- ✓$H_2S < H_2Se < H_2Te$
- B$H_2Se < H_2S < H_2Te$
- C$H_2Te < H_2S < H_2Se$
- D$H_2Se < H_2Te < H_2S$
of central atom increases which weakens the
$\mathrm{M}-\mathrm{H}$ bond. Since, the size increases from $\mathrm{S}$ to $\mathrm{Te}$
thus acidic strength follows the order.
$H_{2} S < H_{2} S e < H_{2} T e$
Acidic nature $\propto \frac{1}{\text { Bond disspciation enthalpy }}$
$S$ to $Te$ size increases, bond dissociation enthalpy
decreases and acidic nature increases.
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Statement $I:$ In $Cl _{2}$ molecule the covalent radius is double of the atomic radius of chlorine.
Statement $II:$ Radius of anionic species is always greater than their parent atomic radius.
Choose the most appropriate answer from options given below :

Here $A$ is nothing but :
$CH_3-CH_2-OH$ $\mathop {\xrightarrow{{(i)\,KMn{O_4}/\mathop O\limits^\Theta H/\Delta }}}\limits_{(ii)\,{H^ \oplus }} (A)\mathop {\xrightarrow{{(i)\,SOC{l_2}}}}\limits_{(ii)\,N{H_3}/\Delta } (B)$ $\xrightarrow{{B{r_2}/KOH}}(C)$
$(C)$ will be :