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(A) The reaction $2Na_2CrO_4 + 2H^{+} \rightarrow Na_2Cr_2O_7 + 2Na^{+} + H_2O$ shows that the product $B$ is sodium dichromate $(Na_2Cr_2O_7)$.The st

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$6$

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(A) The reaction $2Na_2CrO_4 + 2H^{+} \rightarrow Na_2Cr_2O_7 + 2Na^{+} + H_2O$ shows that the product $B$ is sodium dichromate $(Na_2Cr_2O_7)$.The structure of the dichromate ion $(Cr_2O_7^{2-})$ consists of two $CrO_4$ tetrahedra sharing a common oxygen atom at one corner.In this structure,there are $6$ terminal oxygen atoms (three bonded to each chromium atom) and $1$ bridging oxygen atom.Therefore,the number of terminal oxygen atoms is $6$.

The number of terminal oxygen atoms present in the product $B$ obtained from the following reaction is $....$
$4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2$
$2Na_2CrO_4 + 2H^{+} \rightarrow B + H_2O + 2Na^{+}$

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The number of terminal oxygen atoms present in the product $B$ obtained from the following reaction is $....$$4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2$$2Na_2CrO_4 + 2H^{+} \rightarrow B + H_2O + 2Na^{+}$