Chapter 8, Problem 8.89P

(a)

Interpretation Introduction

Interpretation:

The element with the smallest atomic radius in group $6\text{A}\left(16\right)$ is to be determined.

Concept introduction:

The atomic radius of an element is the measure of the size of its atoms. It is the mean distance from the center of the nucleus to the boundary of the surrounding shells of the electrons. But this boundary is not well defined so the atomic radius cannot be measured.

Answer to Problem 8.89P

Oxygen has the smallest size in the group $6\text{A}\left(16\right)$.

Explanation of Solution

Group $6\text{A}\left(16\right)$ consists of oxygen, sulfur, selenium, tellurium, and polonium. Oxygen is the first member of the group and atomic size increases from top to bottom in a group. So the atomic radius of oxygen is the smallest among the elements of the group $6\text{A}\left(16\right)$.

Conclusion

The atomic radius of the element increases from top to bottom in a group and decreases from left to right in a period.

(b)

Interpretation Introduction

Interpretation:

The element with the largest atomic radius in the period $6$ is to be determined.

Concept introduction:

The atomic radius of an element is the measure of the size of its atoms. It is the mean distance from the center of the nucleus to the boundary of the surrounding shells of the electrons. But this boundary is not well defined so the atomic radius cannot be measured.

Answer to Problem 8.89P

Cesium has the largest atomic radius in the period $6$.

Explanation of Solution

Cesium is present at the leftmost region of the 6^th period in the periodic table. The atomic radius decreases from left to right due to the increase in the effective nuclear charge. So the atomic radius of cesium is the largest in the 6^th period.

Conclusion

The atomic radius of the element increases from top to bottom in a group and decreases from left to right in a period.

(c)

Interpretation Introduction

Interpretation:

The smallest metal in period 3 is to be determined.

Concept introduction:

The metallic character is characterized by the tendency of metals to lose their outermost valence shell electrons. Greater the ease of electron removal, higher will be the electropositivity of the corresponding elements and vice versa.

The metallic character increases from top to bottom in a group because the electrons are less tightly held by the nucleus of the atom and therefore removed easily. It decreases from left to right in a period because the electrons are more tightly held with the nucleus of the atom and therefore the removal becomes difficult.

Answer to Problem 8.89P

Aluminium is the smallest metal in the 3^rd period.

Explanation of Solution

The third period contains sodium, magnesium, aluminium, silicon, phosphorus, sulfur, chlorine and argon. Out of these eight elements, sodium, magnesium, and aluminium are metals, silicon is metalloid while phosphorus, sulfur, chlorine, and argon are non-metals. Among sodium, magnesium, and aluminium, aluminium lies to the right of the periodic table so its atomic radius is the smallest and therefore it is the smallest metal in the 3^rd period.

Conclusion

The atomic radius of the element increases from top to bottom in a group and decreases from left to right in a period.

(d)

Interpretation Introduction

Interpretation:

The element with the highest ${\text{IE}}_1$ in the group $4\text{A}\left(14\right)$ is to be determined.

Concept introduction:

The ionization energy is the amount of energy required to remove the most loosely bound valence electrons from an isolated neutral gaseous atom. It is represented by $\text{IE}$.

Its value varies with the ease of removal of the outermost valence electron. If the outermost electron is removed very easily then the value of ionization energy is very small. If the electron is removed with quite a difficulty then the value of ionization energy will be very high.

When the first electron is removed from a neutral, isolated gaseous atom then the ionization energy is known as the first ionization energy $\left({\text{IE}}_1\right)$. Similarly, when the second electron is removed from the positively charged cation the ionization energy is called the second ionization energy $\left({\text{IE}}_2\right)$ and so on.

Answer to Problem 8.89P

Carbon has the highest ${\text{IE}}_1$ in the group $4\text{A}\left(14\right)$.

Explanation of Solution

The group $4\text{A}\left(14\right)$ contains carbon, silicon, germanium, tin, and lead. Carbon is the first member of this group so its atomic size is the smallest and therefore ${\text{IE}}_1$ is the highest.

Conclusion

The ionization energy is inversely related to the atomic size of the element. Larger the atomic size, more will its ionization energy and vice-versa.

(e)

Interpretation Introduction

Interpretation:

The element with the lowest ${\text{IE}}_1$ in period 5 is to be determined.

Concept introduction:

The ionization energy is the amount of energy required to remove the most loosely bound valence electrons from an isolated neutral gaseous atom. It is represented by $\text{IE}$.

Its value varies with the ease of removal of the outermost valence electron. If the outermost electron is removed very easily then the value of ionization energy is very small. If the electron is removed with quite a difficulty then the value of ionization energy will be very high.

When the first electron is removed from a neutral, isolated gaseous atom then the ionization energy is known as the first ionization energy $\left({\text{IE}}_1\right)$. Similarly, when the second electron is removed from the positively charged cation the ionization energy is called the second ionization energy $\left({\text{IE}}_2\right)$ and so on.

Answer to Problem 8.89P

Rubidium has the lowest ${\text{IE}}_1$ in period 5.

Explanation of Solution

Rubidium is present at the leftmost region of 5^th period. Its atomic size is the largest and therefore ${\text{IE}}_1$ is the lowest among the elements of the 5^th period.

Conclusion

The ionization energy is inversely related to the atomic size of the element. Larger the atomic size, more will its ionization energy and vice-versa.

(f)

Interpretation Introduction

Interpretation:

The most metallic element in group $5\text{A}\left(15\right)$ is to be determined.

Concept introduction:

The metallic character is characterized by the tendency of metals to lose their outermost valence shell electrons. Greater the ease of electron removal, higher will be the electropositivity of the corresponding elements and vice versa.

The metallic character increases from top to bottom in a group because the electrons are less tightly held by the nucleus of the atom and therefore removed easily. It decreases from left to right in a period because the electrons are more tightly held with the nucleus of the atom and therefore the removal becomes difficult.

Answer to Problem 8.89P

Bismuth is the most metallic element of the group $5\text{A}\left(15\right)$.

Explanation of Solution

Group $5\text{A}\left(15\right)$ consists of nitrogen, phosphorus, arsenic, antimony, and bismuth. Bismuth is present at the bottom of this group so its atomic size is the largest in the group and therefore the electrons are removed with much ease. So bismuth is the most metallic in the group $5\text{A}\left(15\right)$.

Conclusion

The metallic character is directly related to the atomic size of the atoms. Larger the size of the atom, more will be its metallic character and vice-versa.

(g)

Interpretation Introduction

Interpretation:

The element of the group $\text{3A}\left(13\right)$ that forms the most basic oxide is to be determined.

Concept introduction:

Metal oxides are the chemical compounds formed between metals and oxygen. These oxides generally contain an oxide ion of $-\text{2}$ charge. These oxides are usually basic in nature and exist in a solid state.

Nonmetals react with oxygen to form nonmetal oxides. These are generally present in a gaseous state. These are usually acidic in nature because they release ${\text{H}}^{+}$ ions when dissolved in water.

Answer to Problem 8.89P

Thallium forms the most basic oxide in the group $\text{3A}\left(13\right)$.

Explanation of Solution

Thallium is present at the bottom of the group $\text{3A}\left(13\right)$ so its atomic size and metallic character is the largest among all the elements. The basic nature of the oxides is directly related to the metallic character and therefore thallium forms the most basic oxide.

Conclusion

Basic oxides are formed by the metals whereas the acidic oxides are formed by the non-metals. More the metallic character of the element, more will be the basic nature of the oxides and vice-versa.

(h)

Interpretation Introduction

Interpretation:

The element of period 4 that has the highest filled energy level is to be determined.

Concept introduction:

The electrons that are present inside an atom occupy only certain allowed orbitals with a specific energy. The energy corresponding to each of the allowed orbitals are called energy levels.

Answer to Problem 8.89P

Krypton has the highest filled energy level.

Explanation of Solution

Krypton is a stable noble gas with the electronic configuration $\left[\text{Ar}\right]3d^{10}4s^{2}4p^6$. It has 36 electrons in it and the highest energy level is $4p$. It has 6 electrons that are its maximum capacity. So Krypton has the highest filled energy level among the elements of period 4.

Conclusion

The elements with the filled energy levels are more stable than those with incompletely filled levels.

(i)

Interpretation Introduction

Interpretation:

The element with the condensed ground-state electronic configuration $\left[\text{Ne}\right]3s^{2}3p^2$.

Concept introduction:

The electronic configuration tells about the distribution of electrons in the various atomic orbitals of the element. It is used to predict the chemical properties of the element. The valence shell electronic configuration refers to the distribution of electrons in the outermost shell only.

The condensed electronic configuration is a way to write the electronic configuration where the inner shell configurations are compressed to the nearest noble gas configuration and only the valence shell configuration is written in the expanded form.

Answer to Problem 8.89P

Silicon has the ground-state electronic configuration $\left[\text{Ne}\right]3s^{2}3p^2$.

Explanation of Solution

The electronic configuration $\left[\text{Ne}\right]3s^{2}3p^2$ indicates that the element is present in the 3^rd period and has 4 valence electrons. The element of the 3^rd period with 4 valence electrons is silicon and it has the configuration of $\left[\text{Ne}\right]3s^{2}3p^2$.

Conclusion

The period of the element is identified by the highest filled energy level in its electronic configuration.

(j)

Interpretation Introduction

Interpretation:

The element with the condensed ground-state electronic configuration $\left[\text{Kr}\right]5s^{2}4d^6$ is to be determined.

Concept introduction:

The electronic configuration tells about the distribution of electrons in the various atomic orbitals of the element. It is used to predict the chemical properties of the element. The valence shell electronic configuration refers to the distribution of electrons in the outermost shell only.

The condensed electronic configuration is a way to write the electronic configuration where the inner shell configurations are compressed to the nearest noble gas configuration and only the valence shell configuration is written in the expanded form.

Answer to Problem 8.89P

Ruthenium has the condensed ground-state electronic configuration $\left[\text{Kr}\right]5s^{2}4d^6$.

Explanation of Solution

The electronic configuration $\left[\text{Kr}\right]5s^{2}4d^6$ indicates that the element is present in the 5^th period and has 8 valence electrons. The element of the 5^th period with 8 valence electrons is ruthenium and it has the configuration of $\left[\text{Kr}\right]5s^{2}4d^6$.

Conclusion

The period of the element is identified by the highest filled energy level in its electronic configuration.

(k)

Interpretation Introduction

Interpretation:

The element that forms $+2$ ion with the electronic configuration $\left[\text{Ar}\right]3d^3$ is to be determined.

Concept introduction:

The electronic configuration tells about the distribution of electrons in the various atomic orbitals of the element. It is used to predict the chemical properties of the element. The valence shell electronic configuration refers to the distribution of electrons in the outermost shell only.

Answer to Problem 8.89P

Vanadium forms $+2$ ion with the electronic configuration $\left[\text{Ar}\right]3d^3$.

Explanation of Solution

The electronic configuration of the cation formed is $\left[\text{Ar}\right]3d^3$. So the electronic configuration of the element is $\left[\text{Ar}\right]3d^{3}4s^2$ and the element with this configuration is vanadium. It easily loses its two $4s$ electrons to form $+2$ ion.

Conclusion

The formation of ions is directly related to the ease of gain or loss of electrons.

(l)

Interpretation Introduction

Interpretation:

The element of the 5^th period that forms $+3$ ion with pseudo-noble gas configuration is to be determined.

Concept introduction:

The noble gas configuration refers to the stable electronic configuration of the elements with their complete octet. The elements with the noble gas configuration have no or very less tendency to react with the other species and therefore these are extremely stable. This is also used to write the condensed electronic configuration of the elements.

The elements that possess stable electronic configurations in their outermost valence shell but are not actually noble gas configurations. Such electronic configurations are called pseudo noble gas configuration.

The $\left(n-1\right)d^{10}n{s}^{0}n{p}^0$ electronic configuration is known as pseudo noble gas configuration.

The stability of a pseudo noble gas configuration is comparable to that of the noble gas configuration.

Answer to Problem 8.89P

Indium forms $+3$ ion with the pseudo-noble gas configuration.

Explanation of Solution

The electronic configuration of indium is $\left[\text{Kr}\right]4d^{10}5s^{2}5p^1$. It forms $+3$ ion very easily and therefore acquires a pseudo noble gas configuration of $\left[\text{Kr}\right]4d^{10}$.

Conclusion

The elements with pseudo noble gas configuration are more stable than the other ones.

(m)

Interpretation Introduction

Interpretation:

The transition element of period 4 that forms $+3$ diamagnetic ion is to be determined.

Concept introduction:

Diamagnetism is the property of materials due to which they are slightly repelled by an externally applied magnetic field. It occurs due to the presence of paired electrons so the atoms with all the filled orbitals are diamagnetic.

Answer to Problem 8.89P

Scandium forms $+3$ diamagnetic ion.

Explanation of Solution

The electronic configuration of scandium is $\left[\text{Ar}\right]3d^{1}4s^2$. It can easily lose three electrons to achieve the noble gas configuration of argon that has completely filled electronic configuration so it is diamagnetic.

Conclusion

Diamagnetism is directly related to the absence of unpaired electrons.

(n)

Interpretation Introduction

Interpretation:

The transition element of period 4 that forms $+2$ ion with a half-filled $d$ sublevel is to be determined.

Concept introduction:

The electronic configuration tells about the distribution of electrons in a various atomic orbital. The condensed electronic configuration is a way to write the electronic configuration where the inner shell configurations are compressed to the nearest noble gas configuration and only the valence shell configuration is written in the expanded form.

When a neutral atom gains electrons, it acquires a negative charge on it due to the presence of excess electrons as compared to the protons. This results in the formation of the anion. When a neutral atom loses electrons, it acquires a positive charge on it due to the presence of excess protons as compared to the electrons. This results in the formation of the cation.

Answer to Problem 8.89P

Manganese can form $+2$ ion with a half-filled $d$ sublevel.

Explanation of Solution

The electronic configuration of manganese is $\left[\text{Ar}\right]3d^{5}4s^2$. It loses two $4s$ electrons easily and the resultant configuration becomes $\left[\text{Ar}\right]3d^5$. So manganese forms $+2$ ion that has the half-filled $d$ sublevel.

Conclusion

The half-filled electronic configurations are more stable than the other ones.

(o)

Interpretation Introduction

Interpretation:

The heaviest lanthanide is to be determined.

Concept introduction:

The lanthanide consists of 15 chemical elements from atomic numbers 57 to 71, from lanthanum to lutetium. The elements present in this series have $+3$ as their common oxidation state. There is a large similarity in the atomic radii of the lanthanide elements and therefore these are very difficult to separate.

Answer to Problem 8.89P

Lutetium is the heaviest lanthanide.

Explanation of Solution

The atomic weight of lutetium is $174.967\text{u}$ that is the largest among all the lanthanide elements. So lutetium is the heaviest element.

Conclusion

Heaviness is the measure of the atomic weight of the elements. More the atomic weight, heavier will be the element and vice-versa.

(p)

Interpretation Introduction

Interpretation:

The element of the 3^rd period that forms $-2$ ion that is isoelectronic with $\text{Ar}$ is to be determined.

Concept introduction:

The isoelectronic species are the atoms, molecules or ions that have the same number of electrons. They differ in their chemical and physical properties. ${\text{Na}}^{+}$, ${\text{Mg}}^{2+}$ and ${\text{Al}}^{3+}$ all have ten electrons so they are isoelectronic species.

Answer to Problem 8.89P

The $-2$ ion that is isoelectronic with $\text{Ar}$ is formed by sulfur.

Explanation of Solution

The electronic configuration of sulfur is $\left[\text{Ne}\right]3s^{2}3p^4$. It is two electrons short of the stable noble gas configuration so it gains two electrons and form ${\text{S}}^{2-}$ that has the configuration of $\left[\text{Ne}\right]3s^{2}3p^6$. The number of electrons in both argon and ${\text{S}}^{2-}$ are 18 and therefore both are isoelectronic species.

Conclusion

The isoelectronic species are identified by the same number of electrons.

(q)

Interpretation Introduction

Interpretation:

The alkaline earth metal whose cation is isoelectronic with $\text{Kr}$ is to be determined.

Concept introduction:

The isoelectronic species are the atoms, molecules or ions that have the same number of electrons. They differ in their chemical and physical properties. ${\text{Na}}^{+}$, ${\text{Mg}}^{2+}$ and ${\text{Al}}^{3+}$ all have ten electrons so they are isoelectronic species.

Answer to Problem 8.89P

Strontium forms the cation that is isoelectronic with $\text{Kr}$.

Explanation of Solution

The electronic configuration of strontium is $\left[\text{Kr}\right]5s^2$. It loses two electrons and form ${\text{Sr}}^{2+}$ that has the configuration similar to that of Krypton. The number of electrons in ${\text{Sr}}^{2+}$ and $\text{Kr}$ is 36 and therefore both are isoelectronic species.

Conclusion

The isoelectronic species are identified by the same number of electrons.

(r)

Interpretation Introduction

Interpretation:

The metalloid of the group $5\text{A}\left(15\right)$ that forms the most acidic oxide is to be determined.

Concept introduction:

Metal oxides are the chemical compounds formed between metals and oxygen. These oxides generally contain an oxide ion of $-\text{2}$ charge. These oxides are usually basic in nature and exist in a solid state.

Nonmetals react with oxygen to form nonmetal oxides. These are generally present in a gaseous state. These are usually acidic in nature because they release ${\text{H}}^{+}$ ions when dissolved in water.

Answer to Problem 8.89P

Arsenic forms the most acidic oxide.

Explanation of Solution

Arsenic and antimony show the properties of both metals and non-metals so both are metalloids. The acidic character of the oxides decreases from top to bottom and arsenic lies above antimony so its oxide is more acidic than that of antimony. Arsenic is the metalloid that forms the most acidic oxide in the group $5\text{A}\left(15\right)$.

Conclusion

The acidic nature of the oxides decreases from top to bottom in a group and increases from left to right in a period.