Chapter 6, Problem 83SCQ

(a)

Interpretation Introduction

Interpretation: The group and period of technetium has to be identified.

Concept introduction:

Periodic Table: The available chemical elements are arranged considering their atomic number, the electronic configuration and their properties. The elements placed on the left of the table are metals and non-metals are placed on right side of the table.

In periodic table the horizontal rows are called periods and the vertical column are called group. There are seven periods and 18 groups present in the table and some of those groups are given special name as follows,

  $\begin{array}{l}Group1\to Alkalimetal\\ Group2\to Alkalinemetal\\ Group16\to Chalcogens\\ Group17\to Halogens\\ Group18\to Noblegases\end{array}$

Answer to Problem 83SCQ

Technetium is in Period $5$ and Group $7$

Explanation of Solution

The atomic number of technetium is $43$. The electronic configuration is $\left[\text{Kr}\right]{\text{4d}}^{\text{5}}{\text{5s}}^{\text{2}}$. Here $\text{Tc}$ has $5$ shells so is in Period $5$. The number of valence electron is $7$ which implies that $\text{Tc}$ is in Group $7$ $\left(\text{7B}\right)$.

(b)

Interpretation Introduction

Interpretation: The possible set of quantum number for $\text{5s}$ subshell of technetium is to be determined.

Concept introduction:

Quantum numbers are numbers, which explains the existence and the behavior of electron in an atom.

1. a) Principle quantum number is represented by n and this number describes the energy of the orbital and the size of an atom.
2. b) Angular momentum quantum number (or azimuthal quantum number) is represented by l and this number indicates the shape of the orbitals.
3. c) Magnetic quantum number is represented by ${\text{m}}_{\text{l}}$ and this number indicates the orientation of the orbital.
4. d) Spin quantum number is represented by ${\text{m}}_{\text{s}}$ and this number indicates the spin of the electron.

Principle quantum number is represented by n and this number describes the energy of the orbital and the size of an atom.

The values of l when the principal quantum number is n are from $0$ to $\left(\text{n}-1\right)$. Each l value indicates subshell.

The values of $m_l$ when the orbital angular quantum number is l are from $-l$ to $+l$.

Answer to Problem 83SCQ

The possible set of quantum number for $\text{5s}$ subshell of technetium is $n=5$, $l=0$, $m_l=0$ and $m_s=\text{+1/2}$

Explanation of Solution

Principle quantum number is represented by n and this number describes the energy of the orbital and the size of an atom. The values of l when the principal quantum number is n are from $0$ to $\left(\text{n}-1\right)$. Each l value indicates subshell. The values of $m_l$ when the orbital angular quantum number is l are from $-l$ to $+l$.

For the orbital $\text{5s}$, the principal quantum number,$n=5$ . For s subshell, $l=0$.

When $l=0$, the values of $m_l$ are, $m_l=0$ because the values of $m_l$ are from $-l$ to $+l$.

The values of $m_s$ can be either $\text{+1/2}$ or $-\text{1/2}$. So the set of quantum numbers is possible.

$n=5$, $l=0$, $m_l=0$ and $m_s=\text{+1/2}$

(c)

Interpretation Introduction

Interpretation: The wavelength and frequency of the $\gamma$-ray emitted by technetium has to be calculated.

Concept introduction:

• Planck’s equation,

    $\begin{array}{c}\text{E}\text{=}\text{hν}\\ \text{where, E}\text{=}\text{energy}\\ \text{h}\text{=}\text{Planck's}\text{constant}\\ \text{ν}=\text{frequency}\end{array}$

The energy increases as the wavelength of the light decrease. Also the energy increases as the frequency of the light increases.

• The frequency of the light is inversely proportional to its wavelength.

  $\begin{array}{c}\text{ν}\text{=}\frac{\text{c}}{\text{λ}}\\ \text{where, c}\text{=}\text{speed}\text{of}\text{light}\\ \text{ν}=\text{frequency}\\ \text{λ}=\text{wavelength}\end{array}$

Answer to Problem 83SCQ

The wavelength of the $\gamma$-ray emitted by technetium is $8.79\times {10}^{-12}\text{m}$.

The frequency of the $\gamma$-ray emitted by technetium is $3.41\times {10}^{19}{s}^{-1}$.

Explanation of Solution

Given,

Planck’s constant is $\text{h}\text{=}\text{6}\text{.626}\text{×}{\text{10}}^{-\text{34}}\text{J}\text{.s}$

The speed of light is $\text{2}\text{.998}\text{×}\text{10}{}^{\text{8}}\text{m/s}$

The energy of the $\gamma$-ray emitted by technetium is,

  $\begin{array}{c}0.141MeV=\left(0.141\times {10}^6\times 1.6022\times {10}^{-19}\right)J\\ =2.259\times {10}^{-14}J\end{array}$

• The frequency of the $\underset{\_}{\gamma }$-ray emitted by technetium is calculated by the equation below

The energy is calculated by the equation below,

    $\text{E}\text{=}\text{hν}$

Therefore,

The frequency of the $\gamma$-ray emitted by technetium is,

  $\begin{array}{c}\nu =\frac{E}{h}\\ \\ =\frac{2.259\times {10}^{-14}J}{\text{6}\text{.626}\text{×}{\text{10}}^{-\text{34}}\text{J}\text{.s}}\\ \\ =3.41\times {10}^{19}{s}^{-1}\end{array}$

The frequency of the $\gamma$-ray emitted by technetium is $3.41\times 10^{19}{s}^{-1}$

• The wavelength of the $\underset{\_}{\gamma }$-ray emitted by technetium is calculated by the equation below

  $\begin{array}{c}\text{λ}\text{=}\frac{\text{c}}{\text{ν}}\\ \\ =\frac{\text{2}\text{.998}\text{×}\text{10}{}^{\text{8}}\text{m/s}}{3.41\times {10}^{19}{s}^{-1}}\\ \\ =8.79\times {10}^{-12}\text{m}\end{array}$

The wavelength of the $\gamma$-ray emitted by technetium is $8.79\times 10^{-12}m$

(d)

Interpretation Introduction

Interpretation:. The balanced equation of the reaction between $HTc{O}_4$ and $NaOH$ has to be determined. The mass of $NaTc{O}_4$ produced and $NaOH$ converted have to be calculated.

Concept introduction:

• Chemical equation is the representation of a chemical reaction where the reactants and products of the reactions are represented on left and right side of an arrow respectively by using their respective chemical formulas. It follows conservation law of mass where the number of atoms of each element is equal for reactants and products.
• The number of moles of any substance can be determined  using the equation,

  $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

Answer to Problem 83SCQ

1. i. ${\text{HTcO}}_4{}_{\left(aq\right)}\text{+}{\text{NaOH}}_{\left(aq\right)}\underset{}{\to }{\text{NaTcO}}_4{}_{\left(aq\right)}\text{+}{\text{H}}_2{\text{O}}_{\left(l\right)}$
2. ii. The mass of $NaTc{O}_4$ produced and $NaOH$ converted is $\text{8}\text{.5}\times {\text{10}}^{-3}g$ and $1.8\times {10}^{-3}g$ respectively.

Explanation of Solution

(i)

$NaTc{O}_4$ is formed when $HTc{O}_4$ is treated by $NaOH$. $H_{2}O$ is formed as the byproduct.

The number of elements in reactants and products are equal, so no appropriate coefficient is needed.

Therefore the balanced equation for the given reaction is,

  ${\text{HTcO}}_4{}_{\left(aq\right)}\text{+}{\text{NaOH}}_{\left(aq\right)}\underset{}{\to }{\text{NaTcO}}_4{}_{\left(aq\right)}\text{+}{\text{H}}_2{\text{O}}_{\left(l\right)}$

(ii)

The mass of $NaTc{O}_4$ produced and $NaOH$ converted is calculated

Given,

Mass of $Tc$ is $4.5mg=0.0045g$

Molar mass of $NaTc{O}_4$ is $\text{184}\text{.99}\text{g/mol}$

Molar mass of $NaOH$ is $39.997\text{g/mol}$

Balanced equation for the reaction is,

$\text{7}{\text{HNO}}_3{}_{\left(aq\right)}\text{+}{\text{Tc}}_{\left(s\right)}\underset{}{\to }{\text{HTcO}}_4{}_{\left(aq\right)}\text{+}\text{7}{\text{NO}}_{2\left(g\right)}\text{+}3{\text{H}}_2{\text{O}}_{\left(l\right)}$                (a)

${\text{HTcO}}_4{}_{\left(aq\right)}\text{+}{\text{NaOH}}_{\left(aq\right)}\underset{}{\to }{\text{NaTcO}}_4{}_{\left(aq\right)}\text{+}{\text{H}}_2{\text{O}}_{\left(l\right)}$                    (b)

• The mass of $NaTc{O}_4$ produced is calculated,

The amount of $HTc{O}_4$ require to complete the reaction can be calculated from the amount of $Tc$

The amount of $Tc$ available = $\frac{0.0045g}{98\text{g/mol}}=4.59\times {10}^{-5}molTc$

From the balanced equation (a) it is clear that $HTc{O}_4$ and $Tc$ are reacting in the ratio of $1:1$ which indicates that number of moles of $HTc{O}_4$ and $Tc$ is same

The amount of $HTc{O}_4$ is $4.59\times {10}^{-5}molHTc{O}_4$

From the balanced equation (b) $HTc{O}_4$ and $NaTc{O}_4$ are reacting in the ratio of $1:1$ which indicates that number of moles of $HTc{O}_4$ and $NaTc{O}_4$ is same. Therefore, the amount of $NaTc{O}_4$ available is also $4.59\times {10}^{-5}molNaTc{O}_4$

Number of moles is calculated by the given equation,

  $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

Therefore,

$\begin{array}{c}MassofNaTc{O}_4=Numberofmoles\times molarmass\\ \\ =4.59\times {10}^{-5}molNaTc{O}_4\times \text{184}\text{.99}\text{g/mol}NaTc{O}_4\\ \\ \text{=}\text{8}\text{.5}\times {\text{10}}^{-3}gNaTc{O}_4\end{array}$

The mass of $NaTc{O}_4$ produced is $8.5\times 10^{-3}g$

• The mass of $NaOH$ converted is calculated

From the balanced equation (b) it is clear that $NaOH$ and $NaTc{O}_4$ are reacting in the ratio of $1:1$ which indicates that number of moles of $NaOH$ and $NaTc{O}_4$ is same

The amount of $NaOH$ needed to convert $HTc{O}_4$ to $NaTc{O}_4$ is,

  $\begin{array}{l}=4.59\times {10}^{-5}molHTc{O}_4\times \frac{1molofNaOH}{1molofHTc{O}_4}\times 39.997\text{g/mol}NaOH\\ \\ =1.8\times {10}^{-3}gNaOH\end{array}$

The mass of $NaOH$ converted is $1.8\times 10^{-3}g$

(e)

Interpretation Introduction

Interpretation: Mass and the concentration of $NaTc{O}_4$ has to be calculated if it is dissolved in $10.0mL$ solution.

Concept introduction:

• Molarity (M): Molarity is number of moles of the solute present in the one liter of the solution.

  $\text{Molarity (M) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1}\text{liter}\text{of}\text{solution}}$

• The number of moles of any substance can be determined  using the equation,

  $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

Answer to Problem 83SCQ

Mass of $NaTc{O}_4$ is $0.28mg$ and the concentration of $NaTc{O}_4$ if it is dissolved in $10.0mL$ solution is $0.00015M$

Explanation of Solution

Number of moles of $NaTc{O}_4$ is $1.5micromoles=1.5\times {10}^{-6}moles$

Molar mass of $NaTc{O}_4$ is $\text{184}\text{.99}\text{g/mol}$

Number of moles is calculated by the given equation,

$\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

Therefore,

$\begin{array}{c}MassofNaTc{O}_4=Numberofmoles\times molarmass\\ \\ =1.5\times {10}^{-6}molesNaTc{O}_4\times \text{184}\text{.99}\text{g/mol}NaTc{O}_4\\ \\ \text{=}0.28mgNaTc{O}_4\end{array}$

Mass of $NaTc{O}_4$ is $0.28mg$.

Concentration of $NaTc{O}_4$ if it is dissolved in $10.0mL=0.01L$ is calculated

$\begin{array}{c}\text{Molarity (M) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1}\text{liter}\text{of}\text{solution}}\\ \\ =\frac{1.5\times {10}^{-6}moles}{0.01L}\\ \\ =0.00015M\end{array}$

Concentration of $NaTc{O}_4$ if it is dissolved in $10.0mL$ solution is $0.00015M$