- A$35$
- ✓$36$
- C$37$
- D$38$
Hence no. of ${e^ - } = $ no. of protons $ = 36 = Z$.
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Toluene $\xrightarrow{{KMn{O_4}}}A\xrightarrow{{SOC{l_2}}}$ $B\xrightarrow[{BaS{O_4}}]{{{H_2}/Pd}}C$
the product $C$ is :
| Column $I$ | Column $II$ | ||
| $(A)$ | Kohlrausch law can calculate | $(P)$ | $\frac{{\Lambda _m^c}}{{\Lambda _m^o}}$ |
| $(B)$ | Molar conductance ${\Lambda _m}$ | $(Q)$ | $\frac{1}{R} \times \frac{l}{A}$ |
| $(C)$ | Specific conductance Kappa $\to (k)$ | $(R)$ | $\Lambda _m^o\,of\,c{a_3}{(P{O_4})_2}$ |
| $(D)$ | Degree of ionization of weak electrolyte | $(S)$ | $\frac{{k \times 1000}}{M}$ |
Which of the following option show correct matches
${C_2}{H_5}Br\xrightarrow{X}{\text{product}}\xrightarrow{{\text{Y}}}{C_3}{H_7}N{H_2}$
$(i)$ Pure solvent $\to$ separated solvent molecules $\Delta$ $H_1$
$(ii)$ Pure solute $\to$ separated solute molecules$\Delta$ $H_2$
$(iii)$ Separated solvent and solute molecules $\to$ solution $\Delta H_3$ Solution so formed will be ideal if
$\begin{array}{*{20}{c}}
{\,\,O} \\
{\,\,||} \\
{C{H_3} - C - OH}
\end{array}$ $\xrightarrow[\Delta ]{{N{H_3}}}A\,\xrightarrow[\Delta ]{{{P_2}{O_5}'}}B$