- A$H_2O$
- B$F_2O$
- ✓$Cl_2O$
- D$H_2S$
Oxygen difluoride is a chemical element having the symbol $F_2 O$. The bond angle of water is equal to $102^{\circ}$ and it does not have the largest bond angle. Hence, option $(B)$ is incorrect.
Among the given compounds, the $Cl _2 O$ has the largest bond angle and which is equal to $109.5^{\circ}$. The largest bond angle of chlorine monoxide is because of the presence of large lone pair - bond pair repulsion. And it will increase the bond angle of chlorine monoxide. And this type of repulsion also depends upon the electronegativity. Hence, option $(C)$ is correct.
Hydrogen sulphide is a chemical element having the symbol $H _2 S$. The bond angle of water is equal to $90^{\circ}$ and it does not have the largest bond angle. Hence, option $(D)$ is incorrect.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Reason : Oxidation state of $Cl$ in $HClO_4$ is $+7$ and in $HClO_3$, it is $+ 5$.

Given : $K _{ sp } Cu ( OH )_2=1 \times 10^{-20}$
$\operatorname{Take} \frac{2.303 RT }{ F }=0.059 \,V$
The reduction potential at $pH =14$ for the above couple is $(-) x \times 10^{-2}\,V$. The value of $x$ is $........$.

($1$) $W$ and $\mathbf{X}$ are, respectively
$[A]$ $O_3$ and $P_4 O_6$ $[B]$ $O_2$ and $P_4 O_6$ $[C]$ $O_2$ and $P_4 O_{10}$ $[D]$ $O_3$ and $P_4 O_{10}$
($2$) $Y$ and $Z$ are, respectively
$[A]$ $N_2 O_3$ and $\mathrm{H}_3 \mathrm{PO}_4$
$[B]$ $N_2 O_5$ and $\mathrm{HPO}_3$
$[C]$ $N_2 O_4$ and $HPO_3$
$[D]$ $N_2 O_4$ and $H_3 PO_3$
Give the answer of quetion ($1$) and ($2$)