Choose the correct decreasing order of the basic properties of oxides:

T l_{2} O > A l_{2} O_{3} > T l_{2} O_{3} > G a_{2} O_{3}

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A l_{2} O_{3} > T l_{2} O_{3} > G a_{2} O_{3} > T l_{2} O

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T l_{2} O > T l_{2} O_{3} > G a_{2} O_{3} > A l_{2} O_{3}

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G a_{2} O_{3} > T l_{2} O > A l_{2} O_{3} > T l_{2} O_{3}

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Solution

The correct option is C $T l_{2} O > T l_{2} O_{3} > G a_{2} O_{3} > A l_{2} O_{3}$
Here, the anions are the same so we have to check charges on the cation. The lower the charge on the ions, the more ionic or more basic is the nature.
Here, the charge of aluminium, tellurium $(T l_{2} O_{3})$ , gallium ion is the same. Larger the cation, more ionic and more basic would be the oxide. All of them belong to group 13. Therefore, size of $T l^{3 +} > G a^{3 +} > A l^{3 +}$ .
Hence correct decreasing order basic oxides:
$T l_{2} O > T l_{2} O_{3} > G a_{2} O_{3} > A l_{2} O_{3}$

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