Chapter 12, Problem 141AE

(a)

Interpretation Introduction

Interpretation: Atomic number for element that has 2 electrons with $n$ equal to 1, 8 with $n$ equal to 1, and 10 electrons with $n$ equal to 3 should be identified.

Concept introduction:To remove the electron situated in outermost shell certain minimum energy must be imparted so as to convert an atom to gaseous species. The energy thus imparted represents ionization energy.

The magnitude of ionization energy is determined by how effectively valence electron is held by the nucleus. If the outermost shell has, for instance, one or two electronsthat require very minimum ionization energy because they can attain the noble gas configuration upon loss of those electrons.

Explanation of Solution

With 2 electrons with $n$ equal to 1, 8 with $n$ equal to 1, and 10 electrons with $n$ equal to 3the configuration can be written as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^2$ . Since this ion has resulted from loss of five electrons so expected atomic nmber corresponding to this configuration is 24.

(b)

Interpretation Introduction

Interpretation:Electrons in $s$ -orbital for this element should be identified.

Concept introduction:To remove the electron situated in outermost shell certain minimum energy must be imparted so as to convert an atom to gaseous species. The energy thus imparted represents ionization energy.

The magnitude of ionization energy is determined by how effectively valence electron is held by the nucleus. If the outermost shell has, for instance, one or two electronsthat require very minimum ionization energy because they can attain the noble gas configuration upon loss of those electrons.

Explanation of Solution

With 2 electrons with $n$ equal to 1, 8 with $n$ equal to 1, and 10 electrons with $n$ equal to 3the configuration can be written as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^2$ .The configuration $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^2$ indicates 6 electrons in $s$ -orbital.

(c)

Interpretation Introduction

Interpretation:Electrons in $p$ -orbital for this element should be identified.

Concept introduction:To remove the electron situated in outermost shell certain minimum energy must be imparted so as to convert an atom to gaseous species. The energy thus imparted represents ionization energy.

The magnitude of ionization energy is determined by how effectively valence electron is held by the nucleus. If the outermost shell has, for instance, one or two electronsthat require very minimum ionization energy because they can attain the noble gas configuration upon loss of those electrons.

Explanation of Solution

With 2 electrons with $n$ equal to 1, 8 with $n$ equal to 1, and 10 electrons with $n$ equal to 3the configuration can be written as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^2$ .Theconfiguration $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^2$ indicates12 electrons in $p$ -orbital.

(d)

Interpretation Introduction

Interpretation:Electrons in $d$ -orbital for this element should be identified.

Concept introduction:To remove the electron situated in outermost shell certain minimum energy must be imparted so as to convert an atom to gaseous species. The energy thus imparted represents ionization energy.

The magnitude of ionization energy is determined by how effectively valence electron is held by the nucleus. If the outermost shell has, for instance, one or two electronsthat require very minimum ionization energy because they can attain the noble gas configuration upon loss of those electrons.

Explanation of Solution

With 2 electrons with $n$ equal to 1, 8 with $n$ equal to 1, and 10 electrons with $n$ equal to 3the configuration can be written as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^2$ .The configuration $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^2$ indicates twoelectrons in $d$ -orbital.

(e)

Interpretation Introduction

Interpretation:Number of neutrons in this element should be identified.

Concept introduction:To remove the electron situated in outermost shell certain minimum energy must be imparted so as to convert an atom to gaseous species. The energy thus imparted represents ionization energy.

The magnitude of ionization energy is determined by how effectively valence electron is held by the nucleus. If the outermost shell has, for instance, one or two electronsthat require very minimum ionization energy because they can attain the noble gas configuration upon loss of those electrons.

Explanation of Solution

With 2 electrons with $n$ equal to 1, 8 with $n$ equal to 1, and 10 electrons with $n$ equal to 3the configuration can be written as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^2$ . Since this ion has resulted from loss of five electrons so expected atomic number corresponding to this configuration is 24.

The formula to compute neutrons from mass number is as follows:

  $\text{Number of neutrons}=\text{Mass number}-\text{Atomic number}$

Atomic number is 24.

Mass number is 52.

Substitute the value in above formula.

  $\begin{array}{c}\text{Number of neutrons}=\text{Mass number}-\text{Atomic number}\\ =\text{52}-\text{24}\\ =\text{28}\end{array}$

So there are 28 neutrons in chromium.

(f)

Interpretation Introduction

Interpretation:Mass of $3.01\times {10}^{23}$ atoms is to be identified.

Concept introduction:To remove the electron situated in outermost shell certain minimum energy must be imparted so as to convert an atom to gaseous species. The energy thus imparted represents ionization energy.

The magnitude of ionization energy is determined by how effectively valence electron is held by the nucleus. If the outermost shell has, for instance, one or two electronsthat require very minimum ionization energy because they can attain the noble gas configuration upon loss of those electrons.

Explanation of Solution

With 2 electrons with $n$ equal to 1, 8 with $n$ equal to 1, and 10 electrons with $n$ equal to 3the configuration can be written as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^2$ . Since this ion has resulted from loss of five electrons so expected atomic number corresponding to this configuration is 24. So, element is identified as chromium.

Since molar mass of chromium ion is $51.99\text{ g/mol}$ , so mass of $3.01\times {10}^{23}$ atoms is calculated as follows:

  $\begin{array}{c}\text{Mass}=\left(3.01\times {10}^{23}\text{ atoms}\right)\left(\frac{1\text{ mol}}{6.022\times {10}^{23}\text{ atoms}}\right)\left(\frac{51.99\text{ g}}{1\text{ mol}}\right)\\ =25.9\text{ g}\end{array}$

Thus, mass of $3.01\times {10}^{23}$ atoms is $25.9\text{ g}$ .

(g)

Interpretation Introduction

Interpretation:Ground-state electron configuration of neutral chromium should be written.

Concept introduction:Aufbau rule states that electrons must be filled in lowest energy levels first. For instance, electrons first occupy shells that are lower in energies illustrated as follows:

  

Pauli’s exclusion principle states thatno two or more than two electrons of a poly electron atom can have same values of 4 quantum numbers that are $n$ , $l$ , $m$ and $s$ .It also states that only two electrons can occupy same orbital with opposite spins.

Hund’s rule of maximum multiplicity states that electrons cannot be allowed to pair until each orbital gets singly filled with one electron. These 3 principles form basis for determination of electronic configuration.However, certain elements that are able to achieve nearest half-filled or fully filed configuration show exceptional configurations.

Explanation of Solution

With 2 electrons with $n$ equal to 1, 8 with $n$ equal to 1, and 10 electrons with $n$ equal to 3the configuration can be written as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^2$ . Since this ion has resulted from loss of five electrons so expected atomic number corresponding to this configuration is 24.

With atomic number as 24, expected configuration for $\text{Cr}$ is $\left[\text{Ar}\right]4s^{2}3d^4$ while actual ground electronic configuration for $\text{Cr}$ is $\left[\text{Ar}\right]4s^{1}3d^5$ . This is because of greater stability that arises in latter case due to half-filled configuration.