh(acac)(CO)2] + PPh3  [Rh(acac)(CO1)(PPh3)]