Certainly! Let's analyze the given data and provide a detailed solution for each part of the question.
Question 3
(a) Study the table and answer the following questions:
[tex]\[
\begin{array}{|c|c|}
\hline
\text{Atom} & \text{Atom No.} \\
\hline
\text{A} & 11 \\
\text{B} & 17 \\
\hline
\end{array}
\][/tex]
(i) Compare the positions of A and B in the Periodic Table.
Atom A has the atomic number 11, which corresponds to Sodium (Na). Sodium is located in Group 1 and Period 3 of the Periodic Table.
Atom B has the atomic number 17, which corresponds to Chlorine (Cl). Chlorine is located in Group 17 and Period 3 of the Periodic Table.
Therefore, the position of A (Sodium) is "Group 1, Period 3" and the position of B (Chlorine) is "Group 17, Period 3".
(ii) Which is more metallic?
Metallic character in elements decreases as you move from left to right across a period. Since Sodium (Atom A) is in Group 1 and Chlorine (Atom B) is in Group 17 of the same period, Sodium is more metallic than Chlorine.
Hence, the more metallic element is A (Sodium).
(iii) Write equations for the formation of ions of [tex]\(A\)[/tex] and [tex]\(B\)[/tex].
Sodium (A) forms ions by losing one electron, leading to the formation of a sodium ion [tex]\(\text{Na}^+\)[/tex]. The equation for this process is:
[tex]\[ \text{Na} \rightarrow \text{Na}^+ + e^- \][/tex]
Chlorine (B) forms ions by gaining one electron, leading to the formation of a chloride ion [tex]\(\text{Cl}^-\)[/tex]. The equation for this process is:
[tex]\[ \text{Cl} + e^- \rightarrow \text{Cl}^- \][/tex]
(iv) What type of bond is formed between [tex]\(A\)[/tex] and [tex]\(B\)[/tex]? Mention its physical state and solubility in water.
Sodium (a metal) and Chlorine (a non-metal) form an ionic bond. In an ionic bond, electrons are transferred from the metal atom to the non-metal atom, resulting in the formation of oppositely charged ions that attract each other.
Therefore, the bond type between Sodium (Na) and Chlorine (Cl) is an ionic bond. The compound formed is sodium chloride ([tex]\(\text{NaCl}\)[/tex]).
The physical state of sodium chloride is solid at room temperature, and it is soluble in water.
In summary:
[tex]\[
\begin{align*}
\text{Positions in the Periodic Table:} & \\
\text{A:} & \text{"Group 1, Period 3"} \\
\text{B:} & \text{"Group 17, Period 3"} \\
\text{More metallic:} & \text{A (Sodium)} \\
\text{Ion formation equations:} & \\
\text{A:} & \text{Na} \rightarrow \text{Na}^+ + e^- \\
\text{B:} & \text{Cl} + e^- \rightarrow \text{Cl}^- \\
\text{Bond type:} & \text{Ionic bond} \\
\text{Physical state:} & \text{Solid} \\
\text{Solubility in water:} & \text{Soluble}
\end{align*}
\][/tex]
