Which of the following complex ions is/are diamagnetic in nature? I. [Mn(Br 4 )] 2– II. [V(NO 2 ) 4 ] 4– III. [Zn(NH 3 ) 4 ] 2+ IV. [Sc(H 2 O) 6 ] 3+

Which of the following complex ions is are diamagnetic in nature? Alright. The key to answering this question is to understand what the charge of the cation is first, and then to determine how many d orbital electrons it possesses. Alright. So for the first one, we have manganese connected to 4 bromide ions. The overall charge is 2 minus. This would only result if manganese was plus 2 in nature. So neutral manganese is going to be argon4s2 3d5. 2 plus means we've lost 2 electrons from our highest shell number, so manganese 2 plus is argon3d5. Now we don't have to talk about spin in terms of this or anything because we have an odd number of d orbital electrons. We have 5. When we have an odd number of electrons, there's gonna be at least one of them that's unpaired. So there's no point in figuring out if this is going to be high spin or low spin. So we're gonna say this one is paramagnetic. Next, we have vanadium connected to 4 and o twos. So these are nitrate ions. They have a charge of 4 minus. So that actually means that vanadium here is neutral. So this is neutral vanadium. So neutral vanadium has an electron configuration of argon 4s2, 3d3. Again, we have an odd number of electrons for our 3d orbitals, so that means that this has to be paramagnetic as well. Here we have zinc connected to 4 ammonia molecules, and the overall charge is 2 +. Ammonia is a neutral ligand, so the charge must be coming from the zinc itself. So zinc here is 2 +. Zinc neutral is argon 4s23d10. 2 plus means we've lost our electrons in the 4s orbital, so this becomes argon 3d10. Our d orbital electrons are completely filled because 10 is the maximum they can hold, so this would be diamagnetic in nature, so 3 is an option. Then finally here we have 4. Water is a neutral ligand, so the 3 plus is coming from scandium, so it's 3 plus. Neutral scandium is argon4s23d1. 3 plus means we've lost the electrons from the 4s orbital, and also the one electron in the 3 d orbital. So this just becomes argon. It's isoelectronic with a noble gas. Noble gases are perfect because all their electrons are perfectly matched up with one another, so this would also be diamagnetic. So here, we'd say that options 34 represent diamagnetic complex ions.

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24. Transition Metals and Coordination Compounds

Magnetic Properties of Complex Ions: Octahedral Complexes

General Chemistry

24. Transition Metals and Coordination Compounds

Magnetic Properties of Complex Ions: Octahedral Complexes

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Multiple Choice

Which of the following complex ions is/are diamagnetic in nature?
I. [Mn(Br₄)]^2– II. [V(NO₂)₄]^4– III. [Zn(NH₃)₄]²⁺ IV. [Sc(H₂O)₆]³⁺

I, II, and IV

II and III

I and II

I, III, and IV

iii and Iv

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Verified step by step guidance

Step 1: Understand the concept of diamagnetism. Diamagnetic substances have all their electrons paired and do not have any unpaired electrons. This results in no net magnetic moment.

Step 2: Determine the electron configuration of the central metal ion in each complex. For [Mn(Br4)]2–, Mn is in the +2 oxidation state, for [V(NO2)4]4–, V is in the +4 oxidation state, for [Zn(NH3)4]2+, Zn is in the +2 oxidation state, and for [Sc(H2O)6]3+, Sc is in the +3 oxidation state.

Step 3: Calculate the number of unpaired electrons for each metal ion. Mn2+ has 5 unpaired electrons, V4+ has 1 unpaired electron, Zn2+ has 0 unpaired electrons, and Sc3+ has 0 unpaired electrons.

Step 4: Identify which complexes are diamagnetic. Since diamagnetic complexes have no unpaired electrons, [Zn(NH3)4]2+ and [Sc(H2O)6]3+ are diamagnetic.

Step 5: Conclude that the complexes [Zn(NH3)4]2+ and [Sc(H2O)6]3+ are diamagnetic, corresponding to options III and IV.

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