Answer :
It is true that HI cannot be prepared by the reaction of Sodium iodide (NaI) and Sulphuric acid (H₂SO₄).
When solid sodium chloride (NaCl) is treated with concentrated Sulphuric acid (H₂SO₄), sodium bisulphate (NaHSO₄) and hydrochloric acid (HCl) is produced.
The equation for the above reaction is as follows-
NaCl (s) + H₂SO₄ (s) → NaHSO₄ (g) + HCl (g) ↑
If, NaCl is substituted with NaI and treated with H₂SO₄, the following compounds are produced.
The equation for the above reaction is as follows-
NaI (aq) + H₂SO₄ (aq) → I₂ (s) + H₂S (g) + H₂O (l)
In the above reaction HI is not formed because H₂SO₄ is an oxidising agent and so it oxidises HI but HCl is not oxidised by H₂SO₄.
Hence, HCl is prepared by heating NaCl with conc. H₂SO₄ but HI cannot be prepared by treating NaI with conc. H₂SO₄. NaI being a reducing agent, H₂SO₄ undergoes precipitation reaction with NaI.
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