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An ion X^(n+) of element with atomic number 26 has a magnetic moment of sqrt(24) B.M. Find the number of unpaired electrons and the value of n respectively.

1. 4 and 3
2. 3 and 5
3. 4 and 2
4. 4 and 1

Correct answer: 4 and 2

Solution

Magnetic moment = sqrt(n*(n+2)) = sqrt(24). So n*(n+2) = 24 => n² + 2n - 24 = 0 => (n+6)(n-4) = 0 => n = 4 unpaired electrons. Element Z = 26 is Fe: [Ar] 3d⁶ 4s². Fe²+ (n=2): remove 2 electrons from 4s -> [Ar] 3d⁶. In 3d⁶: 4 unpaired (one pair + 4 singles by Hund's rule in 5 d-orbitals: 2 orbitals share 1 pair, 4 have 1 electron each - actually 3d⁶ gives 4 unpaired: t2g has 4 and eg has 2 in high-spin, giving 4 unpaired). So Fe²+ has 4 unpaired electrons. n = 2. Answer: 4 unpaired electrons and n = 2.

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