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A magnetic moment of 1.73 BM will be shown by one among the following:

1. [Ni(CN)4]^2–
2. [TiCl4]
3. [CoCl4]^2–
4. [Cu(NH3)4]^2+

Correct answer: [Cu(NH3)4]^2+

Solution

The magnetic moment of 1.73 BM corresponds to one unpaired electron. Among the given options, [Cu(NH3)4]^2+ has one unpaired electron in the d-orbital of Cu(II), leading to this magnetic moment.

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