You are watching: Co+ express the bond order numerically to one decimal place.
I am unable to recognize why it is $3.5$ together I am in class 11.
For a long time it to be taught in school and also universities the the HOMO of carbon monoxide is anti-bonding. Without much more context that was also often taught that the bond order in CO is three, due to the fact that there space eight electron in bonding orbitals and also two in anti-bonding orbitals. $$ extBond order = frac12( extbonding - extanti-bonding)$$By assuming the the HOMO is anti-bonding (it is not!) and taking away one electron, the link order needs to increase to 3.5. This is wrong.
When we have a look at the MO diagram, a calculation version have the right to be found here, we understand that the HOMO, i.e. 3σ, is a bonding orbital, if the anti-bonding orbital is the 2σ. Upon ionisation, us would certainly remove one bonding electron and therefore the bond order needs to decrease come 2.5 as you suggested.However, the is not that easy. Strictly speaking the below MO system is, and MO theory itself, one approximation, and only one feasible configuration. When we perform not need to use resonance frameworks with MO theory, we have to think about other configuration (analogous to excited states). So normally the bond order the CO is not strictly 3. And removing an electron go not mean we are removing that from only one orbital, fairly than diminish the electron density. Because of this we can not accurately suspect the bond order v these basic considerations.Experimental observations and also theoretical calculations suggest that the bond certainly becomes stronger when removing an electron. See the linked question and Philipp"s answer in ~ for much more detail. (Don"t look at the various other answers, they space as wrong as they could be.)
In short: The link order that $ceCO$ is not exactly 3 and removing one electron will certainly not boost the bond order come 3.5. In both cases, the it was observed bond bespeak is most likely closer to 2.5, when experiments imply that the shortcut is more powerful in $ceCO+$.
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An orbital through bonding character has actually no node perpendicular come the link axis; one orbital with anti-bonding character contends least one node perpendicular to the bond axis (electron density is zero). Strictly speaking there room no non-bonding orbitals.