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H 2 CuF An overview of H 2 MX complexes G.S. Grubbs II, D.J. Frohman, Z. Yu, S.E. Novick TH13, Columbus 2013 Inorg. Chem. 52, 816-822 (2013).

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Presentation on theme: "H 2 CuF An overview of H 2 MX complexes G.S. Grubbs II, D.J. Frohman, Z. Yu, S.E. Novick TH13, Columbus 2013 Inorg. Chem. 52, 816-822 (2013)."— Presentation transcript:

1 H 2 CuF An overview of H 2 MX complexes G.S. Grubbs II, D.J. Frohman, Z. Yu, S.E. Novick TH13, Columbus 2013 Inorg. Chem. 52, 816-822 (2013).

2 The motivation for these studies is the H 2 storage capacity of Metal Organic Frameworks (MOFs). Shown if MOF-5 from a review article of Yaghi et. al. The yellow sphere is unfilled and shows the porosity Of the MOF. The phenyl rings are shown connecting the Zn 4 O(CO 2 ) 6 units. These units are seen to consist of four ZnO 4 tetrahedra (in blue) with a common vertex. The zinc is hidden in the center of the tetrahedra, oxygen atoms are red, carbons are black. MOFs can hold up to 18 weight-percent hydrogen molecules. We hope to mimic the “active sites” in MOFs in order to help design MOFs for better H 2 storage. MOF-5 H 2 CuF was produced by laser ablation of a copper rod with a gas pulse of 0.6% SF 6 and 3% H 2 in argon with a backing pressure of 5 atm. Standard FTMW spectroscopy was employed.

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6 para-H 2 63 CuF 1500 gas pulses

7 ortho-D 2 63 CuF 2500 gas pulses

8 natural abundance of 63 Cu = 69%; I( 63 Cu)= 3/2 natural abundance of 65 Cu = 31%; I( 65 Cu)= 3/2 Spin statistics The total wavefunction for the interchange of the two identical fermions, 1 H, is of negative parity The total wavefunction for the interchange of the two identical bosons, 2 H, is of positive parity The measured a-type transitions all have K a = 0, and rotational wavefunction is of positive parity a The ortho states are symmetric with respect to interchange of the two identical nuclei  rot  (  1) K a

9 Spectroscopic parameters for para-H 2 CuF, ortho-D 2 CuF, and HD CuF

10 χ aa MHz D e kJ/mol α A 3 D e / α 63 CuF a 21.9562-------------- Ar CuF a 38.0556471.6329 Kr CuF b 41.77452.46518 Xe CuF c 47.76634.0116 p-H 2 CuF61.2751050.659159 N 2 CuF d 61.88251392.1963 OC CuF e 75.4061502.3165 a.C.J. Evans, M.C.L. Gerry, J. Chem. Phys. 112, 9363 (2000). b.J.M. Michaud, S.A. Cooke, M.C.L. Gerry, Inorg. Chem. 43, 3871 (2004). c.J.M. Michaud, M.C.L. Gerry, J. Am. Chem. Soc. 128, 7613 (2006). d.S.G. Francis, S.L. Matthews, O.K. Poleshchuk, N.R. Walker, A.C. Legon, Angew. Chem. Int. Ed. 45, 6341 (2006). e. N.R. Walker, M.C.L. Gerry, Inorg. Chem. 40, 6158 (2001).

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12 FCu FCu - H 2 H 2

13 r(H 2 Cu) = 1.52 A,  r(CuF) = -0.008 A,  r(HH) = +0.087 A, electric field gradient at copper increases by a factor of 3 APFD/ECP10MDF & aug-cc-pVQZ

14 Zinc plasma Tin plasma Copper plasma r(H 2 Cu) = 1.52 A,  r(CuF) = -0.008 A,  r(HH) = +0.087 A, electric field gradient at copper increases by a factor of 3 this talk FC02 (Grubbs) FC03 (Obenchain) TH16 (Pickett) In other presentations in this Symposium we will talk about other complexes, ortho and para spin statistics, and our suggestions on spin-spin coupling APFD/ECP10MDF & aug-cc-pVQZ

15 Acknowledgements Pete Pringle - sage advice & withering criticism Steve Cooke - co-running the research group Dan Obenchain – being a wise-guy (in both senses) Brittany Long – adding new science Herb Pickett – spectroscopic expertise (SPFITish) Jens Grabow – keeping the spectrometers humming Bob Bohn – not breaking the spectrometers Chinh Duong, Brooke Baker, Lucas Hansen, Derek Frank, Samantha Savitch, Tyler Holzschuh – being undergraduates

16 nuclear spin – nuclear spin interaction tensor D(free H 2 ) kHzD(H 2 in H 2 AgCl) kHz D aa = -576.8D bb = -439.02 D bb = +288.4D aa = +165.6 D cc = +288.4D cc = +273.42 a b c a b

17 para-H 2 CuCl 63 Cu 35 Cl 65 Cu 35 Cl 63 Cu 37 Cl 65 Cu 37 Cl (B+C)/2 /MHz4781.8759(3) a 4740.4245(2)4621.9139(3)4580.1110(4) D /kHz2.58(4)2.52(3)2.44(3)2.39(6) eQq Cl /MHz-24.430(4)-24.428(2)-19.249(1)-19.215(6) eQq Cu /MHz54.057(3)50.017(2)54.056(2)50.035(5) C I Cu /kHz12.1(2)13.1(2)12.1(2)13.5(3) rms b /kHz1.61.11.01.1 NcNc 30151410

18 ortho-H 2 CuCl 63 Cu 35 Cl (B+C)/2 /MHz4781.572(2) a (B-C)/4 /MHz4.01(2) D /kHz2.58 b eQq Cl /MHz-24.7(1) eQq Cu /MHz54.7(1) D aa /kHz190(10) D bb -D cc /MHz-791.996 C I,Cu /kHz12.1 b rms c /kHz19.2 NdNd 11

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20 Treating the two coupled hydrogen nuclei as one As is well-known, H 2 can be in the para, I tot = 0, or ortho, I tot = 1, states The spin-spin splitting in the para state is zero For the ortho state the two nuclei can either be treated as two interacting spin-1/2 nuclei or as one combined nucleus with an effective spin of 1 the results are identical Here are two var files for Pickett’s SPCAT ortho test 4, single I = 1 nucleus 10 1000 5 0 0.1760E-11 0.10E+07 0.1E+01 1.0 'a' 3 1 0 10 0 1 1 1 0 1 1 10000 10000. 0.000 / A 20000 2000. 0.000 / B 30000 1000. 0.000 / C 110010000 45. 0.000 / 3/2 Daa 110040000 15. 0.000 / 1/4 (Dbb - Dcc) ortho test 5, two spins of I = ½ 10 1000 5 0 0.1760E-11 0.10E+07 0.1E+01 1.0 'a' 22 1 0 10 0 1 1 1 0 1 1 10000 10000. 0.000 / A 20000 2000. 0.000 / B 30000 1000. 0.000 / C 120010000 45. 0.000 / 3/2Daa 120040000 15. 0.000 / 1/4(Dbb - Dcc) For both tests, D aa = 30 MHz, D bb = 15 MHz, and D cc = -45 MHz

21 one spin 1two spin ½ The extra transitions in the file on the right are the para state transitions and are D independent the two cat files

22 Internuclear distances from nuclear spin – nuclear spin interaction tensor D aa (H 2 ) = -576.8(4) kHz, Ramsey and coworkers, Phys. Rev. 87, 395 (1952) note: D = -10 d, as given in the Ramsey article g 1 = g 2 = 5.5857 gives r(H-H) = 0.7468 A r e (H 2 ) = 0.7414 A, r 0 (H 2 ) = 0.7510 A, r s (H 2 ) = 0.7462 A Huber and Herzberg

23 The nuclear spin – nuclear spin tensor a notation recommendation H ss = I i  D ij  I j where D is traceless H ss = 3S aa I i a I j a + 3S bb I i b I j b + 3S cc I i c I j c where S is not necessarily traceless when only one component of S is fit, D aa = 2S aa when two components of S are fit D aa = 2S aa - S bb D bb = 2S bb - S aa D cc = -S aa - S bb when all three components of S are given D aa = 3S aa - (S aa + S bb + S cc ) D bb = 3S bb - (S aa + S bb + S cc ) D cc = 3S cc - (S aa + S bb + S cc ) We recommend, to avoid confusion, that D be fit and presented


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