Question:

Zn gives $H_2$ gas with $H_2SO_4$ and HCI but not with $HNO_3$ because :-

Updated On: Jul 12, 2022
  • $Zn$ act as oxidising agent when react with $ HNO_3$
  • $HBO_3$ is weaker acid than $H_2SO_4$ & HCI
  • In electrochemical series Zn is placed above hydrogen
  • $NO^{??_{3}$ is reduced in preference to hydronium ion
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The Correct Option is D

Solution and Explanation

Zn have lower value of $E_{cell}^* $ and easily gives oxidation. Zn is present above $H_2$ in electrochemical series. So, it liberates hydrogen gas from dilu. $HCI/H_2SO_4$. But $ HNO_3$ is an oxidising agent. The hydrogen obtained in this reaction is converted into $H_2O.$ In $HNO_3, \, NO_3^-$ ion is reduced and give $NH_4 \, NO_3, \, N_2O, \, NO \, and \, NO_2$ (based upon the concentration of $HNO_3$ ) [ $ Zn + 2HNO_3 \longrightarrow Zn \, ( NO_3)_2 + 2 H] \times 4 $ (Nearly 6% ) $ HNO_3 + BH \longrightarrow NH_3 + 3H_2O $ $ \frac{ NH_3 + HNO_2 \longrightarrow NH_4 \, NO_3}{ 4Zn + 10HNO_3 \longrightarrow 4 Zn ( NO_3)_2 + NH_4 \, NO_3 + 3H_2O} $
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Concepts Used:

D and F Block Elements

The d-block elements are placed in groups 3-12 and F-block elements with 4f and 5f orbital filled progressively. The general electronic configuration of d block elements and f- block elements are (n-1) d 1-10 ns 1-2 and (n-2) f 1-14 (n-1) d1 ns2 respectively. They are commonly known as transition elements because they exhibit multiple oxidation states because of the d-d transition which is possible by the availability of vacant d orbitals in these elements. 

They have variable Oxidation States as well as are good catalysts because they provide a large surface area for the absorption of reaction. They show variable oxidation states to form intermediate with reactants easily. They are mostly lanthanoids and show lanthanoid contraction. Since differentiating electrons enter in an anti-penultimate f subshell. Therefore, these elements are also called inner transition elements.

Read More: The d and f block elements