Practicing Success

Target Exam

CUET

Subject

Chemistry

Chapter

Inorganic: Coordination Compounds

Question:

Match Column I with Column II

Column I

(Metal ion configuration in the weak ligand field)

Column II

(CFSE, \(\Delta _0\) value)

(A) d4

(p) 0.0

(B) d5

(q) –0.4

(C) d6

(r) – 0.6

(D) d7

(s) – 0.8

 

 

Options:

A-r, B-s, C-p, D-q

A-r, B-p, C-q, D-s

A-p, B-r, C-q, D-s

A-q, B-r, C-p, D-p

Correct Answer:

A-r, B-p, C-q, D-s

Explanation:

The correct answer is option (2) A-r, B-p, C-q, D-s

Crystal Field Stabilization Energy refers to the energy difference between the higher-energy set of d-orbitals \((e_g)\) and the lower-energy set of d-orbitals \((t_{2g})\) in an octahedral coordination complex. CFSE is a key concept in Crystal Field Theory and is used to explain the stability and color of transition metal complexes.

In an octahedral complex, the d-orbitals split as follows:

1. Three d-orbitals \((d_{x^2 - y^2}, d_{z^2})\) are raised in energy, forming the "\((e_g)\)" set.

2. The remaining two d-orbitals \((d_{xy}, d_{xz}, d_{yz})\)are lowered in energy, forming the "\((t_{2g})\)" set.



The Crystal Field Stabilization Energy (CFSE) for an octahedral complex can be calculated using the following formula:

\(\text{CFSE = number of electrons in } t_{2g} × \text{ -0.4}\Delta _0 \text{ + number of electrons in  }e_g × \text{ 0.6 }\Delta _0)\)

The crystal field splitting parameter, \(\Delta _0\), is determined experimentally and can vary depending on the metal ion and the nature of the ligands surrounding it. The magnitude of \(\Delta _0\) determines whether the complex has a low-spin or high-spin electronic configuration. If \(\Delta _0\) is large enough to overcome the pairing energy, the complex will adopt a low-spin configuration, with electrons filling the \(t_{2g}\) orbitals before pairing up in the \(e_g\) orbitals. If \(\Delta _0\) is small, the complex will adopt a high-spin configuration, with electrons populating both the \(t_{2g}\) and \(e_g\) orbitals before pairing up.

The CFSE has important implications for the stability and reactivity of transition metal complexes. It is a major factor influencing the thermodynamics of complex formation, and it also affects the absorption and emission spectra, magnetic properties, and catalytic activities of the complexes. Additionally, CFSE plays a significant role in explaining the color of transition metal complexes, as the energy difference between the \(e_g\) and \(t_{2g}\) sets corresponds to specific wavelengths of light that are absorbed or transmitted, giving rise to characteristic colors.

In the case of a weak field, high spin complexes are formed where electrons are distributed in the \(t_{2g}\) level first then in the \(e_g\) level. After the 5th electron enters the \(e_g\) level and then the next electron enters the \(t_{2g}\) level resulting in the pairing of electrons The CFSE for low-spin complexes is given below:

Total d-electrons

Configuration

CFSE

d0

\(t_{2g}^0e_g^0\)

\(\text{CFSE = } 0 × -0.4 \Delta _0  +  0 × 0.6 \Delta _0) = 0\Delta _0 \)

d1

\(t_{2g}^1e_g^0\)

\(\text{CFSE = } 1 × -0.4 \Delta _0  +  0 × 0.6 \Delta _0) = -0.4 \Delta _0 \)

d2

\(t_{2g}^2e_g^0\)

\(\text{CFSE = } 2 × -0.4 \Delta _0  +  0 × 0.6 \Delta _0) = -0.8 \Delta _0  \)

d3

\(t_{2g}^3e_g^0\)

\(\text{CFSE = } 3 × -0.4 \Delta _0  +  0 × 0.6 \Delta _0) = -01.2 \Delta _0  \)

d4

\(t_{2g}^3e_g^1\)

\(\text{CFSE = } 3 × -0.4 \Delta _0  +  1 × 0.6 ×\Delta _0) = -0.6 \Delta _0  \)

d5

\(t_{2g}^5e_g^0\)

\(\text{CFSE = } 3 × -0.4 \Delta _0  +  2 × 0.6 \Delta _0) = 0.0 \Delta _0  \)

d6

\(t_{2g}^6e_g^0\)

\(\text{CFSE = } 4 × -0.4 \Delta _0  +  2 × 0.6 \Delta _0) = -2.4 \Delta _0  \)

d7

\(t_{2g}^6e_g^1\)

\(\text{CFSE = } 5 × -0.4 \Delta _0  +  2 × 0.6 \Delta _0) = -0.8 \Delta _0  \)

d8

\(t_{2g}^6e_g^2\)

\(\text{CFSE = } 6 × -0.4 \Delta _0  +  2 × 0.6 \Delta _0) = -1.2 \Delta _0  \)

d9

\(t_{2g}^6e_g^3\)

\(\text{CFSE = } 6 × -0.4 \Delta _0  +  3 × 0.6 \Delta _0) = -0.6 \Delta _0  \)

d10

\(t_{2g}^6e_g^4\)

\(\text{CFSE = } 6 × -0.4 \Delta _0  +  4 × 0.6 \Delta _0) = 0 \Delta _0  \)