In organic compound with molecular formula ($X$) $C_{8}H_{9}NO$ when reacts with $KOH/Br_{2}$ forms '$P$' which on diazotization forms '$Q$' followed by its reaction with $CuCN$ forms '$R$' which on hydrolysis (acidic) formed '$S$' ($S$ can also be made by hydrolysis of original compound ($X$) ($C_{8}H_{9}NO$). $S$ on reaction with $KMnO_{4}/H^{+}$ forms '$T$' which has two types of hydrogen $X$ will be :

$C_{6}H_{12}O_{3}$ gives positive iodoform test. On hydrolysis with dilute acid, the product formed gives both Tollens' test and iodoform test. Find the correct structure.

Given below are two statements . Statement-1: $HX$ bond length is higher in $HCl$ than $HF$ . Statement-2: The lowest boiling point in hydride of group 15 element is having covalency 4 .

Find the change in internal energy of a gas if its temperature changes by 10K. Moles=10, $C_{p}=7 cal K^{-1}mol^{-1}$, $R=2~cal~K^{-1}mol^{-1}$.

Arrange the given metal ion in increasing order of unpaired electron in low spin complexes formed by $Mn^{3+}, Cr^{3+}, Fe^{3+}, Co^{3+}$ :