The moment of inertia of HCl molecule about an axis passing through its centre of mass and perpendicular to the line joining the H⁺ and Cl⁻ ions will be, if the interatomic distance is 1 Å

The correct option is 1.61 x 10⁻⁴⁷ kg m².

To find the moment of inertia ($I$) of a diatomic molecule like HCl (Hydrogen Chloride) about an axis passing through its center of mass and perpendicular to the interatomic axis, we can model it as a two-particle system.

For a system of two masses $m_1$ and $m_2$ separated by a distance $r$, the moment of inertia about the center of mass is given by:
$I=\mu r^2$
where $\mu$ is the reduced mass of the system.

The reduced mass $\mu$ is calculated as:
$\mu =\frac{m_1{m}_2}{m_1+m_2}$

Let the mass of the hydrogen atom ($H^{+}$ ion) be $m_1\approx 1\text{amu}$ and the mass of the chlorine atom (${\mathrm{Cl}}^{-}$ ion) be $m_2\approx 35.5\text{amu}$.
Converting the mass unit from atomic mass units (amu) to kilograms, where $1\text{amu}\approx 1.66\times {10}^{-27}\text{kg}$:
$m_1=1\times 1.66\times {10}^{-27}\text{kg}$
$m_2=35.5\times 1.66\times {10}^{-27}\text{kg}$

Now, calculate the reduced mass in terms of amu first:
$\mu =\frac{1\times 35.5}{1+35.5}=\frac{35.5}{36.5}\approx 0.9726\text{amu}$

Converting this reduced mass to kg:
$\mu \approx 0.9726\times 1.66\times {10}^{-27}\text{kg}\approx 1.614\times {10}^{-27}\text{kg}$

The interatomic distance $r$ is given as 1 Å, which is:
$r=1\text{Å}={10}^{-10}\text{m}$

Now, substitute the values of $\mu$ and $r$ into the moment of inertia formula:
$I=(1.614\times {10}^{-27}\text{kg})\times {\left({10}^{-10}\text{m}\right)}^2$
$I=1.614\times {10}^{-27}\times {10}^{-20}\text{kg m}{²}^{}$
$I\approx 1.61\times {10}^{-47}\text{kg m}{²}^{}$