Chapter 12, Problem 166CP

a.

Interpretation Introduction

Interpretation: The sketch of first four periods of the periodic table needs to be drawn that will look like in the given universe.

Concept Introduction: The electrons are arranged around the nucleus of an atom in an increasing order of energy levels and this description of atomic orbitals of atom occupied by electrons is known as electronic configuration. The size, shape and energy of an atomic orbital is indicated by a set of four quantum numbers.

Answer to Problem 166CP

The sketch of first four periods of the periodic table in the given universe is:

1							2
3							4
5	6	7	8	9	10	11	12
13	14	15	16	17	18	19	20

Explanation of Solution

Given:

The four quantum numbers, $\text{p, q, r and s}$ . The rules related to them are:

  $\text{p = 1, 2, 3, 4, }\mathrm{.......}$ ; $\text{q }$ takes positive integer values and $\text{q }\le \text{ p}$ ; $\text{r}$ takes on all the even integer values from $\text{-q to +q}$ ; and $\text{s = +}\frac{1}{2}\text{ or -}\frac{1}{2}$ .

The four quantum numbers are:

• Principal quantum number, $\text{p}$ which indicates the size of the orbital.
• Azimuthal quantum number, $\text{q}$ which indicates the shape and angular momentum of the orbital.
• Magnetic quantum number, $\text{r}$ which indicates the orientation of orbitals.
• Spin quantum number, $\text{s}$ which indicates the angular momentum of the electron.

The meaning of given four quantum numbers and actual quantum numbers is same so, the orbital representation will be:

  $\begin{array}{l}\text{Period Orbitals}\\ \text{ 1 s}\\ \text{ 2 s}\\ \text{ 3 s, p}\\ \text{ 4 s, p}\end{array}$

Since, each orbital occupies two electrons so, the periodic table can be constructed as:

• Two elements ( s -orbital) are present in first period.
• Two elements ( s -orbital) are present in second period.
• Eight elements ( s and p -orbital) are present in third period.
• Eight elements ( s and p -orbital) are present in fourth period.

Hence, the periodic table is:

1							2
3							4
5	6	7	8	9	10	11	12
13	14	15	16	17	18	19	20

b.

Interpretation Introduction

Interpretation: The atomic number of first four elements that are least reactive should be determined.

Concept Introduction: The electrons are arranged around the nucleus of an atom in an increasing order of energy levels and this description of atomic orbitals of atom occupied by electrons is known as electronic configuration. The size, shape and energy of an atomic orbital is indicated by a set of four quantum numbers.

Answer to Problem 166CP

The atomic number of first four elements that are least reactive is $\text{2, 4, 12, 20}$ .

Explanation of Solution

The elements containing completely filled sub-shells are least reactive as they are stable and inert. The atomic number of first four elements that are least reactive is:

  $\begin{array}{l}{\text{2: 1s}}^{\text{2}}\\ {\text{4: 1s}}^{\text{2}}{\text{2s}}^{\text{2}}\\ {\text{12: 1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}\\ {\text{20: 1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{4s}}^{\text{2}}\end{array}$

As all the subshells of the above atomic number elements are completely filled so, the atomic number of first four elements that are least reactive is $\text{2, 4, 12, 20}$ .

c.

Interpretation Introduction

Interpretation: The examples of the ionic compounds needs to be determined for the first four rows with the formula ${\text{XY, XY}}_{\text{2}}{\text{, X}}_2{\text{Y, XY}}_3{\text{ and X}}_2{\text{Y}}_3$ .

Concept Introduction: The electrons are arranged around the nucleus of an atom in an increasing order of energy levels and this description of atomic orbitals of atom occupied by electrons is known as electronic configuration. The size, shape and energy of an atomic orbital is indicated by a set of four quantum numbers.

Explanation of Solution

Ionic compounds are formed by complete transfer of electron(s) from one atom to another in order to attain stability. The overall charge in the compound should be neutralized.

The compound $\text{XY}$ is formed by combining element having atomic number 1 and element with atomic number 11 so, that the charges on both the ions will be +1 for X and -1 for Y.

Hence, the combination is based on the atomic number of elements so:

  $\begin{array}{l}\text{Compound type combination of atomic numbers}\\ \text{XY 1+11 and 3+19}\\ {\text{XY}}_{\text{2}}\text{ 6+11 and 14+19}\\ {\text{X}}_{\text{2}}\text{Y 1+10 and 3+18 }\\ {\text{XY}}_{\text{3}}\text{ 7+11 and 15+19 }\\ {\text{X}}_{\text{2}}{\text{Y}}_{\text{3}}\text{ 7+10 and 15+18}\end{array}$

d.

Interpretation Introduction

Interpretation: The number of electrons that can have $\text{p = 3}$ needs to be determined.

Concept Introduction: The electrons are arranged around the nucleus of an atom in an increasing order of energy levels and this description of atomic orbitals of atom occupied by electrons is known as electronic configuration. The size, shape and energy of an atomic orbital is indicated by a set of four quantum numbers.

Answer to Problem 166CP

The number of electrons that can have $\text{p = 3}$ is $\text{8}$ .

Explanation of Solution

For $\text{p = 3}$ , the values of q will be:

  $\text{q = 1 and 3}$ , that means $\text{s and p-}$ orbitals are there. So, the number of electrons that can have $\text{p = 3}$ is:

  $\text{2 (from s-orbital) + 6 (from p-orbital) = 8}$

e.

Interpretation Introduction

Interpretation: The number of electrons that can have $\text{p = 4, q = 3, r = 2}$ needs to be determined.

Concept Introduction: The electrons are arranged around the nucleus of an atom in an increasing order of energy levels and this description of atomic orbitals of atom occupied by electrons is known as electronic configuration. The size, shape and energy of an atomic orbital is indicated by a set of four quantum numbers.

Answer to Problem 166CP

The number of electrons that can have $\text{p = 4, q = 3, r = 2}$ is $\text{6}$ .

Explanation of Solution

For $\text{p = 4}$ , the values of q will be:

  $\text{q = 1 and 3}$ , that means $\text{s and p-}$ orbitals are there.

For $\text{q = 3}$ , the values of r will be:

  $\text{r = 0, 2 (-2 to +2)}$

So, for $\text{p = 4, q = 3, r = 2}$ the orbital present is p-orbital.

Hence, the number of electrons that can have $\text{p = 4, q = 3, r = 2}$ is:

  $\text{6 (from p-orbital)}$

f.

Interpretation Introduction

Interpretation: The number of electrons that can have $\text{p = 4, q = 3}$ needs to be determined.

Concept Introduction: The electrons are arranged around the nucleus of an atom in an increasing order of energy levels and this description of atomic orbitals of atom occupied by electrons is known as electronic configuration. The size, shape and energy of an atomic orbital is indicated by a set of four quantum numbers.

Answer to Problem 166CP

The number of electrons that can have $\text{p = 4, q = 3}$ is $8$ .

Explanation of Solution

For $\text{p = 4}$ , the values of q will be:

  $\text{q = 1 and 3}$ , that means $\text{s and p-}$ orbitals are there.

For $\text{q = 3}$ , the values of r will be:

  $\text{r = 0, 2 (-2 to +2)}$

So, for $\text{p = 4, q = 3}$ the orbitals present are s- and p-orbital.

Hence, the number of electrons that can have $\text{p = 4, q = 3}$ is:

  $\text{2 (from s-orbital) + 6 (from p-orbital) = 8}$

g.

Interpretation Introduction

Interpretation: The number of electrons that can have $\text{p = 3, q = 0, r = 0}$ needs to be determined.

Concept Introduction: The electrons are arranged around the nucleus of an atom in an increasing order of energy levels and this description of atomic orbitals of atom occupied by electrons is known as electronic configuration. The size, shape and energy of an atomic orbital is indicated by a set of four quantum numbers.

Answer to Problem 166CP

The number of electrons that can have $\text{p = 3, q = 0, r = 0}$ is $0$ .

Explanation of Solution

For $\text{p = 3}$ , the values of q will be:

  $\text{q = 1 and 3}$ .

The value of q is positive integer values and $\text{q }\le \text{ p}$ . Hence, the number of electrons that can have $\text{p = 3, q = 0, r = 0}$ is $0$ .

h.

Interpretation Introduction

Interpretation: The possible values of q and r for $\text{p = 5}$ needs to be determined.

Concept Introduction: The electrons are arranged around the nucleus of an atom in an increasing order of energy levels and this description of atomic orbitals of atom occupied by electrons is known as electronic configuration. The size, shape and energy of an atomic orbital is indicated by a set of four quantum numbers.

Answer to Problem 166CP

The possible values of q and r for $\text{p = 5}$ is:

  $\begin{array}{l}\text{q r}\\ \text{1 0}\\ \text{3 -2, 0 , +2}\\ \text{5 -4, -2, 0, +2, +4}\end{array}$

Explanation of Solution

The value of q is positive integer values and $\text{q }\le \text{ p}$ .

For $\text{p = 5}$ , the values of q will be:

  $\text{q = 1, 3 and 5}$ .

For r, the value of r is even integer from $\text{-q to +q}$ .

The values of r will be:

  $\begin{array}{l}\text{q r}\\ \text{1 0}\\ \text{3 -2, 0 , +2}\\ \text{5 -4, -2, 0, +2, +4}\end{array}$

i.

Interpretation Introduction

Interpretation: The number of electrons that can have $\text{p = 6}$ needs to be determined.

Concept Introduction: The electrons are arranged around the nucleus of an atom in an increasing order of energy levels and this description of atomic orbitals of atom occupied by electrons is known as electronic configuration. The size, shape and energy of an atomic orbital is indicated by a set of four quantum numbers.

Answer to Problem 166CP

The number of electrons that can have $\text{p = 6}$ is $18$ .

Explanation of Solution

The value of q is positive integer values and $\text{q }\le \text{ p}$ .

For $\text{p = 6}$ , the values of q will be:

  $\text{q = 1, 3 and 5}$ .

For r, the value of r is even integer from $\text{-q to +q}$ .

The values of r will be:

  $\begin{array}{l}\text{q r}\\ \text{1 0}\\ \text{3 -2, 0 , +2}\\ \text{5 -4, -2, 0, +2, +4}\end{array}$

From the values of r, the total number of orbitals is 9. Since, each orbital can have 2 electrons so, the number of electrons that can have $\text{p = 6}$ is:

  $\text{2}\times 9=18$