Chapter 3, Problem 35P

For each of the general electron-dot formulas for elements, give the following information : [1] the number of valence electrons; [2] the group number of the element; [3] how many electrons would be gained or lost to achieve a noble gas configuration; [4] the charge on the resulting ion; [5] an example of the element.
a. X.

Interpretation Introduction

(a)

Interpretation:

Number of electrons must be gained/lost by cesium to achieve a noble gas configuration should be determined.

Concept Introduction:

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

Here,

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.

Cations are formed by losing electrons. They are positively charged.

Anions are formed by gaining electrons. They are negatively charged.

For example, 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 35P

1 electron must be lost.

Explanation of Solution

Cesium (Cs) has 55 electrons. The electronic configuration of Cs is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^1$.

The nearest noble gas to Cs is Xenon (Xe), whose electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^6$.

Therefore, Cs must lose one electron to achieve Xe atom's electronic configuration.

Because an electron is lost, the result is Cs⁺ cation.

Interpretation Introduction

(b)

Interpretation:

Number of electrons must be gained/lost by barium to achieve a noble gas configuration should be determined.

Concept Introduction:

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

Here,

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.

Cations are formed by losing electrons. They are positively charged.

Anions are formed by gaining electrons. They are negatively charged.

For example, 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 35P

2 electrons must be lost.

Explanation of Solution

Barium (Ba) has 56 electrons. The electronic configuration of Cs is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^2$.

The nearest noble gas to Cs is Xenon (Xe), whose electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^6$.

Therefore, Ba must lose two electrons to achieve Xe atom's electronic configuration.

Because two electrons are lost, the result is Ba²⁺ cation.

Interpretation Introduction

(c)

Interpretation:

Number of electrons must be gained/lost by selenium to achieve a noble gas configuration should be determined.

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.

Here,

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.

Cations are formed by losing electrons, thus they have fewer electrons than protons and are positively charged.

Anions are formed by gaining electrons, thus they have more electrons than protons and are negatively charged.

For example, fluorine (F) atom has 9 electrons ( $1s^{2}2s^{2}2p^5$ ) and the last electronic shell ( $2s^{2}2p^5$ ) is not fully filled. In order to achieve the stable electronic configuration of the nearest noble gas, F must gain one electron and become F^- cation. Now, the electronic configuration of F^- anion is similar to Ne, which is $1s^{2}2s^{2}2p^6$.

Answer to Problem 35P

2 electrons must be gained.

Explanation of Solution

The electronic configuration of selenium (Se) is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^4$.

The nearest noble gas to Sr is krypton (Kr), whose electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^6$.

Therefore, Se must gain two electrons to achieve Kr atom's electronic configuration.

Because two electrons are gained, the result is Se^2- anion.

Interpretation Introduction

(d)

Interpretation:

Number of electrons must be gained/lost by aluminum (Al) to achieve a noble gas configuration should be determined.

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.

Here,

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.

Cations are formed by losing electrons, thus they have fewer electrons than protons and are positively charged.

Anions are formed by gaining electrons, thus they have more electrons than protons and are negatively charged.

For example, 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 35P

3 electrons must be lost.

Explanation of Solution

Aluminum (Al) has 13 electrons. The electronic configuration of Al is $1s^{2}2s^{2}2p^{6}3s^{2}3p^1$.

The nearest noble gas to Al is neon (Ne), whose electronic configuration is $1s^{2}2s^{2}2p^6$.

Therefore, Al must lose three electrons to achieve Ne atom's electronic configuration.

Because three electrons are lost, the result is Al³⁺ cation.