In the reaction particle diagram $CO + O N O \rightarrow N O + O C O$ (carbon monoxide reacting with $ONO$), which collision orientation (1) $OC \rightarrow ONO$, (2) $CO \rightarrow ONO$, or (3) $C \rightarrow NOO$ is most likely to break an $N\text{-}O$ bond and form a new $C\text{-}O$ bond?

What you are looking for in an effective collision

To break an $N\text{-}O$ bond and form a new $C\text{-}O$ bond, the collision has to bring the atom that will make the new bond (carbon) into contact with the atom it will attach to (oxygen). This is the “right orientation” part of collision theory.

Connecting the particle diagram to the chemistry

The overall change shown by

$$CO + ONO \rightarrow NO + OCO$$

matches the common reaction idea: carbon monoxide reacts by taking an oxygen, producing $CO_2$ (shown as $OCO$) and leaving $NO$.

That means the key bond-forming step is: carbon in $CO$ bonds to an oxygen in $ONO$, while an $N\text{-}O$ bond in $ONO$ breaks.

Checking each orientation

• 1. $OC \rightarrow ONO$: the oxygen end of $CO$ leads the collision. That does not directly promote forming a new bond between carbon and oxygen.
• 2. $CO \rightarrow ONO$: the carbon end leads. This places carbon where it can form the new $C\text{-}O$ bond while an $N\text{-}O$ bond breaks.
• 3. $C \rightarrow NOO$: carbon approaches the nitrogen side (or at least not clearly an oxygen). That would favor a $C\text{-}N$ interaction, not the needed $C\text{-}O$ bond.

So, orientation 2 is the best choice.

Quick self-check

Ask: “Which collision puts C next to O?” The diagram that does is the one most likely to produce $OCO$ ($CO_2$).