NORMAL COORDINATE ANALYSIS OF XY 2 BENT MOLECULE – PART 1 Dr.D.UTHRA Head, Dept.of Physics DG Vaishnav College, Chennai-106

This presentation has been designed to serve as a self- study material for Postgraduate Physics students pursuing their programme under Indian Universities, especially University of Madras and its affiliated colleges. If this aids the teachers too who deal this subject, to make their lectures more interesting, the purpose is achieved. -D.Uthra

I acknowledge my sincere gratitude to my teacher Dr.S.Gunasekaran, for teaching me group theory with so much dedication and patience & for inspiring me and many of my friends to pursue research. My acknowledgement to all my students who inspired me to design this presentation. - D.Uthra - D.Uthra

Steps… ► Assign internal coordinates of the molecule ► Assign unit vectors and find their components along the three cartesian coordinates ► Obtain the orthonormalised SALCs ► Use the SALCs and obtain  U - matrix  S - matrix  B - matrix  G - matrix ► Apply Wilson’s FG matrix method

The matrices U - matrix has the form U jk S - matrix has the form S kt B - matrix has the form ∑ k U jk S kt j - order of the symmetry coordinate k- internal coordinate t- atom

In this presentation… Learn to form U-matrix and S-matrix for an XY 2 bent molecule for an XY 2 bent molecule

For XY 2 bent molecule Orthonormalised SALCs are S 1 =(1/√2)[d 1 + d 2 ] S 2 = α S 3 =(1/√2)[d 1 - d 2 ] Internal coordinates are 1- d 1 2- d 2 3- α Atoms are assigned as 1 - Y Y X Note : Order of assigning atoms and internal coordinates is according to the user and it can vary between person to person. But remember to follow that order and must not vary between person to person. But remember to follow that order and must not change it through out the analysis. change it through out the analysis.

To assign unit vectors… ► Unit vectors are assigned along every bond of the molecule ► They have unit magnitude ► Consider (by convention) they are positive if they point towards the atom and negative if they point away from the atom under consideration

For XY 2 bent molecule ► There are two unit vectors v 1 and v 2 along the two bonds d 1 and d 2 respectively ► They point towards the end atoms Y 1 and Y 2 respectively ► We assume that the molecule is lying in XY plane and Z axis is normal to plane containing the molecule ► Y- axis bisects the angle α between the bonds, so that α/2=θ Note : Recap your knowledge in trigonometry and then only proceed!!!

Z X This matrix has ► 3 col equal to 3 cartesian coordinates ► Rows equal to number of unit vectors Magnitude of v 1 and v 2 is one X component Ycomponent Z component V1V1V1V1 -v 1 sin θ = - v 1 s = -s -v 1 cos θ = = -v 1 c = -c 0 V2V2V2V2 v 2 sin θ = v 2 s = s -v 2 cos θ = - v 2 c = -c 0 Components of unit vectors Y v1v1 v2v2

U-matrix U-matrix is formed with the help of symmetry coordinates This matrix has ► Columns equal to number of internal coordinates ► Rows equal to number of symmetry coordinates ► Entry U jk of U matrix implies coefficient of k th internal coordinate of j th symmetry coordinate of the molecule

U matrix for XY 2 bent molecule Number of Rows = SALCs S 1 =(1/√2)[d 1 + d 2 ] S 2 = α S 3 =(1/√2)[d 1 - d 2 ] Number of columns =Internal coordinates 1- d 1 2- d 2 3- α d 1 d2d2 α S1S1 1/√2 0 S2S2001 S3S3 -1/√2 0

S-matrix S-matrix matrix has ► Columns equal to number of atoms ► Rows equal to number of internal coordinates Entry S kt of S matrix indicates the unit vector associated with the vibration involving ► t th atom of the molecule and ► k th internal coordinate of the molecule Use symmetry coordinates to form S-matrix

For XY 2 bent molecule No. of columns = No. of Atoms =3 1 - Y Y X No. of rows = No. of Internal coordinates =3 1- d 1 2- d 2 3- α Y 1 Y2Y2 X d1d1 d2d2 α

► Entry of S kt matrix indicates the vector that is involved the change in k th internal coordinate, corresponding to t th atom ► Rules to form S kt matrix are clearly described by Wilson, Decius and Cross

How to write S kt matrix entries for stretching vibrations?  It should be noted, during any stretching of any bond, two atoms are involved.  The atom towards which the unit vector representing the bond points at is called as end atom, while the other atom from which vector starts is called apex atom.  When the symmetry coordinate represent stretching, then entry in S kt matrix for the atoms involved in that vibration is equal to the unit vector representing the bond that is involved in that stretching.  By convention, unit vector for the end atom (involved in stretching) in S kt matrix takes +sign and unit vector for the other atom involved takes –ve sign.

► For an XY 2 bent atom,  when d 1 changes, atom Y 1 is involved - vector v 1 is +ve as v 1 points towards Y 1, ie Y 1 is end atom atom Y 2 is not involved – so no vector is involved w.r.t Y 2 atom Y 2 is not involved – so no vector is involved w.r.t Y 2 atom X is involved –vector v 1 is –ve as v 1 points away from X, ie X is apex atom  when d 2 changes atom Y 1 is not involved – so no vector is involved w.r.t Y 1 atom Y 1 is not involved – so no vector is involved w.r.t Y 1 atom Y 2 is involved - vector v 2 is +ve as v 2 points towards Y 2, ie Y 2 is end atom atom Y 2 is involved - vector v 2 is +ve as v 2 points towards Y 2, ie Y 2 is end atom atom X is involved –vector v 2 is –ve as v 2 points away from X, ie X is apex atom

Y1 Y1 Y2Y2 X d1d1 v 1 0-v 1 d2d2 0v2v2 -v 2 α

► How to write S kt matrix entries for bending vibrations?  In a bending, 3 atoms – two end atoms and one apex atom are involved  Also, the angle between two bonds d 1 and d 2 are involved (in this case, d 1 =d 2 =d )  Hence, two vectors are involved  For the end atom towards which v 1 points (here, Y 1 ), use the expression (v 1 cosα – v 2 )/ (d 1 d 2 ) ½ sinα (here, d 1 =d 2 =d and so, (d 1 d 2 ) ½ = d.  For the end atom towards which v 2 points (here, Y 2 ), use the expression (v 2 cosα – v 1 )/ (d 1 d 2 ) ½ sinα  For the apex atom (here, X), sum up the expression for end atoms and prefix it with –ve sign, ie., -[(v 1 cosα – v 2 )+ (v 2 cosα – v 1 )]/ (d 1 d 2 ) ½ sinα

Y1 Y1 Y2Y2 X d1d1 v 1 0-v 1 d2d2 0v2v2 -v 2 α (v 1 cosα – v 2 )/ d sinα(v 2 cosα – v 1 )/ d sinα -[(v 1 cosα – v 2 )+ (v 2 cosα – v 1 )] / d sinα

Now find the components of S kt matrix entries along the three cartesian coordinates ► Now your S matrix contains  No.of rows = no.of internal coordinates, in this case, 3  No.of columns = 3x no.of atoms= 3x3=9, in this case ► Use the table containing entries of components of unit vectors. X component Ycomponent Z component V1V1V1V1 -v 1 sin θ = - v 1 s = -s -v 1 cos θ = = -v 1 c = -c 0 V2V2V2V2 v 2 sin θ = v 2 s = s -v 2 cos θ = - v 2 c = -c 0

How to proceed ? ► From the table containing entries of components of unit vectors, note the components of vectors and in S kt matrix in respective positions ► X component of v 1 is –s (S x d 1 Y 1 = -s, S Y d 1 Y 1 = -c, S x d 2 Y 1 = 0, S x d 2 Y 2 = s, S Y d 2 Y 2 = -c ) and so on ► Now the entry corresponding to α for atom Y 1  S X αY 1 = (v 1 cosα – v 2 ) / d sinα = (-s cosα –s) /d sinα = -s(cosα + 1) /d sinα = (-s cosα –s) /d sinα = -s(cosα + 1) /d sinα = -s[2cos 2 (α/2) -1+1] /[2d sin(α/2) cos(α/2)] = -s[2cos 2 (α/2) -1+1] /[2d sin(α/2) cos(α/2)] = -c /d [as cos(α/2)=c and sin(α/2) =s ] S X αY 1 = (v 1 cosα – v 2 ) / d sinα = -c /d [as cos(α/2)=c and sin(α/2) =s ] S X αY 1 = (v 1 cosα – v 2 ) / d sinα  S Y αY 1 = (-s cosα –s) /d sinα = -s(cosα + 1) /d sinα = -s[2cos 2 (α/2) -1+1] /[2d sin(α/2) cos(α/2)] = -s[2cos 2 (α/2) -1+1] /[2d sin(α/2) cos(α/2)] = -c /d [as cos(α/2)=c and sin(α/2) =s ] = -c /d [as cos(α/2)=c and sin(α/2) =s ]

► Similarly, find x and y components of S kt matrix for atom Y 2 ► For atom X, sum up the entries of end atoms and prefix with –ve sign

S kt matrix Y1 Y1 Y2Y2 X XYZXYZXYZ d1d1 -s-c0000+s+c0 d2d2 000+s-c0-s+c0 α -c/ds/d0c/ds/d002s/d0

In this presentation, you have learnt to form U matrix and S matrix for a bent XY 2 molecule. C U in the next presentation to learn to form B-matrix -uthra mam