
For an octahedral complex, which of the following d-electron configurations will give maximum CFSE?
(A) High spin ${{d}^{6}},-0.4{{\Delta }_{o}}$ ${{d}^{6}},-0.4{{\Delta }_{o}}$
(B) Low spin ${{d}^{4}},-1.6{{\Delta }_{o}}$ ${{d}^{4}},-1.6{{\Delta }_{o}}$
(C) Low spin ${{d}^{5}},-2.0{{\Delta }_{o}}$
(D) High spin ${{d}^{7}},-0.8{{\Delta }_{o}}$
Answer
542.4k+ views
Hint: The value of CFSE is more in the given options for the option (C). However it can be understood from the calculation.
Complete step by step answer:
Crystal field splitting energy, CFSE for octahedral complexes is calculated from the following formula,
$\left[ \left( -0.4\times 5 \right)+\left( +0.6\times 2 \right) \right]{{\Delta }_{o}}$
Let us calculate the CFSE for all the given electronic configurations. Let us consider,
- Option (A), in the high spin ${{d}^{6}}$ configuration, there will be 4 ${{t}_{2g}}$ electrons and 2 ${{e}_{g}}$ electrons. The CFSE is $\left[ \left( -0.4\times 4 \right)+\left( +0.6\times 2 \right) \right]{{\Delta }_{o}}$
=$\left[ -1.6+1.2 \right]{{\Delta }_{o}}$
=$-0.4{{\Delta }_{o}}$
- Option (B), in low spin ${{d}^{4}}$ configuration, there will be 4 ${{t}_{2g}}$ electrons and 0 ${{e}_{g}}$ electrons and its CFSE is = $\left[ -0.4\times 4 \right]{{\Delta }_{o}}$
= $-1.6{{\Delta }_{o}}$
- Option (C), for low spin ${{d}^{5}}$ configuration, the number of ${{t}_{2g}}$ electrons are 5 and ${{e}_{g}}$ are 0, its CFSE is,$\left[ -0.4\times 5 \right]{{\Delta }_{o}}$=$-2.0{{\Delta }_{o}}$
- Option (D), in the high spin ${{d}^{7}}$ configuration, the number of ${{t}_{2g}}$ electrons are 5 and ${{e}_{g}}$ are 2, its CFSE is,
=$\left[ \left( -0.4\times 5 \right)+\left( +0.6\times 2 \right) \right]{{\Delta }_{o}}$
=$\left[ -2.0+1.2 \right]{{\Delta }_{o}}$
=$-0.8{{\Delta }_{o}}$
From the above obtained values of Crystal field stabilization energies, the value for low spin ${{d}^{5}}$ configuration is maximum which is, $-2.0{{\Delta }_{o}}$.
The correct answer is option “C” .
Additional Information : The stabilization energy gained by the complex by filling electrons in lower energy ${{t}_{2g}}$ orbitals of octahedral complexes is known as crystal field stabilization energy of octahedral complexes. ${{t}_{2g}}$ and ${{e}_{g}}$ orbitals are sets of orbitals resulting from splitting of the degenerate d-orbitals. The d-orbitals in which their lobes are oriented along x and y axes are named as ${{e}_{g}}$ orbitals; these are also called double degenerate orbitals. The d-orbitals whose lobes are oriented in between x, y, z axes are named as ${{t}_{2g}}$ orbitals. These are also called triple degenerate orbitals. These two sets of orbitals have differences in their energies.
Note: The electrons are filled in the orbitals for calculating CFSE by considering whether the given complex is a low spin complex or a high spin complex. In a low spin complex pairing of electrons takes place whereas in a high spin complex, all the degenerate orbitals are filled by one electron each and then pairing occurs.
Complete step by step answer:
Crystal field splitting energy, CFSE for octahedral complexes is calculated from the following formula,
$\left[ \left( -0.4\times 5 \right)+\left( +0.6\times 2 \right) \right]{{\Delta }_{o}}$
Let us calculate the CFSE for all the given electronic configurations. Let us consider,
- Option (A), in the high spin ${{d}^{6}}$ configuration, there will be 4 ${{t}_{2g}}$ electrons and 2 ${{e}_{g}}$ electrons. The CFSE is $\left[ \left( -0.4\times 4 \right)+\left( +0.6\times 2 \right) \right]{{\Delta }_{o}}$
=$\left[ -1.6+1.2 \right]{{\Delta }_{o}}$
=$-0.4{{\Delta }_{o}}$
- Option (B), in low spin ${{d}^{4}}$ configuration, there will be 4 ${{t}_{2g}}$ electrons and 0 ${{e}_{g}}$ electrons and its CFSE is = $\left[ -0.4\times 4 \right]{{\Delta }_{o}}$
= $-1.6{{\Delta }_{o}}$
- Option (C), for low spin ${{d}^{5}}$ configuration, the number of ${{t}_{2g}}$ electrons are 5 and ${{e}_{g}}$ are 0, its CFSE is,$\left[ -0.4\times 5 \right]{{\Delta }_{o}}$=$-2.0{{\Delta }_{o}}$
- Option (D), in the high spin ${{d}^{7}}$ configuration, the number of ${{t}_{2g}}$ electrons are 5 and ${{e}_{g}}$ are 2, its CFSE is,
=$\left[ \left( -0.4\times 5 \right)+\left( +0.6\times 2 \right) \right]{{\Delta }_{o}}$
=$\left[ -2.0+1.2 \right]{{\Delta }_{o}}$
=$-0.8{{\Delta }_{o}}$
From the above obtained values of Crystal field stabilization energies, the value for low spin ${{d}^{5}}$ configuration is maximum which is, $-2.0{{\Delta }_{o}}$.
The correct answer is option “C” .
Additional Information : The stabilization energy gained by the complex by filling electrons in lower energy ${{t}_{2g}}$ orbitals of octahedral complexes is known as crystal field stabilization energy of octahedral complexes. ${{t}_{2g}}$ and ${{e}_{g}}$ orbitals are sets of orbitals resulting from splitting of the degenerate d-orbitals. The d-orbitals in which their lobes are oriented along x and y axes are named as ${{e}_{g}}$ orbitals; these are also called double degenerate orbitals. The d-orbitals whose lobes are oriented in between x, y, z axes are named as ${{t}_{2g}}$ orbitals. These are also called triple degenerate orbitals. These two sets of orbitals have differences in their energies.
Note: The electrons are filled in the orbitals for calculating CFSE by considering whether the given complex is a low spin complex or a high spin complex. In a low spin complex pairing of electrons takes place whereas in a high spin complex, all the degenerate orbitals are filled by one electron each and then pairing occurs.
Recently Updated Pages
Why is there a time difference of about 5 hours between class 10 social science CBSE

In cricket, what is a "pink ball" primarily used for?

In cricket, what is the "new ball" phase?

In cricket, what is a "death over"?

What is the "Powerplay" in T20 cricket?

In cricket, what is a "super over"?

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

