Chapter 2, Problem 37E

(a)

Interpretation Introduction

Interpretation:The element having atomic number $17$ is to be identified. Its classification as either anoble gas, a halogen, an alkali metal, an alkaline earth metal,a transition metal,a lanthanide metal, or an actinide metal is to be done.

Concept introduction:In the periodic table, group $18$ elements are termed as noble gases or inert gases or rare gases. All these gases have complete valence shell. Thus, they are called noble gases with $n{s}^2,n{p}^6$ general electronic configuration.The elements of group $17$ having $n{s}^2,n{p}^5$ general electronic configuration are called halogens.

The elements of first and second group are usually called alkali metalsand alkaline earth metals with configurations $n{s}^1$ and $n{s}^2$ respectively. They belong to the s-block of periodic table. The d-block elements together comprise the transition series. Three series of elements are formed by the filing of $3d,4d$ and $5d$ shells of electrons.

The lanthanide series consists of $14$ elements which are called lanthanides or lanthanones. They are characterized by the filing up of the antepenultimate $4f$ energy levels, whereas in actinides, the filing up of $5f$ energy levels takes place.

Answer to Problem 37E

The element having atomic number $17$ is chlorine. It is a halogen.

Explanation of Solution

The atomic number of the given elementis $17$.

The electronic configuration of the element having atomic number $17$ is $1s^2,2s^2,2p^6,3s^2,3p^5$.

The outermost shell electronic configuration $3s^2,3p^5$ indicates that the element belongs to the third period of the periodic table and the seven valence electrons shows that it is a ${17}^{\text{th}}$ group element. Therefore, the element is chlorine which is a halogen.

(b)

Interpretation Introduction

Interpretation:The element having atomic number $4$ is to be identified. Its classification as either a noble gas, a halogen, an alkali metal, an alkaline earth metal, a transition metal, a lanthanide metal, or an actinide metal is to be done.

Concept introduction:In the periodic table, group $18$ elements are termed as noble gases or inert gases or rare gases. All these gases have complete valence shell. Thus, they are called noble gases with $n{s}^2,n{p}^6$ general electronic configuration.The elements of group $17$ having $n{s}^2,n{p}^5$ general electronic configuration are called halogens.

The elements of first and second group are usually called alkali metals and alkaline earth metals with configurations $n{s}^1$ and $n{s}^2$ respectively. They belong to the s-block of periodic table. The d-block elements together comprise the transition series. Three series of elements are formed by the filing of $3d,4d$ and $5d$ shells of electrons.

The lanthanide series consists of $14$ elements which are called lanthanides or lanthanones. They are characterized by the filing up of the antepenultimate $4f$ energy levels, whereas in actinides, the filing up of $5f$ energy levels takes place.

Answer to Problem 37E

The element having atomic number $4$ isberyllium. It is an alkaline earth metal.

Explanation of Solution

The atomic number of the given element is $4$.

The electronic configuration of the element having atomic number $4$ is $1s^2,2s^2$.

The outermost shell electronic configuration $2s^2$ indicates that the element belongs to the second period of the periodic table and the two valence electrons shows that it is a $2^{\text{nd}}$ group element. Therefore, the element is beryllium which is an alkaline earth metal.

(c)

Interpretation Introduction

Interpretation:The element having atomic number $63$ is to be identified. Its classification as either a noble gas, a halogen, an alkali metal, an alkaline earth metal, a transition metal, a lanthanide metal, or an actinide metal is to be done.

Concept introduction:In the periodic table, group $18$ elements are termed as noble gases or inert gases or rare gases. All these gases have complete valence shell. Thus, they are called noble gases with $n{s}^2,n{p}^6$ general electronic configuration.The elements of group $17$ having $n{s}^2,n{p}^5$ general electronic configuration are called halogens.

The elements of first and second group are usually called alkali metals and alkaline earth metals with configurations $n{s}^1$ and $n{s}^2$ respectively. They belong to the s-block of periodic table. The d-block elements together comprise the transition series. Three series of elements are formed by the filing of $3d,4d$ and $5d$ shells of electrons.

The lanthanide series consists of $14$ elements which are called lanthanides or lanthanones. They are characterized by the filing up of the antepenultimate $4f$ energy levels, whereas in actinides, the filing up of $5f$ energy levels takes place.

Answer to Problem 37E

The element having atomic number $63$ is europium. It is a lanthanide metal.

Explanation of Solution

The atomic number of the given element is $63$.

The electronic configuration of the element having atomic number $63$ is $1s^2,2s^2,2p^6,3s^2,3p^6,4s^2,3d^{10},4p^6,5s^2,4d^{10},5p^6,6s^2,4f^7$.

The outermost shell electronic configuration $6s^2,4f^7$ indicates that the element belongs to the lanthanide series of the periodic table and the seven valence electrons in $4f$ orbital shows that it is europium. Therefore, the element is europium which is a lanthanide metal.

(d)

Interpretation Introduction

Interpretation:The element having atomic number $72$ is to be identified. Its classification as either a noble gas, a halogen, an alkali metal, an alkaline earth metal, a transition metal, a lanthanide metal, or an actinide metal is to be done.

Concept introduction:In the periodic table, group $18$ elements are termed as noble gases or inert gases or rare gases. All these gases have complete valence shell. Thus, they are called noble gases with $n{s}^2,n{p}^6$ general electronic configuration.The elements of group $17$ having $n{s}^2,n{p}^5$ general electronic configuration are called halogens.

The elements of first and second group are usually called alkali metals and alkaline earth metals with configurations $n{s}^1$ and $n{s}^2$ respectively. They belong to the s-block of periodic table. The d-block elements together comprise the transition series. Three series of elements are formed by the filing of $3d,4d$ and $5d$ shells of electrons.

The lanthanide series consists of $14$ elements which are called lanthanides or lanthanones. They are characterized by the filing up of the antepenultimate $4f$ energy levels, whereas in actinides, the filing up of $5f$ energy levels takes place.

Answer to Problem 37E

The element having atomic number $72$ is hafnium. It is a transition metal.

Explanation of Solution

The atomic number of the given element is $72$.

The electronic configuration of the element having atomic number $72$ is $1s^2,2s^2,2p^6,3s^2,3p^6,4s^2,3d^{10},4p^6,5s^2,4d^{10},5p^6,6s^2,4f^{14},5d^2$.

The outermost shell electronic configuration $4f^{14},5d^2$ indicates that the element belongs to the $5d$ series of transition metals of the periodic table and the two electrons in $5d$ orbital shows that it is hafnium. Therefore, the element is hafnium which is a transition metal.

(e)

Interpretation Introduction

Interpretation:The element having atomic number $2$ is to be identified. Its classification as either a noble gas, a halogen, an alkali metal, an alkaline earth metal, a transition metal, a lanthanide metal, or an actinide metal is to be done.

Concept introduction:In the periodic table, group $18$ elements are termed as noble gases or inert gases or rare gases. All these gases have complete valence shell. Thus, they are called noble gases with $n{s}^2,n{p}^6$ general electronic configuration.The elements of group $17$ having $n{s}^2,n{p}^5$ general electronic configuration are called halogens.

The elements of first and second group are usually called alkali metals and alkaline earth metals with configurations $n{s}^1$ and $n{s}^2$ respectively. They belong to the s-block of periodic table. The d-block elements together comprise the transition series. Three series of elements are formed by the filing of $3d,4d$ and $5d$ shells of electrons.

The lanthanide series consists of $14$ elements which are called lanthanides or lanthanones. They are characterized by the filing up of the antepenultimate $4f$ energy levels, whereas in actinides, the filing up of $5f$ energy levels takes place.

Answer to Problem 37E

The element having atomic number $2$ is helium. It is a noble gas.

Explanation of Solution

The atomic number of the given element is $2$.

The electronic configuration of the element having atomic number $2$ is $1s^2$.

The outermost shell electronic configuration $1s^2$ indicates that the element belongs to the first period of the periodic table and the two valence electrons shows that it is a ${18}^{\text{th}}$ group element. Therefore, the element is helium which is a Snoble gas.

(f)

Interpretation Introduction

Interpretation:The element having atomic number $92$ is to be identified. Its classification as either a noble gas, a halogen, an alkali metal, an alkaline earth metal, a transition metal, a lanthanide metal, or an actinide metal is to be done.

Concept introduction:In the periodic table, group $18$ elements are termed as noble gases or inert gases or rare gases. All these gases have complete valence shell. Thus, they are called noble gases with $n{s}^2,n{p}^6$ general electronic configuration.The elements of group $17$ having $n{s}^2,n{p}^5$ general electronic configuration are called halogens.

The elements of first and second group are usually called alkali metals and alkaline earth metals with configurations $n{s}^1$ and $n{s}^2$ respectively. They belong to the s-block of periodic table. The d-block elements together comprise the transition series. Three series of elements are formed by the filing of $3d,4d$ and $5d$ shells of electrons.

The lanthanide series consists of $14$ elements which are called lanthanides or lanthanones. They are characterized by the filing up of the antepenultimate $4f$ energy levels, whereas in actinides, the filing up of $5f$ energy levels takes place.

Answer to Problem 37E

The element having atomic number $92$ is uranium. It is an actinide metal.

Explanation of Solution

The atomic number of the given element is $92$.

The electronic configuration of the element having atomic number $92$ is $1s^2,2s^2,2p^6,3s^2,3p^6,4s^2,3d^{10},4p^6,5s^2,4d^{10},5p^6,6s^2,4f^{14},5d^{10},6p^6,7s^2,5f^3,6d^1$.

The outermost shell electronic configuration $5f^3,6d^1$ indicates that the element belongs to the actinide series of the periodic table and the four valence electrons in $5f^3,6d^1$ orbitals show that it is uranium. Therefore, the element is uranium which is an actinide metal.

(g)

Interpretation Introduction

Interpretation:The element having atomic number $55$ is to be identified. Its classification as either a noble gas, a halogen, an alkali metal, an alkaline earth metal, a transition metal, a lanthanide metal, or an actinide metal is to be done.

Concept introduction:In the periodic table, group $18$ elements are termed as noble gases or inert gases or rare gases. All these gases have complete valence shell. Thus, they are called noble gases with $n{s}^2,n{p}^6$ general electronic configuration.The elements of group $17$ having $n{s}^2,n{p}^5$ general electronic configuration are called halogens.

The elements of first and second group are usually called alkali metals and alkaline earth metals with configurations $n{s}^1$ and $n{s}^2$ respectively. They belong to the s-block of periodic table. The d-block elements together comprise the transition series. Three series of elements are formed by the filing of $3d,4d$ and $5d$ shells of electrons.

The lanthanide series consists of $14$ elements which are called lanthanides or lanthanones. They are characterized by the filing up of the antepenultimate $4f$ energy levels, whereas in actinides, the filing up of $5f$ energy levels takes place.

Answer to Problem 37E

The element having atomic number $55$ iscesium. It is an alkali metal.

Explanation of Solution

The atomic number of the given element is $55$.

The electronic configuration of the element having atomic number $55$ 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 outermost shell electronic configuration $6s^1$ indicates that the element belongs to the sixth period of the periodic table and one valence electron shows that it is a $1^{\text{st}}$ group element. Therefore, the element is cesium which is an alkali metal.