Question

Increasing order of basic character of $MgO,SrO,K_{2}O,NiO\text{and}C{s}_{2}O$

• A. $C{s}_{2}O<K_{2}O<SrO<MgO<NiO$
• B. $C{s}_{2}O>K_{2}O>SrO>MgO>NiO$
• C. $C{s}_{2}O>K_{2}O>MgO>SrO>NiO$
• D. $K_{2}O>C{s}_{2}O>SrO>MgO>NiO$

Answer 1

The correct option is B $C{s}_{2}O>K_{2}O>SrO>MgO>NiO$

The proper increasing order of basic character is
$NiO<MgO<SrO<K_{2}O<C{s}_{2}O$

Generally metal oxides are basic in nature.
The basic character of oxides increases when we move down the group. So, $K_{2}O<C{s}_{2}O$ and $MgO<SrO$.

Again, basic character of the oxides decreases as we move from left to right across a period, thus $NiO$ has the least basicity.