Chapter 3, Problem 33P

Interpretation Introduction

(a)

Interpretation:

Number of electrons must be gained/lost by lithium 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. The electronic configuration of these noble gases are as follows:

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 33P

1 electron must be lost.

Explanation of Solution

The electronic configuration of Li is $1s^{2}2s^1$.

The nearest noble gas to Li is helium (He), whose electronic configuration is $1s^2$.

Therefore, Li must lose one electron to achieve He atom's electronic configuration.

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

Interpretation Introduction

(b)

Interpretation:

Number of electrons must be gained/lost by iodine 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. The electronic configuration of these noble gases are as follows:

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 33P

1 electron must be gained

Explanation of Solution

The electronic configuration of I is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^5$.

The nearest noble gas to Li 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, I must gain one electron to achieve Xe atom's electronic configuration.

Because an electron is lost, the result is I^- anion.

Interpretation Introduction

(c)

Interpretation:

Number of electrons must be gained/lost by sulfur 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. The electronic configuration of these noble gases are as follows:

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 33P

2 electrons must be gained

Explanation of Solution

The electronic configuration of S is $1s^{2}2s^{2}2p^{6}3s^{2}3p^4$.

The nearest noble gas to S is argon (Ar), whose electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$.

Therefore, S must gain two electrons to achieve Ar atom's electronic configuration.

Because two electrons are lost, the result is S^2-anion.

Interpretation Introduction

(d)

Interpretation:

Number of electrons must be gained/lost by strontium (Sr) 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. The electronic configuration of these noble gases are as follows:

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 33P

2 electrons must be lost

Explanation of Solution

The electronic configuration of Sr is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^2$.

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, Sr must lose two electrons to achieve Kr atom's electronic configuration.

Because two electrons are lost, the result is Sr²⁺cation.