Chapter 6, Problem 67QAP

Write the symbol of each element described below.

(a) largest atomic radius in Group 1

(b) smallest atomic radius in period 3

(c) largest first ionization energy in Group 2

(d) most electronegative in Group 16

(e) element(s) in period 2 with no unpaired p electron

(f) abbreviated electron configuration is [Ar] 4s²3d³

(g) A +2 ion with abbreviated electron configuration [Ar]3d⁵

(h) A transition metal in period 4 forming a +2 ion with no unpaired electrons

Interpretation Introduction

Interpretation:

The symbol of an element which has the largest atomic radius in Group 1 should be determined.

Concept Introduction:

The Modern Periodic law states that the physical and chemical properties of the elements are a periodic function of their atomic numbers. This means that changes in the chemical properties of various elements present in the modern periodic table are due to their atomic numbers. Atomic number of an element gives the total number or electrons or protons present in that atom. The main cause of periodicity is the repetition of similar outer electronic configuration after regular intervals.

Answer to Problem 67QAP

The symbol of the element which has the largest size in Group 1 is Cs.

Explanation of Solution

The atomic radius of an element is a measure of size of its atoms. It can be measured and defined by assuming the atom to be spherical. It refers to one half the distance of closest approach between the atoms present in an elemental substance.

When the electronic configuration of an atom is written, it describes the number of electron present in each sublevel by the superscript.

As the number of shells in an atom increases, the distance of last electron from the nucleus increases, and hence the atomic radius increases.

The atomic radius decreases as we move across the period due to an increase in the effective nuclear charge on the last valence electron.

This results in increase in pull of outermost electrons more tightly towards the nucleus and hence leads to decrease in atomic radius.

Now in case of Li, Na, K and Rb and Cs, the electronic configuration of these atoms are as follows:

Li (3)- ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{1}}$

Na(11)- ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{1}}$

K(19)- ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{4s}}^{\text{1}}$

Rb (37)- ${\text{1s}}^{\text{2 }}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{5s}}^{\text{1}}$

Cs (55)- ${\text{1s}}^{\text{2 }}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{6s}}^{\text{1}}$

With the addition of more number of principal shells down the group, the atomic size will increase and will be highest for Cesium (Cs).

Interpretation Introduction

Interpretation:

To determine the symbol of the element having smallest radius in period 3.

Concept Introduction:

The Modern Periodic law states that the physical and chemical properties of the elements are a periodic function of their atomic numbers. This means that changes in the chemical properties of various elements present in the modern periodic table are due to their atomic numbers. Atomic number of an element gives the total number or electrons or protons present in that atom. The main cause of periodicity is the repetition of similar outer electronic configuration after regular intervals.

Answer to Problem 67QAP

The symbol of the element having smallest radius in period 3 is Ar.

Explanation of Solution

The atomic radius of an element is a measure of size of its atoms. It can be measured and defined by assuming the atom to be spherical. It refers to one half the distance of closest approach between the atoms present in an elemental substance.

When the electronic configuration of an atom is written, it describes the number of electron present in each sublevel by the superscript.

As the number of shells in an atom increases, the distance of last electron from the nucleus increases and hence, the atomic radius increases.

The atomic radius decreases as we move across the period due to increase in effective nuclear charge on the last valence electron.

This results in increase in pull of outermost electrons more tightly towards the nucleus and hence, leads to decrease in atomic radius.

In period 3, as we move from sodium towards argon, electrons are added to the same 3p subshell, due to which the increase in shielding is negligible as each extra electron enters the same subshell. The force of attraction between the nucleus and the electron increases and thus, atomic radius decreases. Since, argon is present at the extreme right of the periodic table in period number 3. Therefore, its atomic radius will be smallest.

Interpretation Introduction

Interpretation:

The symbol of an element which has the largest ionization energy in group 2 should be determined.

Concept Introduction:

The Modern Periodic law states that the physical and chemical properties of the elements are a periodic function of their atomic numbers. This means that changes in the chemical properties of various elements present in the modern periodic table are due to their atomic numbers. Atomic number of an element gives the total number or electrons or protons present in that atom. The main cause of periodicity is the repetition of similar outer electronic configuration after regular intervals.

Answer to Problem 67QAP

The symbol of the element which has the largest ionization energy in group 2 is Be.

Explanation of Solution

The atomic radius of an element is a measure of size of its atoms. It can be measured and defined by assuming the atom to be spherical. It refers to one half the distance of closest approach between the atoms present in an elemental substance.

When the electronic configuration of an atom is written, it describes the number of electron present in each sublevel by the superscript.

As the number of shells in an atom increases, the distance of last electron from the nucleus increases, and hence the atomic radius increases.

The atomic radius decreases as we move across the period due to an increase in the effective nuclear charge on the last valence electron.

This results in an increase in the pull of outermost electrons more tightly towards the nucleus and hence, leads to a decrease in the atomic radius.

The first ionization energy of a chemical element is the energy required to remove the most loosely bound electron from an atom in its gaseous state. It is inversely proportional to the size of the atoms.

This means:

1. On moving down the group in the periodic table, the ionization energy of the elements decreases as the size of the atom increases. Therefore, the removal of an electron becomes easier.
2. On moving across a period, the ionization energy of the elements increases due to the decrease in the size of the atoms. Hence, the effective nuclear charge increases and the removal of electrons become difficult.

Group 2 elements consists of Be, Mg ,Ca, Sr , Ba and Ra. Since, the size of Be is smallest, therefore, its electrons are held more tightly to the nucleus and more energy is required for the removal of electrons from its 2s subshell. Hence, its ionzation energy is highest.

Interpretation Introduction

Interpretation:

To determine the symbol of the most electronegative element in group 16.

Concept Introduction:

The Modern Periodic law states that the physical and chemical properties of the elements are a periodic function of their atomic numbers. This means that changes in the chemical properties of various elements present in the modern periodic table are due to their atomic numbers. Atomic number of an element gives the total number or electrons or protons present in that atom. The main cause of periodicity is the repetition of similar outer electronic configuration after regular intervals.

Answer to Problem 67QAP

The symbol of the most electronegative element in group 16 is O.

Explanation of Solution

The electronegativity of an atom is a chemical property which describes its tendency to attract shared electron pair towards itself and it results in the formation of a covalent bond. The electronegativity value decreases down the group because as we move down the group, the atomic radius increases.

In group 16, electronegativity of oxygen is highest as it has the smallest size and therefore, has the highest tendency to accept electrons in comparison to other group 16 elements.

Interpretation Introduction

Interpretation:

To determine the symbol of the element(s) in period 2 which have no unpaired p electron.

Concept Introduction:

The simplest method for describing the arrangement of electrons in an atom is by writing its electronic configuration. Since the set of four quantum numbers is used to describe the atomic orbitals in an atom, therefore by writing the electronic configuration, one can get details of the number of electrons present in each sublevel.

Answer to Problem 67QAP

The symbol of the element(s) in period 2 which have no unpaired p electron is Ne.

Explanation of Solution

When the electronic configuration of an atom is written, it describes the number of electron present in each sublevel by the superscript. While writing the electronic configuration, it is assumed that atom is present in its isolated gaseous state. Electrons are filled in order of the increasing energies of the various sublevels. Atomic number of an element gives the total number of electrons present in an atom in case of neutral atom.

When the ground state electronic configuration of an atom is written, it is assumed that none of its electrons have been excited to the higher energy level. The number of unpaired electrons in an orbital is the electrons which are singly occupied in the orbitals.

In period 2, the elements are:

Li − ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{1}}$

Be- ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}$

B- ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{1}}$

C- ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{2}}$

N- $1s^{2}2s^{2}2p^3$

O- $1s^{2}2s^{2}2p^4$

F- $1s^{2}2s^{2}2p^5$

Ne- $1s^{2}2s^{2}2p^6$

Only Be and Ne have no unpaired electrons as in case of Be the valence orbital 2s contains two electrons which are a maximum number of electrons that can be accommodated by any s-subshell. In case of Ne, 2p orbital contains 6 electrons which are a maximum number of electrons that can be accommodated by and p-subshell.

Interpretation Introduction

Interpretation:

To determine the symbol of the element whose abbreviated electronic configuration is $\left[\text{Ar}\right]{\text{ 4s}}^{\text{2}}{\text{3d}}^{\text{3}}$ .

Concept Introduction:

The simplest method for describing the arrangement of electrons in an atom is by writing its electronic configuration. Since the set of four quantum numbers is used to describe the atomic orbitals in an atom, therefore by writing the electronic configuration, one can get details of the number of electrons present in each sublevel.

This filling of electrons in the atomic orbitals takes place according to the Aufbau principal which states that when an atom is present in its ground state, electrons are filled in order of increasing energy of the orbitals, which means that firstly lower energy orbitals are filled, and then filling of higher energy orbitals takes place.

Answer to Problem 67QAP

The symbol of the element whose abbreviated electronic configuration is $\left[\text{Ar}\right]{\text{ 4s}}^{\text{2}}{\text{3d}}^{\text{3}}$ is V.

Explanation of Solution

When the electronic configuration of an atom is written, it describes the number of electron present in each sublevel by the superscript. While writing the electronic configuration, it is assumed that atom is present in its isolated gaseous state. Electrons are filled in order of the increasing energies of the various sublevels. Atomic number of an element gives the total number of electrons present in an atom. When the ground state electronic configuration of an atom is written, it is assumed that none of its electrons have been excited to the higher energy level.

The abbreviated configuration of an element depicts its total number of electrons by writing the nearest noble gas configuration in square brackets plus the remaining orbitals in the respective orbitals.

Here, the abbreviated electronic configuration of element is given to be $\left[\text{Ar}\right]{\text{ 4s}}^{\text{2}}{\text{3d}}^{\text{3}}$ .This clearly suggests that it contains 23 electrons. The element having atomic number 23 is Vanadium which is a transition metal element and denoted by the symbol V.

Interpretation Introduction

Interpretation:

To determine the symbol of the element whose +2 ion has the abbreviated electronic configuration is $\left[\text{Ar}\right]{\text{ 3d}}^{\text{5}}$ .

Concept Introduction:

The simplest method for describing the arrangement of electrons in an atom is by writing its electronic configuration. Since the set of four quantum numbers is used to describe the atomic orbitals in an atom, therefore by writing the electronic configuration, one can get details of the number of electrons present in each sublevel.

This filling of electrons in the atomic orbitals takes place according to the Aufbau principal which states that when an atom is present in its ground state, electrons are filled in order of increasing energy of the orbitals, which means that firstly lower energy orbitals are filled, and then filling of higher energy orbitals takes place.

Answer to Problem 67QAP

The symbol of the element whose +2 ion has the abbreviated electronic configuration is $\left[\text{Ar}\right]{\text{ 3d}}^{\text{5}}$ is Mn.

Explanation of Solution

When the electronic configuration of an atom is written, it describes the number of electron present in each sublevel by the superscript. While writing the electronic configuration, it is assumed that atom is present in its isolated gaseous state. Electrons are filled in order of the increasing energies of the various sublevels. Atomic number of an element gives the total number of electrons present in an atom. When the ground state electronic configuration of an atom is written, it is assumed that none of its electrons have been excited to the higher energy level.

A cation is formed by the loss of electrons. +2 ion must have been formed by the loss of two electrons by the atom. The abbreviated configuration of an element depicts its total number of electrons by writing the nearest noble gas configuration in square brackets plus the remaining orbitals in the respective orbitals.

Here, the abbreviated electronic configuration of the +2 ion is $\left[\text{Ar}\right]{\text{ 3d}}^{\text{5}}$ .This means that the atom must have lost two electrons to attain this configuration. If two electrons are added, then the desired electronic configuration of the element is $\left[\text{Ar}\right]{\text{ 3d}}^{\text{5}}{\text{4s}}^{\text{2}}$ which the abbreviated electronic configuration of Mn atom. Therefore, when Mn loses two electrons then the desired electronic configuration is $\left[\text{Ar}\right]{\text{ 3d}}^{\text{5}}$ .

Interpretation Introduction

Interpretation:

To determine the symbol of the transition element in period 4 which forms +2 ion with no unpaired electrons.

Concept Introduction:

The simplest method for describing the arrangement of electrons in an atom is by writing its electronic configuration. Since the set of four quantum numbers is used to describe the atomic orbitals in an atom, therefore by writing the electronic configuration, one can get details of the number of electrons present in each sublevel.

Answer to Problem 67QAP

The symbol of the transition element in period 4 which forms +2 ion with no unpaired electrons is Zn.

Explanation of Solution

When the electronic configuration of an atom is written, it describes the number of electron present in each sublevel by the superscript. While writing the electronic configuration, it is assumed that atom is present in its isolated gaseous state. Electrons are filled in order of the increasing energies of the various sublevels. Atomic number of an element gives the total number of electrons present in an atom. When the ground state electronic configuration of an atom is written, it is assumed that none of its electrons have been excited to the higher energy level. The number of unpaired electrons in an orbital is the electrons which are singly occupied in the orbitals.

The transition elements which are present in period 4 are Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu and Zn .

The abbreviated electronic configuration of Zn is: $\left[\text{Ar}\right]{\text{ 3d}}^{\text{10}}{\text{4s}}^{\text{2}}$

When it forms Zn²⁺, its electronic configuration becomes: $\left[\text{Ar}\right]{\text{ 3d}}^{\text{10}}$

Now as d- orbital can accommodate a maximum of ten electrons, therefore all the electrons are paired in this orbital.