Question

$\begin{array}{cccc}\text{Species}& \text{Protons}& \text{Neutrons}& \text{Electrons}\\ \text{A}& \text{17}& \text{18}& \text{18}\\ \text{B}& \text{11}& \text{12}& \text{10}\\ \text{C}& \text{17}& \text{20}& \text{17}\\ \text{D}& \text{13}& \text{14}& \text{12}\\ \text{E}& \text{15}& \text{16}& \text{15}\\ \text{F}& \text{9}& \text{10}& \text{10}\end{array}$

Choose the correct option which shows neutral, positively and negatively charged species.

• A. $\begin{array}{ccc}\text{Neutral}& \text{Positively charged}& \text{Negatively charged}\\ \text{C and D}& \text{B and E}& \text{A and F}\end{array}$
• B. $\begin{array}{ccc}\text{Neutral}& \text{Positively charged}& \text{Negatively charged}\\ \text{C and E}& \text{B and D}& \text{A and F}\end{array}$
• C. $\begin{array}{ccc}\text{Neutral}& \text{Positively charged}& \text{Negatively charged}\\ \text{C and E}& \text{F and D}& \text{A and B}\end{array}$
• D. $\begin{array}{ccc}\text{Neutral}& \text{Positively charged}& \text{Negatively charged}\\ \text{A and E}& \text{B and D}& \text{C and F}\end{array}$

Answer 1

The correct option is B $\begin{array}{ccc}\text{Neutral}& \text{Positively charged}& \text{Negatively charged}\\ \text{C and E}& \text{B and D}& \text{A and F}\end{array}$

Atomic number = Number of protons = Number of electrons

Mass number = Number of protons + Number of neutrons

Case 1: If the number of protons is equal to the number of electrons, the species is neutral.

Case 2: If the number of protons is more than the number of electrons, the species is a positively charged ion because it is formed by losing one or more electrons in order to achieve nearest noble gas configuration.

Case 3: If the number of protons is less than the number of electrons, the species is a negatively charged ion because it is formed by gaining one or more electrons in order to achieve nearest noble gas configuration.

Therefore, species C and E are neutral, B and D are positively charged and A and F are negatively charged.

Hence, the correct answer is option (b).