Question

The vibrational frequencies of ${}^{23}{\mathbf{Na}}^1\mathbf{H}{,}^{23}{\mathbf{Na}}^{35}\mathbf{Cl}$ and ${}^{23}{\mathbf{Na}}^{127}\mathbf{I}$ are $\mathbf{3}\mathbf{.}\mathbf{51}\mathbf{\times }{\mathbf{10}}^{13}{\mathbf{s}}^{\mathbf{-}\mathbf{1}}\mathbf{,}\mathbf{1}\mathbf{.}\mathbf{10}\mathbf{\times }{\mathbf{10}}^{13}{\mathbf{s}}^{\mathbf{-}\mathbf{1}}\mathbf{,}\mathbf{and}\mathbf{0}\mathbf{.}\mathbf{773}\mathbf{\times }{\mathbf{10}}^{13}{\mathbf{s}}^{\mathbf{-}\mathbf{1}}$, respectively. Their bond lengths are $\mathbf{1}\mathbf{.}\mathbf{89}\mathbf{\AA }\mathbf{,}\mathbf{2}\mathbf{.}\mathbf{36}\mathbf{\AA }\mathbf{,}\mathbf{and}\mathbf{2}\mathbf{.}\mathbf{71}\mathbf{\AA }$. What are their reduced masses? What are their force constants? If $\mathbf{NaH}\mathbf{and}\mathbf{NaD}$ have the same force constant, what is the vibrational frequency of $\mathbf{NaD}\mathbf{?}\mathbf{D}$ is ${}^2\mathbf{H}$.

Short answer

Reduced mass of ${}^{23}{\mathrm{Na}}^{127}\mathrm{l}\mathrm{is}3.301\times {10}^{-26}\mathrm{kg}$.

Reduced mass of ${}^{23}{\mathrm{Na}}^{35}\mathrm{Cl}\mathrm{is}2.303\times {10}^{-26}\mathrm{kg}$.

Reduced mass of ${}^{23}{\mathrm{Na}}^{127}\mathrm{l}\mathrm{is}3.301\times {10}^{-26}\mathrm{kg}$.

For ${}^{\text{23}}{\text{Na}}^{\text{1}}{\text{H,\hspace{0.17em}k\hspace{0.17em}\hspace{0.17em}is\hspace{0.17em}77.89Nm}}^{\text{-1}}$.

For ${}^{\text{23}}{\text{Na}}^{\text{35}}{\text{Cl,\hspace{0.17em}\hspace{0.17em}k\hspace{0.17em}\hspace{0.17em}is109.9Nm}}^{-1}$.

For${}^{\text{23}}{\text{Na}}^{\text{127}}{\text{I,\hspace{0.17em}\hspace{0.17em}k\hspace{0.17em}\hspace{0.17em}is\hspace{0.17em}\hspace{0.17em}7.78Nm}}^{\text{-1}}$.

The vibrational frequency of $\mathrm{NaD}\mathrm{is}2.53\times {10}^{13}{\mathrm{s}}^{-1}$ .

Step-by-step solution

Reduced mass

The depiction of a two-body system as a single-body system is known as reduced mass.

The moving body’s inertial mass with respect to the body at rest can be simplified to a lower mass when the distinct motion of two bodies is only under “common interactions.”

Given information and formula used

• Vibrational frequency of${}^{23}{\mathbf{Na}}^1{\mathbf{H}}^{23}=\mathbf{3}.\mathbf{51}\times {\mathbf{10}}^{13}{\mathbf{s}}^{-\mathbf{1}}$
• Vibrational frequency of ${}^{\mathbf{23}}{\mathbf{Na}}^{\mathbf{35}}\mathbf{Cl}\mathbf{=}\mathbf{1.10}\mathbf{\times }{\mathbf{10}}^{\mathbf{13}}{\mathbf{s}}^{\mathbf{-}\mathbf{1}}$
• Vibrational frequency of ${}^{\mathbf{23}}{\mathbf{Na}}^{\mathbf{127}}\mathbf{I}\mathbf{=}\mathbf{0.773}\mathbf{\times }{\mathbf{10}}^{\mathbf{13}}{\mathbf{s}}^{\mathbf{-}\mathbf{1}}$

• Bond length of${}^{\mathbf{23}}{\mathbf{Na}}^{\mathbf{1}}\mathbf{H}\mathbf{=}\mathbf{1.89}\mathbf{\AA }$
• Bond length of${}^{\mathbf{23}}{\mathbf{Na}}^{\mathbf{35}}\mathbf{Cl}\mathbf{=}\mathbf{2.36}\mathbf{\AA }$
• Bond length of${}^{\mathbf{23}}{\mathbf{Na}}^{\mathbf{127}}\mathbf{I}\mathbf{=}\mathbf{2.71}\mathbf{\AA }$

Reduced mass of body is given by:

$\mathbf{\mu }\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{1}}{\mathbf{m}}_{\mathbf{2}}}{{\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}}$

Vibrational frequency, $\mathbf{\nu }\mathbf{=}\frac{\mathbf{c}}{\overline{\mathbf{\nu }}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi }}\sqrt{\frac{\mathbf{k}}{\mathbf{\mu }}}$

Where,$\mathit{k}$is force constant and$\mathit{\mu }$is reduced mass.

Calculation of reduced masses

Mass of ${}^1\mathrm{H}=1.008\mathrm{u}$.

Mass of data-custom-editor="chemistry" ${}^{23}Na=22.9897u$.

Substitute these values in given equation:

data-custom-editor="chemistry" $\mu =\frac{m_1{m}_2}{m_1+m_2}$

data-custom-editor="chemistry" $\begin{array}{c}=\frac{22.9897\times 1.008}{22.9897+1.008}\\ =0.9657u\\ =1.603\times {10}^{-27}kg\end{array}$

∴Reduced mass of data-custom-editor="chemistry" ${}^{23}N{a}^{1}His1.603\times {10}^{-27}kg$.

Mass of ${}^{35}\mathrm{Cl}=34.9688\mathrm{u}$.

Mass of ${}^{23}\mathrm{Na}=22.9897\mathrm{u}$.

Substitute these values in given equation:

$\begin{array}{c}\mathrm{\mu }=\frac{{\mathrm{m}}_1{\mathrm{m}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}\\ =\frac{34.9688\times 22.987}{34.9688+22.987}\\ =13.87\mathrm{u}\\ =2.303\times {10}^{-26}\mathrm{kg}\end{array}$

∴Reduced mass of ${}^{23}N{a}^{35}Clis2.303\times {10}^{-26}kg$.

Mass of ${}^{127}l=126.9045u$.

Mass of role="math" localid="1668060516390" ${}^{23}Na=22.9897u$.

Substitute these values in given equation:

$\begin{array}{c}\mu =\frac{m_1{m}_2}{m_1+m_2}\\ =\frac{126.904\times 22.987}{126.904+22.987}\\ =19.464u\\ =3.301\times {10}^{-26}kg\end{array}$

∴Reduced mass of ${}^{23}N{a}^{127}lis3.301\times {10}^{-26}kg$.

Calculation of force constant 

For ${}^{23}N{a}^{1}H,\mu is1.603\times {10}^{-27}kg$.

Substitute the value in given equation for force constant (k).

$\begin{array}{l}k=\mu {\left(2\pi \nu \right)}^2\\ =(1.603\times {10}^{-27}){\left(2\pi \times 3.51\times {10}^{13}\right)}^2\\ =77.89N{m}^{-1}\end{array}$

For ${}^{23}N{a}^{35}Cl,\mu is2.303\times {10}^{-26}kg$.

Substitute the value in given equation for force constant (k).

$\begin{array}{l}k=\mu {\left(2\pi \nu \right)}^2\\ =(2.303\times {10}^{-27}){\left(2\mathrm{\pi }\times 1.10\times {10}^{13}\right)}^2\\ =109.9N{m}^{-1}\end{array}$

For${}^{23}N{a}^{127}I,\mu is0.773\times {10}^{-27}kg$.

Substitute the value in given equation for force constant (k).

role="math" localid="1668060921081" $\begin{array}{l}k=\mu {\left(2\pi \nu \right)}^2\\ =(3.301\times {10}^{-27}){\left(2\pi \times 0.773\times {10}^{13}\right)}^2\\ =7.78N{m}^{-1}\end{array}$

Vibrational frequency of NaD

Mass of D is $2.014u.$

Substitute values in equation and calculate reduced mass of role="math" localid="1668060977229" ${}^{23}NaD$.

$\mu =\frac{m_1{m}_2}{m_1+m_2}$

data-custom-editor="chemistry" $\begin{array}{l}=\frac{22.9897\times 2.014}{22.9897+2.014}\\ =1.852u\\ =3.0774\times {10}^{-27}kg\end{array}$

Given that force constant for $NaHandNaD$ are same.

data-custom-editor="chemistry" $\begin{array}{c}{\left({\text{2πν}}_{\text{NaH}}\right)}^{\text{2}}{\text{μ}}_{\text{NaH}}\text{=}{\left({\text{2πν}}_{\text{NaD}}\right)}^{\text{2}}{\text{μ}}_{\text{NaD}}\\ {\text{ν}}_{\text{NaD}}{\text{=ν}}_{\text{NaH}}\sqrt{\frac{{\text{μ}}_{\text{NaH}}}{{\text{μ}}_{\text{NaD}}}}\end{array}$

Substitute the value in above equation.

data-custom-editor="chemistry" $\begin{array}{c}{\nu }_{NaD}=3.51\times {10}^{13}\sqrt{\frac{1.603\times {10}^{-27}}{3.074\times {10}^{-27}}}\\ =3.51\times {10}^{13}\times 0.722\\ =2.53\times {10}^{13}{s}^{-1}\end{array}$

The vibrational frequency of$NaDis2.53\times {10}^{13}{s}^{-1}$.