Match the items of Column I and Column II based on data given below:

Column I	Column II
(i)F2	(a)metal is the strongest reducing agent
(ii)Li	(b)metal ion which is the weakest oxidizing agent
(iii)Au 3+	(c)non-metal which is the best oxidizing agent
(iv)Br −	(d)unreactive metal
(v)Au	(e)anion that can be oxidized by Au 3+
(vi)Li+	(f)anion which is the weakest reducing agent
(vii)F−	(g)metal which is an oxidising agent

E $\frac{\mathit{G}}{\mathit{R}}$ _F2/F=2.87V  E $\frac{\mathit{K}}{\mathit{C}}$ _Li/Li= 3.5V E_Au3+/Au = 1.4V E $\mathit{}\frac{\mathbf{\Lambda }\mathbf{m}}{{\mathbf{\Lambda }}^0\mathbf{m}}$ _Br2/Br^- = 1.09V

<p>This is a Matching Type Questions as classified in NCERT Exemplar</p><p><strong>Ans: </strong>&nbsp; (i) F₂is a non-metal which is the best oxidizing agent since the standard reduction potential of F₂ is 2.87V.</p><p> (ii)Liis a metal and the strongest reducing agent because the standard reduction potential of Li is −3.05V.</p><p> (iii)Au³⁺is a metal ion which is an oxidizing agent as standard reduction potential of Au³⁺ is 1.40V.</p><p> (iv) Br⁻is an anion that can be oxidized by Au³⁺ as standard reduction potential of Au³⁺is 1.40V which is more than that of Br⁻ which is E<span contenteditable="false"> <math> <mi mathvariant="bold-italic">Q</mi> </math> </span>_Br2/Br−0=1.09V.</p><p> (v)Auis an unreactive metal</p><p> (vi)Li⁺&nbsp;is a metal ion having the least value of standard reduction potential of 3.05V. Hence it is the weakest oxidizing agent</p><p> (vii)F⁻is an anion that is the weakest reducing agent as it has less oxidation potential.</p><p>Hence, the answer is:</p><p> (i)-c; (ii)-a; (iii)-g; (iv)-e; (v)-d; (vi)-b; (vii)-f</p>

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