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Self-assembled Transition Metal Coordination Frameworks of ...

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Chapter 3<br />

= -2J1{§, -3*, +.§, -$1,} -2.12{.§*, -s, +3‘, -s4} -2J3{§, -s, +§, -$4} (1)<br />

For compounds 1 to 4 J3 is assumed to be zero, because there is no cross­<br />

coupling connection, and it is reasonable to assume that J = J1 = J2 by the<br />

approximation <strong>of</strong> similar M—X-M angles within each square grid. Assuming all the<br />

four Ni(H)—X—Ni(Il) couplings are equal, i.e. using a single coupling constant J that<br />

takes into account all the exchange pathways to be equal (Scheme 3.2), is a reasonable<br />

assumption in the light <strong>of</strong> the X-ray crystallographic study <strong>of</strong> la which reveals a<br />

similar nature <strong>of</strong> M—S—M angles (~l67°), there is probably a super exchange<br />

mechanism through the intervening sulfur atoms. Therefore, for a square arrangement<br />

<strong>of</strong> four metal centers, D4,, [2 >< 2] grid, eqn. (l) reduces to eqn. (2) [26],<br />

131 =-2J(.§, -s, +$, -s, +§, -$4 +5‘, -$4) (2)<br />

s1<br />

J1<br />

J2 s2 s1 J s2<br />

J3 \/_> J1 J J<br />

I<br />

S4 I J2 $3 S4 J S3<br />

Scheme 3.2. Magnetic exchange models, showing the general case (left) and<br />

approximation used for 1 to 4 (right). _<br />

Here, the values <strong>of</strong> the spin angular momentum operator 3‘, (i = 1- 4) are<br />

the same, and using the conventional spin-vector coupling model [52,53] the<br />

corresponding eigenvalues can be obtained analytically and are given by eqn. (3) [51],<br />

E(S',S,3,S24) = —J[S'(S'+l)—S,3(S,3 +l)—S24 (S24 +l)] (3)<br />

where 5'“ = 3', +.§3; $'24 = 6'2 +.§4; 5" = .§'l +5‘, +.§3 +.§'4. For Ni(lI) molecular<br />

squares S l , S 2 , S3 , S 4 = 1. There are nineteen spin states as given below.<br />

............................................................................................................................. .:......................_...,...........................,...........................:.....................<br />

S‘ 5 i s l4%3%2%1%0l i 2 E 3 E ; 5s<br />

114<br />

No.<strong>of</strong>spinstates 1 3 6 6 3

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