- AIonic character $ \Rightarrow MCl < MCl_2 < MCl_3$
- BPolarizibility $ \Rightarrow F^-< Cl^-< Br^-< I^-$
- ✓Polarising power $ \Rightarrow Ca^{+2} < Mg^{+2} < Al^{+3}$
- DCovalent character $ \Rightarrow LiF < LiCl < LiBr < LiI$
(Absolute value)
$Cl > F > Br > I$
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$I \frac{\left( CH _3 CO \right)_2 O }{ CH _3 COONa } J \xrightarrow[\begin{array}{l}\text { (ii) } SOCl _2 \\ \text { (ii) anhyd. } AlCl _3\end{array}]{\text { (i) } H _2, Pd / C } K$
$J \left( C _9 H _8 O _2\right)$ gives effervescence on treatment with $NaHCO _3$ and positive Baeyer's test
$1.$ The compound $K$ is
mcq $Image$
$2.$ The compound $I$ is
mcq $Image$
Give the answer question $1$ and $2.$
$Z\,\xrightarrow{{PC{l_5}}}\,X\,\xrightarrow[\Delta ]{{{\text{Alc}}{\text{.KOH}}}}\,Y\,\xrightarrow{{{H_2}O/{H^ \oplus }}}\,\mathop Z\limits_{{\text{(Major)}}} \,,\,\,Z$ is
$(i)$ $PCl_3 + 3H_2O \to H_3PO_3 + 3HCl$
$(ii)$ $SF_4 + 3H_2O \to H_3SO_3 + 4HF$
$(iii)$ $BCI_3 + 3H_2O \to H_3BO_3 + 3HCl$
$(IV)$ $XeF_6 + 3H_2O \to XeO_3 + 6HF$
Then according to given information the incorrect statement is
| Rate constant | Activation energy | |
| Step $1$ | $k_1$ | $E_{a1} = 180\ kJ/mol$ |
| Step $2$ | $k_2$ | $E_{a2} = 80\ kJ/mol$ |
| Step $3$ | $k_3$ | $E_{a3} = 50\ kJ/mol$ |
If overall rate constant, $k = {\left( {\frac{{{k_1}{k_2}}}{{{k_3}}}} \right)^{2/3}}$ ,then overall activation energy of the reaction will be .......... $ kJ/mol$