Oxides - Practice Questions & MCQ

$(\mathrm{A})-(\mathrm{IV}),(\mathrm{B})-(\mathrm{III}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{II})$

$(\mathrm{A})-(\mathrm{IV}),(\mathrm{B})-(\mathrm{II}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{III})$

$(\mathrm{A})-(\mathrm{II}),(\mathrm{B})-(\mathrm{IV}),(\mathrm{C})-(\mathrm{III}),(\mathrm{D})-(\mathrm{I})$

$(\mathrm{A})-(\mathrm{I}),(\mathrm{B})-(\mathrm{II}),(\mathrm{C})-(\mathrm{III}),(\mathrm{D})-(\mathrm{IV})$

$\mathrm{Cl}_2 \mathrm{O}_7$ is an anhydride of perchloric acid.

$\mathrm{O}_3$ molecule is bent

$O N F$ is isoelectronic with $\mathrm{O} 2 \mathrm{~N}^{-}$

Oxygen difluoride (OF2\mathrm{OF}_2OF2) is not an oxide of fluorine; rather, it is a fluoride of oxygen. In oxides, oxygen is typically the more electronegative element and carries a −2-2−2 oxidation state, as seen in compounds like H2O\mathrm{H_2O}H2O and CO2\mathrm{CO_2}CO2. However, in OF2\mathrm{OF}_2OF2, fluorine is the most electronegative element and dictates the oxidation states. Since fluorine always has a −1-1−1 oxidation state, the oxidation state of oxygen in OF2\mathrm{OF}_2OF2 is +2. This makes OF2\mathrm{OF}_2OF2 a fluoride of oxygen rather than an oxide of fluorine.

$\mathrm{SO}_3$

$\mathrm{SO}_2$

$\mathrm{TeO}_3$

$\mathrm{SeO}_3$

$\mathrm{Rb}_2 \mathrm{O}$

$P_4 O_{10}$

$\mathrm{SO}_3$

$\mathrm{CP}_2 \mathrm{O}_7$

$\mathrm{BaO}_2$

BeO

Na₂O

SO₃

CO

$\mathrm{CO}_2$

$\mathrm{C}_3 \mathrm{O}_2$

$\mathrm{CO}_3^{-2}$

A

B

C

D

$\mathrm{Na}_2 \mathrm{O}>\mathrm{MgO}>\mathrm{Al}_2 \mathrm{O}_3>\mathrm{SiO}_2>\mathrm{P}_2 \mathrm{O}_5>\mathrm{SO}_3$

$\mathrm{Na}_2 \mathrm{O}<\mathrm{MgO}<\mathrm{Al}_2 \mathrm{O}_3<\mathrm{SiO}_2<\mathrm{P}_2 \mathrm{O}_5<\mathrm{SO}_3$

$\mathrm{Na}_2 \mathrm{O}<\mathrm{MgO}<\mathrm{Al}_2 \mathrm{O}_3>\mathrm{SiO}_2>\mathrm{P}_2 \mathrm{O}_5>\mathrm{SO}_3$

$\mathrm{Na}_2 \mathrm{O}>\mathrm{MgO}>\mathrm{Al}_2 \mathrm{O}_3<\mathrm{SiO}_2<\mathrm{P}_2 \mathrm{O}_5<\mathrm{SO}_3$

$\mathrm{H}_2 \mathrm{~S}$

$\mathrm{SO}_3$

$\mathrm{SO}_2$

$\mathrm{O}_2$