Sum of number of oxygen atoms in S and T is ______.

Step 1 - Index of hydrogen deficiency for $$P$$
For an open-chain compound $$C_9H_{18}O_2$$ the index is
$$\text{IHD}= \dfrac{2\times 9+2-18}{2}=1$$
So $$P$$ possesses one and only one π bond (no ring).

Step 2 - Nature of functional groups in $$P$$
$$P$$ decolorises bromine water ⇒ the single IHD corresponds to a $$C=C$$ bond.
$$P$$ also gives a positive iodoform test; without any carbonyl present (IHD already used up), this can only arise from a secondary alcohol of the type $$CH_3-C^*(OH)-$$. Hence both oxygen atoms of $$P$$ are present as non-carbonyl oxygens (alcohol/ether). In short, $$P$$ is an >C=C< containing di-oxygenated alcohol (or alcohol-ether).

Step 3 - Oxidative ozonolysis of the double bond
Ozonolysis followed by $$H_2O_2/H_2O$$ cleaves the $$C=C$$, converting each double-bond carbon into a -COOH group while leaving every pre-existing -OH intact. Thus both fragments, $$Q$$ and $$R$$, are $$\beta$$-hydroxy carboxylic acids.

Step 4 - Why $$Q$$ shows, but $$R$$ lacks, the iodoform test
The iodoform test in acids is positive only when the acid contains a $$CH_3CO-$$ or $$CH_3CH(OH)-$$ unit. Therefore the -OH in $$Q$$ is on the carbon that already bears a methyl group, giving $$CH_3-C(OH)(\ldots )-COOH$$; $$R$$ lacks this arrangement.

Step 5 - Oxidation with PCC followed by heating
(i) PCC oxidises the secondary -OH in each $$\beta$$-hydroxy acid to a carbonyl group, producing a $$\beta$$-keto acid.
(ii) A classic property of $$\beta$$-keto acids is thermal decarboxylation:
$$R-CO-CH_2-COOH \xrightarrow{\Delta} R-CO-CH_3 + CO_2$$
Thus $$Q$$ → $$S$$ and $$R$$ → $$T$$, where both $$S$$ and $$T$$ are methyl ketones.

Step 6 - Counting oxygen atoms in $$S$$ and $$T$$
A methyl ketone contains only one oxygen atom (the carbonyl oxygen).
Hence $$n_O(S)=1$$ and $$n_O(T)=1$$.

Step 7 - Required sum
$$n_O(S)+n_O(T)=1+1=2$$

Therefore, the sum of the number of oxygen atoms present in $$S$$ and $$T$$ is 2.