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(C) Let us choose the nucleus of the hydrogen atom as the origin for measuring distance.Mass of hydrogen atom,$m_1 = 1$ unit.Since the chlorine atom i

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$1.24$

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(C) Let us choose the nucleus of the hydrogen atom as the origin for measuring distance.Mass of hydrogen atom,$m_1 = 1$ unit.Since the chlorine atom is $35.5$ times as massive as the hydrogen atom,the mass of the chlorine atom,$m_2 = 35.5$ units.The distance between the nuclei is $r = 1.27 \ \mathring{A}$.Let the position of the hydrogen atom be $x_1 = 0$ and the position of the chlorine atom be $x_2 = 1.27 \ \mathring{A}$.The distance of the centre of mass $(X)$ from the hydrogen atom is given by the formula:$X = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$Substituting the values:$X = \frac{1 	imes 0 + 35.5 	imes 1.27}{1 + 35.5} \ \mathring{A}$$X = \frac{35.5 	imes 1.27}{36.5} \ \mathring{A}$$X \approx 0.9726 	imes 1.27 \ \mathring{A} \approx 1.235 \ \mathring{A}$Rounding to two decimal places,we get $X \approx 1.24 \ \mathring{A}$.

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