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 _ _ _ _ _ _ _ _ _ _ _ _ .
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 _ _ _ _ _ _ _ _ _ _ _ _ .
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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.
Updated On: Apr 7, 2026
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Correct Answer: 15
Solution and Explanation
Step 1: Find the number of unpaired electrons for \(d^3\).
For a \(d^3\) configuration in an octahedral complex, the three electrons occupy three different \(d\)-orbitals singly according to Hund's rule.
So, the number of unpaired electrons is:
\[ 3 \]
Step 2: Find the number of unpaired electrons for \(d^4\) (low spin).
In a low spin \(d^4\) complex, electrons pair up in the lower energy \(t_{2g}\) orbitals before entering the higher energy \(e_g\) orbitals.
Thus, the arrangement is \(t_{2g}^4 e_g^0\).
This gives:
\[ 2 \text{ unpaired electrons} \]
Step 3: Find the number of unpaired electrons for \(d^5\) (high spin).
In a high spin \(d^5\) complex, all five electrons occupy the five \(d\)-orbitals singly before any pairing occurs.
Thus, all five electrons remain unpaired.
So, the number of unpaired electrons is:
\[ 5 \]
Step 4: Find the number of unpaired electrons for \(d^7\) (low spin).
In a low spin \(d^7\) complex, electrons first fill and pair in the \(t_{2g}\) orbitals and then start entering the \(e_g\) orbitals.
Thus, the arrangement is \(t_{2g}^6 e_g^1\).
Hence, only one electron remains unpaired.
So, the number of unpaired electrons is:
\[ 1 \]
Step 5: Find the number of unpaired electrons for \(d^6\) (high spin).
In a high spin \(d^6\) complex, the arrangement is \(t_{2g}^4 e_g^2\).
In this case, four electrons remain unpaired.
So, the number of unpaired electrons is:
\[ 4 \]
Step 6: Add all the unpaired electrons.
Now, total number of unpaired electrons is:
\[ 3 + 2 + 5 + 1 + 4 = 15 \]
Step 7: State the final answer.
Hence, the total number of unpaired electrons is:
\[ \boxed{15} \]
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