Arrange the following compounds in order of decreasing basicity: mgo, cs2o, cl2o7, sno2, p4o10
1) mgo > cs2o > p4o10 > sno2 > cl2o7
2) cs2o > sno2 > mgo > p4o10 > cl2o7
3) mgo > cs2o > sno2 > p4o10 > cl2o7
4) cl2o7 > p4o10 > sno2 > cs2o > mgo
5) cs2o > mgo > sno2 > p4o10 > cl2o7

5) Cs2O > MgO > SnO2 > P4O10 > Cl2O7

Explanation:

Hello,

In this case, the decreasing basicity order depends on the metal contained into the oxide, in such a way, on the attached picture you will find how the basicity increases leftwards and downwards on the periodic table, as the metallic character does.

In this manner, as the cesium in the Cs₂O is the leftest atom, it is the most basic, next MgO, afterwards SnO₂, then P₄O₁₀ and finally Cl₂O₇, therefore, the order is: 5) Cs₂O> MgO > SnO₂> P₄O₁₀> Cl₂O₇.

Best regards.