Practicing Success

Target Exam

CUET

Subject

Chemistry

Chapter

Inorganic: Coordination Compounds

Question:

Match List I with List II.

List I List II
A. [CoF6]3- I. d2sp3, low spin
B. [NiCl4]2- II. sp3d2, high spin
C. [Co(NH3)6]3+ III. sp3, high spin
D. [Ni(CO)4] IV. sp3, no unpaired electron
Options:

A-III, B-IV, C-I, D-II

A-II, B-III, C-I, D-IV

A-III, B-II, C-IV, D-I

A-II, B-I, C-IV, D-III

Correct Answer:

A-II, B-III, C-I, D-IV

Explanation:

In the diamagnetic octahedral complex, [Co(NH3)6]3+, the cobalt ion is in +3 oxidation state and has the electronic configuration 3d6.

Six pairs of electrons, one from each NH3 molecule, occupy the six hybrid orbitals. Thus, the complex has octahedral geometry and is diamagnetic because of the absence of unpaired electron. In the formation of this complex, since the inner d orbital (3d) is used in hybridization, the complex, [Co(NH3)6]3+ is called an inner orbital or low spin or spin paired complex.

The paramagnetic octahedral complex, [CoF6]3– uses outer orbital (4d) in hybridization (sp3d2). It is thus called outer orbital or high spin or spin free complex.

In [NiCl4]2-, nickel is in +2 oxidation state and the ion has the electronic configuration 3d8.

Each Cl ion donates a pair of electrons. The compound is paramagnetic since it contains two unpaired electrons.

[Ni(CO)4] has tetrahedral geometry but is diamagnetic since nickel is in zero oxidation state and contains no unpaired electron.