\(\begin{array}{|c|c|c|c|}
\hline
\text{} & \text{Column I} & \text{} & \text{Column II}\\\hline
\text{A} & \text{Ions having positive charge} & \text{(v)} & \text{Cation}\\\hline
\text{B} & \text{The sum of oxidation number of all atoms in a neutral molecule} & \text{(iv)} & \text{0}\\\hline
\text{C} & \text{Oxidation number of hydrogen ion \((H^+)\)} & \text{(iii)} & \text{+1}\\\hline
\text{D} & \text{Oxidation number of fluorine in NaF} & \text{(ii)} & \text{-1}\\\hline
\text{E} & \text{Ions having negative charge} & \text{(vi)} & \text{Anion}\\\hline
\end{array}\)
A) When atoms lose electrons, they exist as positive ions and are called cations.
B) Compound is electrically neutral and thus sum of oxidation number of all atoms is \(0\).
For example:
In \(FeO\): Oxidation number of \(Fe= +2\)
Oxidation number of \(O= −2\)
Sum of oxidation number \(= 2 + (−2)= 2−2=0\)
C) The charge on hydrogen ion is \(+1\), which signifies that hydrogen atom on losing one electron exists as \(H^+\) ion
Thus, oxidation number of \(H^+\) is \(+1\).
D) Oxidation number of fluorine is NaF Let oxidation number of \(F\) be \(x\)
\( +1+x=0\)
\(x=−1\)
Thus, oxidation number of fluorine in NaF is \(-1\).
E) When elements gain electrons, they exist as negative ions and are called anions.
Hence, correct answer is:
\((A)\rightarrow (v), (B) \rightarrow (iv), (C) \rightarrow (iii), (D) \rightarrow (ii), (E) \rightarrow (vi)\)