Chapter 3, Problem 43E

Would you expect each of the following atoms to gain or lose electrons when forming ions? What ion is the most likely in each case?

a. Ra

b. In

c. P

d. Te

e. Br

f. Rb

Interpretation Introduction

Interpretation: The ability to lose or gain electrons of the given atom when forming ions is to be calculated. The ion most likely formed in each case is to be stated.

Concept introduction: Any atom or molecule gain or lose electrons to acquire stable state. An ion is a charged atom formed by the loss or gain of electron.

To determine: The Radium $\left(\text{Ra}\right)$ atom loses or gain electrons while forming ion; the ion formed by Radium $\left(\text{Ra}\right)$ atom.

Answer to Problem 43E

The Radium $\left(\text{Ra}\right)$ atom loses electrons. The Radium $\left(\text{Ra}\right)$ forms ${\text{Ra}}^{\text{2+}}$ ion.

Explanation of Solution

• To determine: The Radium atom loses or gain electrons.

Metals always loose electrons to form positive ion. They lose electron to acquire more stable state.

The electronic configuration of Radium is:

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^{\text{14}}{\text{5d}}^{\text{10}}{\text{6s}}^{\text{2}}{\text{6p}}^{\text{6}}{\text{7s}}^{\text{2}}$

Radium is a metal hence, it lose electrons.

• To determine: The ion formed by Radium $\left(\text{Ra}\right)$ atom.

Explanation:

Radium atom contains two electrons in last s-sub shell. Hence it can loose two electrons to form ${\text{Ra}}^{\text{2+}}$. The electronic configuration of ${\text{Ra}}^{\text{2+}}$ is:

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^{\text{14}}{\text{5d}}^{\text{10}}{\text{6s}}^{\text{2}}{\text{6p}}^{\text{6}}$

Conclusion

Generally metals lose electrons to acquire stable state whereas non- metals gain electrons.

Interpretation Introduction

Interpretation: The ability to lose or gain electrons of the given atom when forming ions is to be calculated. The ion most likely formed in each case is to be stated.

Concept introduction: Any atom or molecule gain or lose electrons to acquire stable state. An ion is a charged atom formed by the loss or gain of electron.

To determine: The Indium $\left(\text{In}\right)$ atom loses or gain electrons; the ion formed by Indium $\left(\text{In}\right)$ atom.

Answer to Problem 43E

The Indium $\left(\text{In}\right)$ atom loses electrons. The Indium $\left(\text{In}\right)$ atom form ${\text{In}}^{+}\text{,}{\text{In}}^{\text{3+}}$ ions.

Explanation of Solution

• To determine: The Indium atom loses or gain electrons.

Metals always loose electrons to form positive ion. They lose electron to acquire more stable state.

The electronic configuration of Indium is:

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{1}}$

Indium is a metal hence, it lose electrons.

• To determine: The ion formed by Indium $\left(\text{In}\right)$ atom.

Indium atom contains one electron in last p-sub shell and two electrons in last s-sub shell. Hence it can loose one electron to form ${\text{In}}^{\text{+}}$ and three electrons to form ${\text{In}}^{\text{3+}}$. The electronic configuration of ${\text{In}}^{\text{+}}$ and ${\text{In}}^{\text{3+}}$ are:

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}$

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}$

Interpretation Introduction

Interpretation: The ability to lose or gain electrons of the given atom when forming ions is to be calculated. The ion most likely formed in each case is to be stated.

Concept introduction: Any atom or molecule gain or lose electrons to acquire stable state. An ion is a charged atom formed by the loss or gain of electron.

To determine: The Phosphorus $\left(\text{P}\right)$ atom loses or gain electrons; the ion formed by Phosphorus $\left(\text{P}\right)$ atom.

Answer to Problem 43E

The Phosphorus $\left(\text{P}\right)$ atom gain electrons. The Phosphorus $\left(\text{P}\right)$ atom forms ${\text{P}}^{\text{3}-}$  ion.

Explanation of Solution

• To determine: The Phosphorus $\left(\text{P}\right)$ atom loses or gain electrons.

Non-metals always gain electrons to form negative ion. They gain electron to acquire more stable state.

The electronic configuration of phosphorus is:

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{3}}$

Phosphorus is a non-metal hence, it gain electrons.

• To determine: The ion formed by Phosphorus $\left(\text{P}\right)$ atom.

Explanation:

Phosphorus atom contains three electrons in last p-sub shell. Hence it gains three electrons to completely fill the p-orbital to form ${\text{P}}^{\text{3}-}$ ion. The electronic configuration of ${\text{P}}^{\text{3}-}$ is:

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^6$

Conclusion

Generally metals lose electrons to acquire stable state whereas non- metals gain electrons.

Interpretation Introduction

Interpretation: The ability to lose or gain electrons of the given atom when forming ions is to be calculated. The ion most likely formed in each case is to be stated.

Concept introduction: Any atom or molecule gain or lose electrons to acquire stable state. An ion is a charged atom formed by the loss or gain of electron.

To determine: The Tellurium $\left(\text{Te}\right)$ atom loses or gain electrons; the ion formed by Tellurium $\left(\text{Te}\right)$ atom.

Answer to Problem 43E

The Tellurium $\left(\text{Te}\right)$ atom gain electrons. The Tellurium $\left(\text{Te}\right)$ atom forms ${\text{Te}}^{\text{2}-}$ ion.

Explanation of Solution

• To determine: The Tellurium $\left(\text{Te}\right)$ atom loses or gain electrons.

Non-metals always gain electrons to form negative ion. They gain electron to acquire more stable state.

The electronic configuration of Tellurium is:

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{4}}$

Tellurium is a non-metal hence, it gains electrons.

• To determine: The ion formed by Tellurium $\left(\text{Te}\right)$ atom.

Tellurium atom contains four electrons in last p-sub shell. Hence it gains two electrons to completely fill the p-orbital to form ${\text{Te}}^{\text{2}-}$ ion. The electronic configuration of ${\text{Te}}^{\text{2}-}$ is:

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^6$

Interpretation Introduction

Interpretation: The ability to lose or gain electrons of the given atom when forming ions is to be calculated. The ion most likely formed in each case is to be stated.

Concept introduction: Any atom or molecule gain or lose electrons to acquire stable state. An ion is a charged atom formed by the loss or gain of electron.

To determine: The bromine $\left(\text{Br}\right)$ atom loses or gain electrons; the ion formed by bromine $\left(\text{Br}\right)$ atom.

Answer to Problem 43E

The bromine $\left(\text{Br}\right)$ atom gain electrons. The bromine $\left(\text{Br}\right)$ atom forms ${\text{Br}}^{-}$ ion.

Explanation of Solution

• To determine: The Bromine $\left(\text{Br}\right)$ atom loses or gain electrons.

Non-metals always gain electrons to form negative ion. They gain electron to acquire more stable state.

The electronic configuration of bromine is:

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{5}}$

Bromine is a non-metal hence, it gain electrons.

• To determine: The ion formed by bromine atom

Bromine atom contains five electrons in last p-sub shell. Hence it gains one electron to completely fill the p-orbital to form ${\text{Br}}^{\text{-}}$. The electronic configuration of ${\text{Br}}^{-}$ is:

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^6$

Interpretation Introduction

Interpretation: The ability to lose or gain electrons of the given atom when forming ions is to be calculated. The ion most likely formed in each case is to be stated.

Concept introduction: Any atom or molecule gain or lose electrons to acquire stable state. An ion is a charged atom formed by the loss or gain of electron.

To determine: The rubidium $\left(\text{Rb}\right)$ atom loses or gain electrons; the ion formed by rubidium $\left(\text{Rb}\right)$ atom.

Answer to Problem 43E

The rubidium $\left(\text{Rb}\right)$ atom loses an electron. The rubidium $\left(\text{Rb}\right)$ atom forms ${\text{Rb}}^{+}$ ion.

Explanation of Solution

• To determine: The rubidium atom loses or gain electrons.

Metals always loose electrons to form positive ion. They lose electron to acquire more stable state.

The electronic configuration of rubidium is:

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{5s}}^{\text{1}}$

Rubidium is a metal hence, it lose electrons.

• To determine: The ion formed by Rubidium $\left(\text{Rb}\right)$ atom

Rubidium atom contains one electron in last s-sub shell. Hence it can loose one electron to form ${\text{Rb}}^{+}$. The electronic configuration of ${\text{Rb}}^{+}$ is:

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}$