If a molecule demonstrates paramagnetism, then: The substance can have both paired and unpaired electrons. The bond order is not a whole number. It can be determined by drawing a Lewis structure. It must be an ion. I. II. IV. O A. I, I O B. I, II, IV O C. I, II D.I only E. All of the above are correct.

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Chapter10: Molecular Geometry And Chemical Bonding Theory
Section: Chapter Questions
Problem 10.23QP
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If a molecule demonstrates paramagnetism, then:
I.
I.
The substance can have both paired and unpaired electrons.
The bond order is not a whole number.
It can be determined by drawing a Lewis structure.
It must be an ion.
II.
IV.
O A. I, II
O B.I, II, IV
OC.II, II
O D.I only
E. All of the above are correct.
Transcribed Image Text:If a molecule demonstrates paramagnetism, then: I. I. The substance can have both paired and unpaired electrons. The bond order is not a whole number. It can be determined by drawing a Lewis structure. It must be an ion. II. IV. O A. I, II O B.I, II, IV OC.II, II O D.I only E. All of the above are correct.
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