- A$s-$ block
- B$p-$ block
- C$d-$ block
- ✓$f-$ block
Since, its last electron enter in $f- $ sub-shell, therefore it is a member of $f-$ block.
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$XeF_6 + H_2O \rightleftharpoons XeOF_4 + 2HF, \,\,\,\,K_1$
$XeO_4 + XeF_6 \rightleftharpoons XeOF_4 + XeO_3F_2,\,\,\,\, K_2$
The equilibrium constant for the reaction
$XeO_4 + 2HF \rightleftharpoons XeO_3F_2 + H_2O$ will be
Reason : $Six F$ atoms in $SF_6$ prevent the attack of $H_2O$ on sulphur atom of $SF_6$.
$\mathrm{NaOH}+\mathrm{Cl}_{2} \rightarrow(\mathrm{A})+$ side products
(hot and conc.)
$\mathrm{Ca}(\mathrm{OH})_{2}+\mathrm{Cl}_{2} \rightarrow(\mathrm{B})+$ side products
$(dry)$
$\mathrm{C}_{7} \mathrm{H}_{8} \stackrel{3 \mathrm{Cl}_{2} / \Delta}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{Br}_{2} / \mathrm{Fe}}{\rightarrow} \mathrm{B} \stackrel{\mathrm{Zn} / \mathrm{HCl}}{\rightarrow} \mathrm{C}$
The product '$C$' is
