4. The planar molecule ClF3 is T-shaped with the F atom at the mid- point of the

Question

4. The planar molecule ClF3 is T-shaped with the F atom at the mid-

point of the cross-piece of the “T”.

(a) If the F-Cl distance in the stem of the T is 160 pm and the F-Cl

distances in the cross-bar of the T are 170 pm, find the position

of the centre of mass. (You may assume a bond angle of 90,

which is almost correct.)

(b) Calculate the moment of inertia (in kg m2) about an axis normal

to the molecular plane which passes through the centre of mass.

[Relative isotopic masses: mF=19.00u; mCl= 34.97u]

c

Explanation / Answer

The chlorine atom is the one at the midpoint of the crosspiece of the
letter T, or inverted letter T, and not fluorine … so now, consider the
inverted T structure and take a 2D rectangular coordinate system
whose origin is at the Cl-atom … the coordinates of the F-atoms are
… left fluorine atom = ( x , y ) = ( - 170 pm , 0 ) …
… right fluorine atom = ( x , y ) = ( +170 pm , 0 ) …
… top fluorine atom = ( x , y ) = ( 0 ,+160 pm ) …
BTW, by taking the origin of the coordinate system at the Cl-atom, we
need not consider its mass … only those of the F-atoms …
By definition of the center of mass …
… Xcm = [ m x + m x + m x + m x ] / [ m + m + m + m ] …
… Ycm = [ m y + m y + m y + m y ] / [ m + m + m + m ] …
since we have a total of four particles … m = m = m = 19 u …
and … m = 35 u … ( x , y ) = ( 0 , 0 ) for the Cl-atom …

(b) Use the formula … I = m r ² … for the moment of inertia of a point
particle with mass m and distance r from the axis of rotation … but since
ClF consists of four particles (atoms), we have …
… I = I + I + I + I = m r ² + m r ² + m r ² + m r ² … where
… r ² = 170 ² + ( Ycm ) ² = r ² … r = 160 Ycm … r = Ycm …
Just substitute the values and compute

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