Question

Select correct order(s).

• A. $\text{Order of density}:H_{2}O<D_{2}O$
• B. $\text{Boiling Point of isomeric amines}:1^o>2^o>3^o$
• C. $\text{Ionization constant}\left[H^{+}\right]\left[O{H}^{–}\right]:H_{2}O>D_{2}O$
• D. $\text{Vapour pressure at R.T.}:N{H}_3<HF<H_{2}O$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\text{Order of density}:H_{2}O<D_{2}O$
B $\text{Boiling Point of isomeric amines}:1^o>2^o>3^o$
C $\text{Ionization constant}\left[H^{+}\right]\left[O{H}^{–}\right]:H_{2}O>D_{2}O$

(a) As we know, $\text{Density}=\frac{\text{Mass}}{\text{Volume}}$
Mass of deuterium is 2 amu. So, density of deuterium will be more.

(b) As number of $H$ atoms bonded to $N$ increases $H–$bonding increases and boilling point also increases. $1^o$ amines form more hydrogen bonds.

(c) Electronegativity of hydrogen atom is more than deuterium due to which bond strength is more stronger and hence, ionization constant of $H_{2}O>D_{2}O$.

(d) More is the extent of H–bonding among the molecules, lower will be the vapour pressure. Though $F$ atom has highest electronegativity, but due to the presence of extra $H$ atom in $H_{2}O$, extent of H- bonding is highest in $H_{2}O$ and least in $N{H}_3$.
$\text{Vapour pressure at R.T.}:H_{2}O<HF<N{H}_3$