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Total number of geometrical isomers for the complex $\left[\mathrm{RhCl}(\mathrm{CO})\left(\mathrm{PPh}_3\right)\left(\mathrm{NH}_3\right)\right]$ is
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The correct answer is:
3
$[M(a b c d)]$ complex is square planar so will have 3 geometrical isomers.
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