Chemistry

Not all dipole – dipole interactions are responsible for hydrogen bonding, so assertion A is false. F is most E. N atom and also hydrogen bonds in HF are symmetrical.

$K_{2}C{r}_2{O}_7+H_{2}S{O}_4+\underset{GasX}{S{O}_2}\to \underset{\begin{array}{l}Green\\ Y\end{array}}{C{r}_2{\left(S{O}_4\right)}_3}+K_{2}S{O}_4+H_{2}O$

Hydrogen peroxide reduces iodine to iodide ion is basic medium as;

$H_2{O}_2+2O{H}^{-}+l_2\to O_2+2l^{-}+2H_{2}O$

Here A is CH₃ - CH₂ – CH₂ – COO – CH₂ – CH₃

The given sequence of reaction is:

$C{H}_3-C{H}_2-C{H}_2-C{H}_2-OH\underset{}{\overset{Oxidation}{\to }}C{H}_3-C{H}_2-C{H}_2-COOH$

Ce, Pr, Nd, Tb and Dy are the only lanthanoids which shows + 4 O.S, so can form MO₂, but Yb does not shows +4 O.S so can't form MO₂.

Only 1° aromatic amines give stable diazonium salt on reaction with nitrous acid

Here 1° aromatic amine is  which gives most stable diazonium salt

N > O > Be > B

Due to half filled configuration N has more I.E than oxygen and due to fully filled configuration Be has more I.E than B.

(a) → (ii)                (b) → (iii)            (c) → (iv)            (d) → (i)