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Magnetic moment 2.83 BM is given by which of the following ions? (At. No. Ti = 22, Cr = 24, Mn = 25, Ni = 28)
medium
The d and f Block Elements
2014
chemistry
Explanation To solve this problem, we need to determine which ion has a magnetic moment of 2.83 BM (Bohr Magneton).
\newline The magnetic moment ( μ ) (\mu) ( μ ) is given by the formula: μ = n ( n + 2 )
\newline \mu = \sqrt{n(n+2)} \, μ = n ( n + 2 )
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BM
where
is the number of unpaired electrons.
Given:
μ = 2.83 \mu = 2.83 \, μ = 2.83 BM
Let's calculate
using the formula:
2.83 = n ( n + 2 )
\newline 2.83 = \sqrt{n(n+2)}
\newline 2.83 = n ( n + 2 ) Squaring both sides, we get:
( 2.83 ) 2 = n ( n + 2 ) 8.0089 = n ( n + 2 )
\newline (2.83)^2 = n(n+2)
\newline 8.0089 = n(n+2)
\newline ( 2.83 ) 2 = n ( n + 2 ) 8.0089 = n ( n + 2 ) Solving the quadratic equation
n 2 + 2 n − 8.0089 = 0 , n^2 + 2n - 8.0089 = 0, n 2 + 2 n − 8.0089 = 0 , we find
n ≈ 2. n \approx 2.
\newline n ≈ 2. Therefore, the ion should have 2 unpaired electrons.
Let's analyze each option:
•
Atomic number of Ti is 22. Electronic configuration:
[ A r ] 3 d 1 4 s 0 . [Ar] \, 3d^1 \, 4s^0.
\newline [ A r ] 3 d 1 4 s 0 . Ti
loses 3 electrons:
[ A r ] 3 d 1 . [Ar] \, 3d^1. [ A r ] 3 d 1 . Number of unpaired electrons
•
Atomic number of Ni is 28. Electronic configuration:
[ A r ] 3 d 8 4 s 2 . [Ar] \, 3d^8 \, 4s^2.
\newline [ A r ] 3 d 8 4 s 2 . Ni
loses 2 electrons:
[ A r ] 3 d 8 . [Ar] \, 3d^8. [ A r ] 3 d 8 . Number of unpaired electrons
•
Atomic number of Cr is 24. Electronic configuration:
[ A r ] 3 d 5 4 s 1 . [Ar] \, 3d^5 \, 4s^1.
\newline [ A r ] 3 d 5 4 s 1 . Cr
loses 3 electrons:
[ A r ] 3 d 3 . [Ar] \, 3d^3. [ A r ] 3 d 3 . Number of unpaired electrons
•
Atomic number of Mn is 25. Electronic configuration:
[ A r ] 3 d 5 4 s 2 . [Ar] \, 3d^5 \, 4s^2.
\newline [ A r ] 3 d 5 4 s 2 . Mn
loses 2 electrons:
[ A r ] 3 d 5 . [Ar] \, 3d^5. [ A r ] 3 d 5 . Number of unpaired electrons
The ion with 2 unpaired electrons is
N i 2 + . Ni^{2+}.
\newline N i 2 + . Therefore, the correct option is Option 2: