Question

Match the items in column I with relevant items in column II.

$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \text{A}& \text{Ions having positive charge}& \text{(i)}& \text{+7}\\ \text{B}& \text{The sum of oxidation number of all atoms in a neutral molecule}& \text{(ii)}& \text{-1}\\ \text{C}& \text{Oxidation number of hydrogen ion}\left(H^{+}\right)& \text{(iii)}& \text{+1}\\ \text{D}& \text{Oxidation number of fluorine in NaF}& \text{(iv)}& \text{+0}\\ \text{E}& \text{Ions having negative charge}& \text{(v)}& \text{Cation}\\ & & \text{(vi)}& \text{Anion}\end{array}$

Answer 1

$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \text{A}& \text{Ions having positive charge}& \text{(v)}& \text{Cation}\\ \text{B}& \text{The sum of oxidation number of all atoms in a neutral molecule}& \text{(iv)}& \text{0}\\ \text{C}& \text{Oxidation number of hydrogen ion}\left(H^{+}\right)& \text{(iii)}& \text{+1}\\ \text{D}& \text{Oxidation number of fluorine in NaF}& \text{(ii)}& \text{-1}\\ \text{E}& \text{Ions having negative charge}& \text{(vi)}& \text{Anion}\end{array}$

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:
$\left(A\right)\to \left(v\right),\left(B\right)\to \left(iv\right),\left(C\right)\to \left(iii\right),\left(D\right)\to \left(ii\right),\left(E\right)\to \left(vi\right)$