Question

For the complex ion with configurations \(d^3\), \(d^4\) (low spin), \(d^5\) (high spin), \(d^7\) (low spin) and \(d^6\) (high spin), the total number of unpaired electrons is _ _ _ _ _ _ _ _ _ _ _ _ .

Hint:

For octahedral complexes, low spin means pairing takes place in \(t_{2g}\) orbitals first, while high spin means electrons occupy higher energy orbitals before pairing. Always write the \(t_{2g}\) and \(e_g\) arrangement to count unpaired electrons correctly.

Answer 1

Correct Answer: 15

Step 1: Understanding the Concept:
According to Crystal Field Theory (CFT), in an octahedral field, the five d-orbitals split into two sets:
\(\quad t_{2g}\) (lower energy): \(d_{xy}, d_{yz}, d_{zx}\)
\(\quad e_g\) (higher energy): \(d_{x^2-y^2}, d_{z^2}\)
In a high spin complex (weak field ligands), electrons occupy the \(e_g\) orbitals before pairing in \(t_{2g}\).
In a low spin complex (strong field ligands), electrons pair up in the \(t_{2g}\) orbitals before occupying \(e_g\).
Step 2: Detailed Explanation:
Let us count the unpaired electrons for each configuration in an octahedral field:
1. d\(^3\): Configuration: \(t_{2g}^3 \, e_g^0\)
Each of the three \(t_{2g}\) orbitals has one electron (singly occupied).
Unpaired electrons = 3.
2. d\(^4\) (low spin): Configuration: \(t_{2g}^4 \, e_g^0\)
One \(t_{2g}\) orbital is doubly occupied (1 pair), and two have 1 electron each.
Unpaired electrons = 2.
3. d\(^5\) (high spin): Configuration: \(t_{2g}^3 \, e_g^2\)
All five d-orbitals are singly occupied (maximum unpaired).
Unpaired electrons = 5.
4. d\(^7\) (low spin): Configuration: \(t_{2g}^6 \, e_g^1\)
All three \(t_{2g}\) orbitals are fully paired; one \(e_g\) orbital has 1 electron.
Unpaired electrons = 1.
5. d\(^6\) (high spin): Configuration: \(t_{2g}^4 \, e_g^2\)
One \(t_{2g}\) orbital is paired; two \(t_{2g}\) and two \(e_g\) have 1 electron each.
Unpaired electrons = 4.
Total unpaired electrons: \[ 3 + 2 + 5 + 1 + 4 = 15 \]
Step 3: Final Answer:
The total number of unpaired electrons is 15.