MCQ
Correct stability order of the given tautomers is


- A$I > II > III$
- B$III > II > I$
- ✓$II > I > III$
- D$II > III > I$


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$C{{H}_{3}}OH\xrightarrow{HI}C{{H}_{3}}I\xrightarrow{KCN}$
$C{{H}_{3}}CN\xrightarrow{\text{reduction}}X\xrightarrow{HN{{O}_{3}}}Y$
$(i)$ $PCl_3 + 3H_2O \to H_3PO_3 + 3HCl$
$(ii)$ $SF_4 + 3H_2O \to H_3SO_3 + 4HF$
$(iii)$ $BCI_3 + 3H_2O \to H_3BO_3 + 3HCl$
$(IV)$ $XeF_6 + 3H_2O \to XeO_3 + 6HF$
Then according to given information the incorrect statement is
$(b)\, NaOH$ in $20\, gm$ will be respectively.
| Column-$I$ | Column-$II$ (Shape) |
| $(A) \,SF_4$ | $(1)$ Tetrahedral |
| $(B)\, BrF_3$ | $(2)$ Pyramidal |
| $(C)\, BrO_3^-$ | $(3)$ Sea-Saw shaped |
| $(D)\, NH_4^+$ | $(4)$ Bent $T-$ shaped |