Nanotechnology Questions and Answers – Fullerene as Ligand

This set of Nanotechnology Multiple Choice Questions & Answers (MCQs) focuses on “Fullerene as Ligand”.

1. Which of the following effect is not observed when C60 forms metal complex?
a) Removal of a double bond from 29 fullerene bonds
b) A polyene system shows decrease in electron affinity of the system
c) Reduction in electron affinity upon transfer of e- density from metal to π* orbitals of double bonds
d) Increased conjugation shown by a polyene system
View Answer

Answer: d
Explanation: C60 upon coordinating with transition metals form metal complexes. There are various effects observed during this process. There is removal of a double bond from the remaining 29 fullerene double bonds. For a polyene system, this decrease in conjugation is believed to raise the energy of the LUMO and eventually reduce the electron affinity of the system. Further, the d-orbital back bonding is responsible for the reduction in electron affinity. This is due to the transfer of electron density from the metal into π* orbitals of the remaining double bonds.

2. Why is binding of C60 to a transition metal in a hexahepato fashion unfavorable?
a) Orbital overlap is weakened
b) Geometry of p-orbitals remain perpendicular to the centre of the ring
c) Orbital overlap is strengthened
d) Geometry of p-orbitals remain parallel to the centre of the ring
View Answer

Answer: a
Explanation: The binding of C60 to a metal in a hexahepato fashion is discouraging since it weakens the orbital overlap. This is evident from the geometry of p-orbitals within a hexagon of the C60 framework which remain tilted away from the centre of the ring.

3. Complete the given reaction.

R-Hg-Pt (PPh3)2-Br+C60

a) C60Pt(PPh2)+R-Hg-Br
b) C60P(PtPh2)3+R-Br-Hg
c) C60Pt(PPh3)2+R-Hg-Br
d) C60Pt(PPh2)3+R-Br-Hg
View Answer

Answer: c
Explanation: The above given reaction is

R-Hg-Pt (PPh3)2-Br+C60 ➔ C60Pt(PPh3)2+R-Hg-Br

This reaction is a pathway for preparing platinum-fullerene complexes. The procedure ensures the absence of free phosphine ligand in the reaction mixture.

4. Choose the incorrect statement regarding (Et3P)2MC60 (M=Ni, Pt, Pd).
a) Harder to reduce than organic monoadducts of type C60RR’
b) Lower selectivity for formation of monoadducts
c) Higher inductive donation of electron density into C60 moiety
d) Tendency of the addition of a second metal fragment to a monoadduct is highly reduced
View Answer

Answer: b
Explanation: In the organometallic complexes (Et3P2)MC60, metal complexes have an additional shift compared to the organic derivatives. This is due to higher inductive donation of electron density into the C60 moiety, resulting in the decrease of electron affinity. This further explains the high selectivity for the formation of monoadducts with low-valent transition metals.

5. How can the ratio of Pd:C60 be changed in organo-palladium polymers of C60?
a) By centrifuging the suspension
b) By heating the suspension
c) By cooling the suspension
d) By desiccating the suspension
View Answer

Answer: b
Explanation: The ratio of Pd:C60 can be changed from 1:1 to 3:1 upon heating the suspension of the initially formed polymer. This is done in toluene under reflux with regeneration of free C60. Depending on this ratio these polymers are believed to form linear “ pearl necklace “ or two or three dimensional arrangements.

6. Olefins coordinated to transition metals lose their planarity.
a) True
b) False
View Answer

Answer: a
Explanation: Olefins bonded to transition metals lose their planarity because the four groups bound to the olefin bend back away from the metal. This deformation is more prominent and effective in substituted ethenes when the electronegativity of the groups attached to it is increased.

7. Why do fullerenes behave as ligands?
a) Globular structure
b) Behave as alkanes
c) [6,6] double bond
d) Electron rich
View Answer

Answer: c
Explanation: Fullerenes behave as olefins with strained double bonds in it. The [6,6] double bond is electron deficient and more reactive than [6,5]. They are great electron acceptors and hence form metallic bonds with η=2 hapticity. Fullerenes might be ligated to metals in different hapticity, from η1 to η6.

8. Which of the following is a Vaska complex?
a) Ir(CO)Cl(PPh3)2
b) (Ph3P)2Pt(η2-C60)
c) [(Et3P)2Pt]6C60
d) Pd2(dba)3.CHCl3
View Answer

Answer: a
Explanation: Vaska complex is the trivial name for the compound Ir(CO)Cl(PPh3)2. Its IUPAC name is transcarbonylchlorobisiridium (I). This complex reacts with electron accepting olefins to form stable adducts. For example: – (η2-C60)Ir(CO)Cl(PPh3)2 is an adduct with C60-fullerene, that behaves as an electron deficient olefin.

9. What is the ratio of the (Ph3P)2Pt(η2-C2H4) and C60 used for the formation of the dark emerald green solution?
a) 2:1
b) 1:3
c) 1:1
d) 3:2
View Answer

Answer: c
Explanation: The η2-binding of C60 with low-valent transition metals is owing to the strain and electron deficiency of the 6-6 double bonds in C60. For example:- X-ray structure analysis revealed that equimolar amounts of (Ph3P2)Pt(η2-C2H4) and C60 when reacted, resulted in a dark emerald green solution of (Ph3P2)Pt(η2-C60) that has η2 binding of platinum to a [6-6] double bond in fullerene.

10. How can organo-palladium polymers be formed?
a) From Pd2(dba)2.CHCl
b) From Pd2.CHCl3
c) From Pd(dba)3.CHCl
d) From Pd2(dba)3.CHCl3
View Answer

Answer: d
Explanation: Organometallic polymers of C60 can be produced by the reaction of palladium complex Pd2(dba)3.CHCl3 (dba=dibenzylideneacetone) with C60 solutions. The dba ligands are replaced and dark brown precipitate of C60Pdn is obtained.

Sanfoundry Global Education & Learning Series – Nanotechnology.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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