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In a certain Patterson map for an orthorhombic crystal, space group
P222, with coordinates of equivalent positions at $x, y, z ; x, y,-z,-x,-y, 1 / 2+z,-x, y, 1 / 2-z$, there
were high peaks at $u, v, w$ of $0,0,0.74 ; 0.6,0,0 ; 0,0.36,0.25$, assuming one heavy atom per
asymmetric unit, find the coordinates of this heavy atom.
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Answer from Sia

Posted 6 months ago

A

Key Concept

Patterson Map Analysis

Explanation

In a Patterson map, peaks correspond to vectors between atom pairs. The peak at $u,v,w = 0,0,0.74$ suggests a vector along the $c$-axis, which is half the unit cell length in an orthorhombic P222 space group, indicating the heavy atom is at $z = 0.37$. The peaks at $0.6,0,0$ and $0,0.36,0.25$ correspond to vectors from the origin to the heavy atom's $x$ and $y$ coordinates, respectively. Thus, the heavy atom's coordinates are approximately $(0.3, 0.18, 0.37)$.

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