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(D) The reaction of lanthanoids $(Ln)$ with oxygen produces the stable oxide $(Ln_2O_3)$:$4Ln + 3O_2 \rightarrow 2Ln_2O_3$ (Compound $A$)When the lant

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$Ln_2(CO_3)_3$

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(D) The reaction of lanthanoids $(Ln)$ with oxygen produces the stable oxide $(Ln_2O_3)$:$4Ln + 3O_2 \rightarrow 2Ln_2O_3$ (Compound $A$)When the lanthanoid oxide $(Ln_2O_3)$ reacts with excess $CO_2$,it forms the corresponding carbonate $(Ln_2(CO_3)_3)$:$Ln_2O_3 + 3CO_2 \rightarrow Ln_2(CO_3)_3$ (Compound $B$)Therefore,the formula of compound $(B)$ is $Ln_2(CO_3)_3$.

Identify the formula of compound $(B)$ formed in the following reaction:
$(i)$ Lanthanoids $(Ln)$ are burnt in oxygen to form compound $(A)$.
$(ii)$ Compound $(A)$ reacts with $CO_2$ in excess to form compound $(B)$.

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Identify the formula of compound $(B)$ formed in the following reaction:$(i)$ Lanthanoids $(Ln)$ are burnt in oxygen to form compound $(A)$.$(ii)$ Compound $(A)$ reacts with $CO_2$ in excess to form compound $(B)$.