Question
Which of the following statement is correct
Since $X ^{-}$(negatively charged species has more electron density, hence it can give easily electron so its $IP$ is less than $X$ )
I$\varepsilon$ is greater than $\varepsilon$.A since in $\varepsilon . A$ of neutral species energy generally releases than in I$\varepsilon$ which require more energy to extract an electron.
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$n - Pr = n -$ propyl
Product' $(A)$ 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