- A${H_5}{P_3}{O_{10}}$
- B${H_6}{P_4}{O_{13}}$
- ✓${H_5}{P_5}{O_{15}}$
- D${H_7}{P_5}{O_{16}}$
Therefore only $H _5 P _5 O _{15}=\left( HPO _3\right)_5$ satisfies this.
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Reason $(R)$ : Oxygen forms $\mathrm{p} \pi-\mathrm{p} \pi$ multiple bonds with itself and other elements having small size and high electronegativity like $\mathrm{C}, \mathrm{N}$, which is not possible for sulphur.
In the light of the above statements, choose the most appropriate answer from the options given below :
How many structural isomers are possible when one of the hydrogens is replaced by a chlorine atom?
Statement $I$: Group $13$ trivalent halides get easily hydrolyzed by water due to their covalent nature.
Statement $II$: $\mathrm{AlCl}_3$ upon hydrolysis in acidified aqueous solution forms octahedral $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ ion.
In the light of the above statements, choose the correct answer from the options given below :