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NEET CHEMISTRYCoordination CompoundsMedium

Question

In [Cr(NH3)6](Br)3[Cr(NH_3)_6](Br)_3, number of unpaired electrons in Cr is:

A

4

B

3

C

1

D

2

Step-by-Step Solution

In the given complex [Cr(NH3)6](Br)3[Cr(NH_3)_6](Br)_3, the central chromium atom is present as Cr3+Cr^{3+} ion. The ground state electronic configuration of Cr (Z = 24) is [Ar]3d54s1[Ar] 3d^5 4s^1. Therefore, the electronic configuration of Cr3+Cr^{3+} is [Ar]3d3[Ar] 3d^3. In an octahedral crystal field, these three electrons will occupy the lower energy t2gt_{2g} orbitals singly (t2g3eg0t_{2g}^3 e_g^0) according to Hund's rule, irrespective of the field strength of the ligand. Thus, there are 3 unpaired electrons in the chromium ion.

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 Compoundscrnhbrnumberunpairedelectrons

More Coordination Compounds Questions

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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
MediumSolve

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.$[Co(NH_3)_6]^{3+}$
MediumSolve

Mg is present in:

A.Chlorophyll
B.Haemoglobin
C.Vitamin-B12
D.Vitamin-B
EasySolve

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
EasySolve

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.$[Ni(C_2O_4)(en)_2]$
D.$[Cr(SCN)_2(NH_3)_4]^+$
HardSolve

$[Cu(NH_3)_4]^{2+}$ 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
MediumSolve

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⁻
MediumSolve

The IUPAC name of the following complex is: $[Ag(H_2O)_2][Ag(CN)_2]$

A.Diaquasilver(I) dicyanidoargentate(I)
B.Dicyanidosilver(II) diaquaargentate(II)
C.Diaquasilver(II) dicyanidoargentate(II)
D.Dicyanidosilver(I) diaquaargentate(I)
EasySolve

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