Chapter 3, Problem 3.46P

Label each of the following elements or regions in the periodic table.

1. a group that forms cations with $\text{a +2}$ charge
2. a group that forms anions with $\text{a - 2}$ charge
3. a group that forms cations with $\text{a + 1}$ charge
4. a group that forms anions with $\text{a - 1}$ charge
5. elements that form ions with the same electronic configuration as Ne
6. elements that form ions with the same electronic configuration as He

Interpretation Introduction

(a)

Interpretation:

Elements or regions representing the cations having +2 charge needs to be labeled.

Concept Introduction:

Electron configurations of noble gasses such as helium (He), neon (Ne) or argon (Ar) are stable because their electronic shells or subshells are completely filled. See below;

the electronic configuration of He = $1s^2$

the electronic configuration of Ne = $1s^{2}2s^{2}2p^6$

the electronic configuration of Ar = $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$

An atom of a main group element loses or gains electrons to obtain the electronic configuration of the noble gas closest to it in the periodic table to form ions.

There are two types of ions, cationic and anionic.

Atoms lose electron/s to form cations thus, they have fewer electrons than protons and are positively charged.

Atoms gain electron/s to form anion thus, they have more electrons than protons and are negatively charged.

E.g. Sodium (Na) atom has 11 electrons ( $1s^{2}2s^{2}2p^{6}3s^1$ ) and the last electronic shell ( $3s^1$ ) is not fully filled. In order to achieve the stable electronic configuration of the nearest noble gas, Na must lose one electron and become Na+ cation. Now, the electronic configuration of Na+ cation is similar to Ne, which is $1s^{2}2s^{2}2p^6$.

Answer to Problem 3.46P

  

Explanation of Solution

Metals form cations by losing electron/s from a valence shell to form completely filled outer shell of electrons.

Group 2A metals have only two electrons in the last electronic shell ( $s^2$ ). In order to achieve the stable electronic configuration of the nearest noble gas, these metals must lose two electrons and become 2+cations. Therefore, group 2A is labeled as a.

Interpretation Introduction

(b)

Interpretation:

Elements or regions representing the cations having -2 charge needs to be labeled.

Concept Introduction:

Electron configurations of noble gasses such as helium (He), neon (Ne) or argon (Ar) are stable because their electronic shells or subshells are completely filled. See below;

the electronic configuration of He = $1s^2$

the electronic configuration of Ne = $1s^{2}2s^{2}2p^6$

the electronic configuration of Ar = $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$

An atom of a main group element loses or gains electrons to obtain the electronic configuration of the noble gas closest to it in the periodic table to form ions.

There are two types of ions, cationic and anionic.

Atoms lose electron/s to form cations thus, they have fewer electrons than protons and are positively charged.

Atoms gain electron/s to form anion thus, they have more electrons than protons and are negatively charged.

Answer to Problem 3.46P

  

Explanation of Solution

Non-metals form anions by gaining one or two electrons, the resultant anions will have filled outer shells of electrons.

Non-metals in group 6A have six electrons in the last electronic shell ( $s^2{p}^4$ ). In order to achieve the stable electronic configuration of the nearest noble gas, these elements must gain two electrons and become -2cations. Therefore, group 6A is labeled as b.

Interpretation Introduction

(c)

Interpretation:

Elements or regions representing the cations having +1 charge needs to be labeled.

Concept Introduction:

Electron configurations of noble gasses such as helium (He), neon (Ne) or argon (Ar) are stable because their electronic shells or subshells are completely filled. See below;

the electronic configuration of He = $1s^2$

the electronic configuration of Ne = $1s^{2}2s^{2}2p^6$

the electronic configuration of Ar = $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$

An atom of the main group element loses or gains electrons to obtain the electronic configuration of the noble gas closest to it in the periodic table to form ions.

There are two types of ions, cationic and anionic.

Atoms lose electron/s to form cations thus, they have fewer electrons than protons and are positively charged.

Atoms gain electron/s to form anion thus, they have more electrons than protons and are negatively charged.

Answer to Problem 3.46P

  

Explanation of Solution

Metals form cations by losing electron/s, the resultant cation will have the completely filled outer shell of electrons.

Group 1A metals have only one electron in the last electronic shell ( $s^1$ ). In order to achieve the stable electronic configuration of the nearest noble gas, these metals must lose that one electron and become 1+cations. Therefore, group 1A is labeled as c.

Interpretation Introduction

(d)

Interpretation:

Elements or regions representing the cations having -1 charge needs to be labeled.

Concept Introduction:

Electron configurations of noble gasses such as helium (He), neon (Ne) or argon (Ar) are stable because their electronic shells or subshells are completely filled. See below;

the electronic configuration of He = $1s^2$

the electronic configuration of Ne = $1s^{2}2s^{2}2p^6$

the electronic configuration of Ar = $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$

An atom of a main group element loses or gains electrons to obtain the electronic configuration of the noble gas closest to it in the periodic table to form ions.

There are two types of ions, cationic and anionic.

Atoms lose electron/s to form cations thus, they have fewer electrons than protons and are positively charged.

Atoms gain electron/s to form anion thus, they have more electrons than protons and are negatively charged.

Answer to Problem 3.46P

  

Explanation of Solution

Non-metals form anions by gaining one or two electrons, the resultant anion will have the completely filled valence electronic shells.

Non-metals in group 7A have seven electrons in the last electronic shell ( $s^2{p}^5$ ). In order to achieve the stable electronic configuration of the nearest noble gas, these elements must gain two electrons and become -1cations. Therefore, group 7A is labeled as d.

Interpretation Introduction

(e)

Interpretation:

Regions representing the elements that form ions with the same electronic configuration as Ne needs to be labeled.

Concept Introduction:

Electron configurations of noble gasses such as helium (He), neon (Ne) or argon (Ar) are stable because their electronic shells or subshells are completely filled. See below;

the electronic configuration of He = $1s^2$

the electronic configuration of Ne = $1s^{2}2s^{2}2p^6$

the electronic configuration of Ar = $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$

An atom of a main group element loses or gains electrons to obtain the electronic configuration of the noble gas closest to it in the periodic table to form ions.

There are two types of ions, cationic and anionic.

Atoms lose electron/s to form cations thus, they have fewer electrons than protons and are positively charged.

Atoms gain electron/s to form anion thus, they have more electrons than protons and are negatively charged.

Answer to Problem 3.46P

  

Explanation of Solution

Elements loss or gain electrons to achieve the electronic configuration of the nearest noble gas.

The electronic configuration of Ne noble gas is $1s^{2}2s^{2}2p^6$ . The elements closer to Ne such as O, F, Na, Mg, and Al either loss or gain electrons to achieve Ne's electronic configuration. Therefore, the places of these elements in the periodic table are marked as e.

Interpretation Introduction

(f)

Interpretation:

Regions representing the elements that form ions with the same electronic configuration as He needs to be labeled.

Concept Introduction:

Electron configurations of noble gasses such as helium (He), neon (Ne) or argon (Ar) are stable because their electronic shells or subshells are completely filled. See below;

the electronic configuration of He = $1s^2$

the electronic configuration of Ne = $1s^{2}2s^{2}2p^6$

the electronic configuration of Ar = $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$

An atom of a main group element loses or gains electrons to obtain the electronic configuration of the noble gas closest to it in the periodic table to form ions.

There are two types of ions, cationic and anionic.

Atoms lose electron/s to form cations thus, they have fewer electrons than protons and are positively charged.

Atoms gain electron/s to form anion thus, they have more electrons than protons and are negatively charged.

Answer to Problem 3.46P

  

Explanation of Solution

Elements loss or gain electrons to achieve the electronic configuration of the nearest noble gas.

The electronic configuration of He noble gas is $1s^2$ . The elements closer to He are Li, Be and B. They lose one, two or three electrons, respectively to achieve He's electronic configuration. Therefore, the places of these elements in the periodic table are marked as f.