Chemistry

Chlorine nitrate as hydrolysis gives hypohalous acid and nitric acid as,

$C\mathcal{l}ON{O}_2+H_{2}O\to \underset{A}{HOC\mathcal{l}}+\underset{B}{HN{O}_3}$                

Chlorine nitrate on reaction with HCl produces Cl₂ and HNO₃ as,

$C\mathcal{l}ON{O}_2+HC\mathcal{l}\to \underset{C}{C{\mathcal{l}}_2}+\underset{B}{HN{O}_3}$            

Electronic configuration of the given ions is :

$E{u}^{2+}-4f^{7}6s^0$                 

$\begin{array}{l}T{m}^{2+}-4f^{13}6s^0\\ S{m}^{2+}-4f^{6}6s^0\\ T{m}^{3+}-4f^{12}6s^0\end{array}$

$\begin{array}{l}T{b}^{4+}-4f^{7}6s^0\\ Y{b}^{2+}-4f^{14}6s^0\\ D{y}^{3+}-4f^{9}6s^0\\ y{b}^{3+}-4f^{13}6s^0\end{array}$              

Hence, Eu²⁺ and Tb⁴⁺ have half | filled f-orbitals and Yb²⁺ have completely filled f-orbitals.

In 3d-series, all metals except Cu have negative value of $E_{M^{2+}/M}^0$ due to low hydration enthalpy and high I.E and sublimation enthalpy.

Fluorine forms only one oxoacid which is hypofluorous acid HOF because it shows only – 1 oxidation state. Which is due to its the smallest size among halogens & the highest electronegativity.

$N{H}_{4}C\mathcal{l}\left(aq\right)+NaN{O}_2\left(aq\right)\to N_2\left(g\right)+2H_{2}O\left(\mathcal{l}\right)+NaC\mathcal{l}\left(aq\right)$  

Hence, N₂ gas is produced.

For isoelectronic species

$Ionicradii\propto \frac{Numberofe}{Numberofp}$              

$\begin{array}{l}N{a}^{+}M{g}^{2+}{N}^{3-}{O}^{2-}{F}^{-}\\ e=1010101010\\ p=1112789\end{array}$                

Hence, correct order of ionic radii is;

$M{g}^{2+}<N{a}^{+}<F^{-}<O^{2-}<N^{3-}$                

Higher the E.N. difference between hydrogen and other atom then higher be the strength of intermolecular H-bond

Here, order of difference in E.N is

O - H > N – H > C - H

Hence, correct order of H bond strength is,

CH₄ < HCN < NH₃

Leaching involves the given reaction,

$4Au\left(s\right)+8C{N}^{-}\left(aq\right)+2H_{2}O\left(aq\right)+O_2\left(g\right)\to 4{\left[Au{\left(CN\right)}_2\right]}^{-}\left(aq\right)+4O{H}^{-}\left(aq\right)$              

Here, O₂ is required for formation of Au (l) cyanide complex but no complex in absence of O₂.

$2{\left[Au{\left(CN\right)}_2\right]}^{-}\left(aq\right)+Zn\left(s\right)\to {\left[Zn{\left(CN\right)}_4\right]}^{2-}\left(aq\right)+2Au\left(s\right)$                

In above displacement reaction, Zn is oxidized.