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NEET CHEMISTRYCoordination CompoundsMedium

Question

The number of unpaired electrons in the complex ion [CoF6]3−[CoF_6]^{3-} is: (Atomic No. Co = 27)

A

2

B

3

C

4

D

Zero

Step-by-Step Solution

In the complex ion [CoF6]3−[CoF_6]^{3-}, the central cobalt atom is in the +3+3 oxidation state. The ground state electronic configuration of Co (Z = 27) is [Ar]3d74s2[Ar] 3d^7 4s^2, so Co3+Co^{3+} has a 3d63d^6 configuration. Fluoride (F−F^-) is a weak field ligand, so it does not cause forced pairing of the inner 3d3d electrons. The complex uses outer 4d4d orbitals and undergoes sp3d2sp^3d^2 hybridization, forming a high-spin (outer orbital) complex. According to Hund's rule, the six 3d3d electrons are arranged to give the maximum number of unpaired electrons (configuration t2g4eg2t_{2g}^4 e_g^2 in CFT). Therefore, there are 4 unpaired electrons.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Coordination Compounds. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYCoordination Compoundsnumberunpairedelectronscomplexatomic

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The complex, $[\text{Pt}(\text{Py})(\text{NH}_3)\text{BrCl}]$ will have how many geometrical isomers?

A.4
B.0
C.2
D.3
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Among the given complexes, the complex having the highest paramagnetic behaviour is: (At. No. Ti = 22, Cr = 24, Co = 27, Zn = 30)

A.$[Zn(NH_3)_6]^{2+}$
B.$[Ti(NH_3)_6]^{3+}$
C.$[Cr(NH_3)_6]^{3+}$
D.$[Co(NH_3)_6]^{3+}$
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Mg is present in:

A.Chlorophyll
B.Haemoglobin
C.Vitamin-B12
D.Vitamin-B
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The complexes $[\text{Co}(\text{NH}_3)_6][\text{Cr}(\text{CN})_6]$ and $[\text{Cr}(\text{NH}_3)_6][\text{Co}(\text{CN})_6]$ are examples of:

A.Ionisation isomerism
B.Coordination isomerism
C.Geometrical isomerism
D.Linkage isomerism
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The coordination compound that gives the maximum number of isomers is:

A.$[Co(NH_3)_4Cl_2]$
B.$[Ni(en)(NH_3)_4]^{2+}$
C.$[Ni(C_2O_4)(en)_2]$
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$[Cu(NH_3)_4]^{2+}$ reacts with $HNO_3$ in excess of water and gives:

A.$Cu(OH)_2$
B.$Cu(NO_3)_2$
C.$[Cu(H_2O)_4]^{2+}$
D.None of the above
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CFSE (in the octahedral field) will be maximum in: (Atomic number Co = 27)

A.[Co(H₂O)₆]³⁺
B.[Co(NH₃)₆]³⁺
C.[Co(CN)₆]³⁻
D.[Co(C₂O₄)₃]³⁻
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The IUPAC name of the following complex is: $[Ag(H_2O)_2][Ag(CN)_2]$

A.Diaquasilver(I) dicyanidoargentate(I)
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C.Diaquasilver(II) dicyanidoargentate(II)
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