Q15 / 160 · WBJEE 2015 · Chemistry
$$
\begin{aligned}
&\text { The decreasing order of basic character of }\\
&\mathrm{K}_{2} \mathrm{O}, \mathrm{BaO}, \mathrm{CaO} \text { and } \mathrm{MgO} \text { is }
\end{aligned}
$$
- Correct answerA$\mathrm{K}_{2} \mathrm{O}>\mathrm{BaO}>\mathrm{CaO}>\mathrm{Mg} \mathrm{O}$
- B$\mathrm{K}_{2} \mathrm{O}>\mathrm{CaO}>\mathrm{BaO}>\mathrm{MgO}$
- C$\mathrm{MgO}>\mathrm{BaO}>\mathrm{CaO}>\mathrm{K}_{2} \mathrm{O}$
- D$\mathrm{MgO}>\mathrm{CaO}>\mathrm{BaO}>\mathrm{K}_{2} \mathrm{O}$
Solution
On the basis of electropositivity of metal, the order of basic character is
$$
\mathrm{K}_{2} \mathrm{O}>\mathrm{BaO}>\mathrm{CaO}>\mathrm{MgO}
$$
Hence, the correct answer is option: 'A'.