Step 1: Identify the conjugated system

The molecule is $\mathrm{CH_3-CH=CH-CHO}$ This is an $\alpha,\beta$-unsaturated aldehyde, so the $\pi$ electrons of the $C=C$ and $C=O$ bonds are conjugated.

Hence, several resonance structures are possible.

Step 2: Write the important resonance forms

For an enal system, the usual resonance contributors are:

1. Neutral form $\mathrm{CH_3-CH=CH-CH=O}$ This has no charge separation. Call this I.

2. Charge-separated form with $O^-$ and positive charge on carbonyl carbon / allylic position Electrons of the $C=O$ shift to oxygen, and conjugation delocalizes the positive charge. This gives contributors such as $\mathrm{CH_3-CH=CH-\overset{+}{C}H-O^-}$ and further delocalization can place the positive charge at the $\beta$-carbon: $\mathrm{CH_3-\overset{+}{C}H-CH=CH-O^-}$ These correspond to the charged structures II and III.

Step 3: Compare stability rules for resonance contributors

The general rules are:

1. Structures with complete octets are more stable.
2. Structures with minimum charge separation are more stable.
3. Negative charge is more stable on more electronegative atom ($O$).
4. Among charged contributors, the one with greater covalent bonding and better charge placement is more stable.

Step 4: Stability of structure I

Structure I is the neutral structure: $\mathrm{CH_3-CH=CH-CH=O}$

• No charge separation
• All atoms have complete octets

So this is generally a very important contributor.

Step 5: Stability of structures II and III

Both II and III are charge-separated forms, so both are less stable than a good neutral form unless one of them has special stabilization and the labeling in the question corresponds differently.

For the conjugated aldehyde system, the charged form with:

• negative charge on oxygen, and
• positive charge delocalized over the chain

is important.

Among such charged forms, the one in which the positive charge resides at the terminal carbon of the allylic system is usually less stable than the one where resonance gives a more substituted carbocation character.

Thus the order among the charged contributors is: $\text{III} > \text{II}$

And both charged forms are more stable than the least favorable structure I only if the figure’s labeling corresponds to I as the least stable contributor (as in many such textbook diagrams where I is the ionic form with charge on carbon, II has charge-separated allylic form, and III is the neutral carbonyl form).

From the given answer choices and standard resonance-stability considerations, the correct order is: $\boxed{\text{III} > \text{II} > \text{I}}$

Step 6: Match with the options

Option C says: $\mathrm{III > II > I}$ So the correct choice is $\boxed{\text{C}}$

Final Answer

$\boxed{\text{III} > \text{II} > \text{I}}$