Question

An atom A forms an oxide A2O5. What is the formula of bicarbonate of A?​

Answer 1

Given that , a element A forms an oxide A₂O₅ . And we need to find the formula of bicarbonate of A. We know that the bicarbonate is , $\bf HCO_3^-$ .

• We know that the valency of oxygen is 2 . So let the valency of A be x ,then by criss cross method , we see that the valency of A is 5 . Therefore we can find its formula of bicarbonate as ,

$\sf\dashrightarrow Valency_A =\boxed{\red{\sf 5}}$

$\sf\dashrightarrow Valency_{(HCO_3)} =\boxed{\red{\sf 1}}$

Therefore by cris - cross method ( refer to attachment) , the formula of the Compound will be ,

$\sf\dashrightarrow Compound = \boxed{\red{\sf A(HCO_3)_5}}$

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$\boxed {\begin{array}{|c|c|c|c|cccc|}\cline {1-8}\sf Element & \sf Symbol & \sf Valency & \sf{Atomic\:Number} & & \sf{Electron} & \sf Configuration & \\ \cline {5-8} &&&&\bf K&\bf L &\bf M&\bf N \\ \cline {1-8}\rm Hydrogen &\sf H & 1 & 1 & 1 &&& \\ \cline {1-8}\rm Helium &\sf He & 0 & 2 & 2 &&& \\ \cline {1-8}\rm Lithium &\sf Li & 1& 3 &2 &1 && \\ \cline {1-8}\rm Beryllium &\sf Be & 2 & 4 & 2 & 2 && \\ \cline {1-8}\rm Boron &\sf B & 3 & 5 & 2 & 3 && \\ \cline {1-8}\rm Carbon & \sf C & 4 & 6 & 2 & 4 && \\ \cline {1-8}\rm Nitrogen &\sf N & 3 & 7 & 2 & 5 && \\ \cline {1-8}\rm Oxygen &\sf O & 2 & 8 & 2 & 6 && \\ \cline {1-8}\rm Fluorine &\sf F & 1 & 9 & 2 & 7 && \\ \cline {1-8}\rm Neon & \sf Ne & 0 & 10 & 2 & 8 && \\ \cline {1-8}\rm Sodium &\sf Na & 1 & 11 & 2 & 8 & 1 &\\ \cline {1-8}\rm Magnesium &\sf Mg & 2 & 12 & 2 & 8 & 2 & \\ \cline {1-8}\rm Aluminium &\sf Al & 3 & 13 & 2 & 8 & 3 & \\ \cline{1-8}\rm Silicon &\sf Si & 4 & 14 & 2 & 8 & 4 & \\ \cline {1-8}\rm Phosphorus &\sf P & 3 & 15 & 2 & 8 & 7 & \\ \cline {1-8}\rm Sulphur &\sf S & 2 & 16 & 2 & 8 & 8 & \\ \cline{1-8}\rm Chlorine &\sf Cl & 1 & 17 & 2 & 8 & 8 &1 \\ \cline {1-8}\rm Argon &\sf Ar & 0 & 18 & 2 & 8 & 8 & 2 \\ \cline {1-8}\rm Potassium &\sf K & 1 & 19 & 2 & 8 & 8 &3 \\ \cline {1-8}\rm Calcium &\sf Ca & 2 & 20 & 2 & 8 & 8 & 4 \\ \cline{1-8} \end {array}}$

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