Q73P - Chemistry: Molecular Nature Of Matter And Change

Question

For each ionic depiction, give the name of the parent atom, its mass number, and its group and period numbers:

Step-by-Step Solution

Answer

Step 1: Identify the atomic number of the elements

The mass number of the element is the sum of the neutrons and the protons. The atomic number of any element is determined by the number of protons present in that element. Hence, the name of the atom or element is depicted by the number of protons. 

In a neutral atom, the number of protons is equal to the number of electrons.

Step 2: Determine the parent atom of the given ionic atoms for (a)

From the ionic depiction of atom (a), 

 $$\begin{gathered} \mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{proton}\mathbf{=}\mathbf{35} \\ \mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{electrons}\mathbf{=}\mathbf{36} \\ \mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{neutrons}\mathbf{=}\mathbf{44} \end{gathered}$$ 

Therefore, the atomic number of the ion is equal to the number of protons, that is, 35.

$$\begin{gathered} \mathbf{Mass}\mathbf{ }\mathbf{number}\mathbf{ }\mathbf{=}\mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{proton}\mathbf{+}\mathbf{ }\mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{neutrons} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{35}\mathbf{+}\mathbf{44} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{79} \end{gathered}$$ 

Hence, the atom with the atomic number 35 is Bromine (Br) and belongs to group 7A. It is found in period 4 in the periodic table.

Step 3: Determine the parent atom of the given ionic atoms for (b)

From the ionic depiction of atom (b), 

  $$\begin{gathered} \mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{proton}\mathbf{=}\mathbf{7} \\ \mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{electrons}\mathbf{=}\mathbf{10} \\ \mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{neutrons}\mathbf{=}\mathbf{8} \end{gathered}$$

Therefore, the atomic number of the ion is equal to the number of protons, that is, 7.

 $$\begin{gathered} \mathbf{Mass}\mathbf{ }\mathbf{number}\mathbf{ }\mathbf{=}\mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{proton}\mathbf{+}\mathbf{ }\mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{neutrons} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{7}\mathbf{+}\mathbf{8} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{15} \end{gathered}$$ 

Hence, the atom with the atomic number 7 is Nitrogen (N) and belongs to group 5A. It is found in period 2 in the periodic table.

Step 4: Determine the parent atom of the given ionic atoms for (c)

From the ionic depiction of atom (c), 

 $$\begin{gathered} \mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{proton}\mathbf{=}\mathbf{37} \\ \mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{electrons}\mathbf{=}\mathbf{36} \\ \mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{neutrons}\mathbf{=}\mathbf{48} \end{gathered}$$ 

Therefore, the atomic number of the ion is equal to the number of protons, that is, 37.

 $$\begin{gathered} \mathbf{Mass}\mathbf{ }\mathbf{number}\mathbf{ }\mathbf{=}\mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{proton}\mathbf{+}\mathbf{ }\mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{neutrons} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{37}\mathbf{+}\mathbf{48} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{85} \end{gathered}$$

Hence, the atom with the atomic number 37 is Rubidium (Rb) and belongs to group 1A. It is found in period 5 in the periodic table.