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why CLO2 do not form backbonding but CO does?

why CLO2 do not form backbonding but CO does?

Grade:12th pass

1 Answers

Vikas TU
14149 Points
5 years ago

The electrons are partially transferred from a d-orbital of the metal to anti-bonding molecular orbitals of CO (and its analogues). This electron-transfer (i) strengthens the metal–C bond and (ii) weakens the C–O bond. The strengthening of the M–CO bond is reflected in increases of the vibrational frequencies for the M–C bond (often outside of the range for the usual IR spectrophotometers). Furthermore, the M–CO bond length is shortened. The weakening of the C–O bond is indicated by a decrease in the frequency of the νCO band(s) from that for free CO (2143 cm−1), for example to 2060 cm−1 in Ni(CO)4 and 1981 cm−1 in Cr(CO)6, and 1790 cm−1 in the anion [Fe(CO)4]2−.[4] For this reason, IR spectroscopy is an important diagnostic technique inmetal–carbonyl chemistry. The article infrared spectroscopy of metal carbonyls discusses this in detail.

Many ligands other than CO are strong "backbonders". Nitric oxide is an even stronger π-acceptor than is CO and νNO is a diagnostic tool in metal–nitrosyl chemistry. Isocyanides, RNC, are another class of ligands that are able of π-backbonding. In contrast with CO, the σ-donor lone pair on the C atom of isocyanides is antibonding in nature and upon complexation the CN bond is strengthened and the νCN increased. At the same time, π-backbonding lowers the νCN. Depending on the balance of σ-bonding versus π-backbonding, the νCN can either be raised (for example, upon complexation with weak π-donor metals, such as Pt(II)) or lowered (for example, upon complexation with strong π-donor metals, such as Ni(0)). [5] For the isocyanides, an additional parameter is the MC=N–C angle, which deviates from 180° in highly electron-rich systems. Other ligands have weak π-backbonding abilities, which creates a labilization effect of CO, which is described by the cis effect.


CLO2 doesnot have any coordinate bond hence.

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