- ✓lead chromate
- Blead sulphate
- Clead iodide
- Dbasic lead acetate
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\mathrm{NaCl}+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{Conc} .) \rightarrow(\mathrm{A})+$ Side products
$(\mathrm{A})+\mathrm{NaOH} \rightarrow(\mathrm{B})+$ side product
$(\mathrm{B})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{dilute})+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow(\mathrm{C})+$ Side product
The sum of the total number of atoms in one molecule each of $(A), (B)$ and $(C)$ is
$M(s) \to M(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\, ........(1)$
$M(s) \to M^{2+} (g) + 2e^-\,\,\,\,\,\,\,\,.......(2)$
$M(g) \to M^+(g) + e^-\,\,\,\,\,\,\,\,\,\,\,.........(3)$
$M^+ (g) \to M^{2+} (g) + e^-\,\,\,\,\,\,\,\,\,.........(4)$
$M(g) \to M^{2+} (g) +2e^-\,\,\,\,\,\,\,\,\,\,\,..........(5)$
The second ionization energy of $M$ could be calculated from the energy values assoclated with
$n - Bu -\equiv\frac{(i) n-BuLi,n - C _{5} H _{11} Cl}{(ii) Lindlar\,\, cat, H _{2}}$