Sample 1Medium
A crystal field splitting diagram for an octahedral complex shows the five degenerate d orbitals of the free metal ion splitting into a lower set t2g, at −0.4Δo, and an upper set eg, at +0.6Δo. A metal ion with four d electrons sits in a ligand field where Δo is greater than the pairing energy P. What is its d-electron configuration?
- A
t2g² eg², high spin
- B
t2g⁰ eg⁴, low spin
- C
t2g⁴ eg⁰, low spin
- D
t2g³ eg¹, high spin
Show the answer
The answer is C.When Δo > P, it costs less energy to pair the fourth electron in t2g than to promote it to eg. The configuration is therefore t2g⁴ eg⁰, a low spin complex.
- A.
Placing two electrons in eg while t2g still has room needs more energy and does not fit either field.
- B.
Electrons fill the lower t2g set first; the eg set is higher in energy.
- D.
t2g³ eg¹ arises when Δo is smaller than P, so the fourth electron goes up to eg.
NCERT: Class 12 Chemistry, Chapter 5
Sample 2NEET level
Consider the following statements: I. [Co(NH3)6]³⁺ is an inner orbital complex with d²sp³ hybridisation. II. [CoF6]³⁻ is diamagnetic. III. [NiCl4]²⁻ is tetrahedral and paramagnetic. IV. [Ni(CN)4]²⁻ is square planar and diamagnetic. Which of the statements are correct?
- A
I and II only
- B
I, III and IV only
- C
II and IV only
- D
I, II, III and IV
Show the answer
The answer is B.In [Co(NH3)6]³⁺ the six 3d electrons pair up, giving an inner orbital d²sp³ diamagnetic complex, whereas [CoF6]³⁻ is an outer orbital sp³d² complex with four unpaired electrons, so it is paramagnetic. [NiCl4]²⁻ is sp³, tetrahedral with two unpaired electrons, and [Ni(CN)4]²⁻ is dsp², square planar and diamagnetic.
- A.
Statement II is false: [CoF6]³⁻ has four unpaired electrons, and III is also correct.
- C.
Statement II is false, and statements I and III are also correct.
- D.
Statement II is false: [CoF6]³⁻ is paramagnetic.
NCERT: Class 12 Chemistry, Unit 5 (Coordination Compounds), Bonding in Coordination Compounds: Valence Bond Theory and Magnetic Properties of Coordination Compounds