Chapter 6, Problem 49QAP

Write the ground state electron configuration for

(a) Mg, Mg²⁺(b) N, N^3-

(c) Ti, Ti⁴⁺

(d) Sn²⁺ Sn⁴⁺

Interpretation Introduction

(a)

Interpretation:

To state the ground state electronic configuration for ${\text{Mg, Mg}}^{2+}$.

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. A cation is formed when the parent atom loose one or more electrons. An anion is formed by the gain of one or more electrons by the parent atom in its valence shell.

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 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.

Answer to Problem 49QAP

The ground state electronic configuration for Mg is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}$

The ground state electronic configuration for Mg²+ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}$

Explanation of Solution

Since the atomic number of Magnesium (Mg) is 12, therefore its ground state electronic configuration is: ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}$

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. Mg²+ is formed from Mg after the loss of two electrons from its valence shell. This means that loss of two electrons take place from 3s orbital of Mg to form Mg²+cation. The electronic configuration of Mg²+becomes:

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

Interpretation Introduction

(b)

Interpretation:

To state the ground state electronic configuration for ${\text{N, N}}^{3-}$.

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. A cation is formed when the parent atom loose one or more electrons. An anion is formed by the gain of one or more electrons by the parent atom in its valence shell.

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 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.

Answer to Problem 49QAP

The ground state electronic configuration for N is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{3}}$

The ground state electronic configuration forN³- is

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

Explanation of Solution

Since the atomic number of Nitrogen atom denoted by N is 7, therefore its ground state electronic configuration is:

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

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. When N atom gains three electrons in order to complete its octet, it leads to the formation of an anion namely, nitride ion denoted by N^3-.The electronic configuration of N³-becomes:

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

Interpretation Introduction

(c)

Interpretation:

To state the ground state electronic configuration for ${\text{Ti, Ti}}^{\text{4+}}$.

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. A cation is formed when the parent atom loose one or more electrons. An anion is formed by the gain of one or more electrons by the parent atom in its valence shell.

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 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.

Answer to Problem 49QAP

The ground state electronic configuration for Ti is 1s²2s²2p⁶ 3s²3p⁶3d²4s²

The ground state electronic configuration for Ti⁴+is 1s²2s²2p⁶ 3s²3p⁶

Explanation of Solution

Since the atomic number of Titanium atom denoted by Ti is 22, therefore its ground state electronic configuration is: ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{2}}{\text{4s}}^{\text{2}}$

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. When Ti loses four electrons it form Ti⁴+cation and the electronic configuration of resultant cation becomes;

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

Interpretation Introduction

(d)

Interpretation:

To state the ground state electronic configuration for ${\text{Sn}}^{2+}{\text{, Sn}}^{4+}$.

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. A cation is formed when the parent atom loose one or more electrons. An anion is formed by the gain of one or more electrons by the parent atom in its valence shell.

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 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.

Answer to Problem 49QAP

The ground state electronic configuration of Sn²+ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6 }}{\text{4s}}^{\text{2}}{\text{3d}}^{\text{10}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}$

The ground state electronic configuration of Sn⁴+ is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6 }}{\text{4s}}^{\text{2}}{\text{3d}}^{\text{10}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}$

Explanation of Solution

Since the atomic number of Tin atom denoted by Sn is 50, therefore its ground state electronic configuration is: ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6 }}{\text{4s}}^{\text{2}}{\text{3d}}^{\text{10}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^2$

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. Sn has the tendency to form both Sn²+ and Sn⁴+cations by losing two and four electrons respectively. When it loses two electrons from outer 5p orbital it forms Sn²+whose electronic configuration is:

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

When it loses four electrons (two from 5p and two from 5s), then it forms Sn⁴⁺, and the electronic configuration becomes: ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6 }}{\text{4s}}^{\text{2}}{\text{3d}}^{\text{10}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}$