MCQ
Product $(B)$ is


- A

- B$CH_3-O-CH_2-CH_2-CH_3$
- ✓$H_2C = CH-CH_2 - O - CH_3$
- D





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Reason : Oxidation state of $Cl$ in $HClO_4$ is $+VII$ and in $HClO_3$ $+V$.


$(a)$ Lewis acidity order : $SiF_4 < SiC_{4} < SiBr_4 < Sil_4$
$(b)$ Melting point : $NH_3 > SbH_3 > AsH_3 > PH_3$
$(c)$ Boiling point : $NH_3 > SbH_3 > AsH_3 > PH_3$
$(d)$ Dipole moment order : $NH_3 > SbH_3 > AsH_3 > PH_3$
$(A)$ The number of $Cl = O$ bonds in $(ii)$ and $(iii)$ together is two
$(B)$ The number of lone pairs of electrons on $Cl$ in $(ii)$ and $(iii)$ together is three
$(C)$ The hybridization of $Cl$ in $(iv)$ is $sp ^3$
$(D)$ Amongst $(i)$ to $(iv)$, the strongest acid is $(i)$
$(iii)$ Size $(iv)$ Energy level