An element of p-block forms a species of type EH$_4^{+}$, which when passed through a basic solution of K$_2$[HgI$_4$], forms a brown ppt. Select the correct option:
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In p-block elements, the formation of brown precipitates with mercury salts is a characteristic reaction of ammonia (NH$_3$).
Element E has maximum electron affinity in its group.
EH$_3$ is phosphine.
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The Correct Option isB
Solution and Explanation
Step 1: Identifying the species and the element.
The given species is EH$_4^{+}$, and when passed through a basic solution of K$_2$[HgI$_4$], it forms a brown ppt. This reaction is characteristic of ammonia (NH$_3$), where the species formed is NH$_4^{+}$, and a brown ppt of HgO. Hg(NH$_3$)I is formed.
Step 2: Analyzing the options. (1) Element E has maximum covalency equal to 5: This is incorrect. The covalency of Nitrogen (N) in NH$_4^{+}$ is 4, not 5. (2) Brown ppt. formed is HgO. Hg(NH$_3$)I: Correct — This is the brown precipitate formed when ammonia reacts with the mercury(I) complex. (3) Element E has maximum electron affinity in its group: This is incorrect. Nitrogen does not have the maximum electron affinity in its group; it is oxygen. (4) EH$_3$ is phosphine: This is incorrect. EH$_3$ would correspond to ammonia (NH$_3$), not phosphine (PH$_3$).
Step 3: Conclusion.
The correct answer is (2) Brown ppt. formed is HgO. Hg(NH$_3$)I, as it correctly matches the reaction behavior of ammonia.
