Arrange the following ions in the decreasing order of their magnetic moment
I. V4+
II. Mn3+
III. Fe3+
IV. Ni2+
[At nos. V=23 Mn=25 Fe=26 Ni=28]
Answers
Given:
I. V4+
II. Mn3+
III. Fe3+
IV. Ni2+
To find:
Arrange the given ions in the decreasing order of their magnetic moment
Solution:
We calculate the magnetic moment using the number of lone pair of electrons present in an ion.
so, we have,
μ = √[n (n + 2)]
where,
μ = magnetic moment
n = number of unpaired electrons
From given, we have,
I. V4+
V4+ has 1 lone pair of electrons,
μ = √[n (n + 2)]
μ = √[1 (1 + 2)]
∴ μ = 1.73 BM
II. Mn3+
Mn3+ has 4 lone pair of electrons,
μ = √[n (n + 2)]
μ = √[4 (4 + 2)]
∴ μ = 4.89 BM
III. Fe3+
Fe3+ has 5 lone pair of electrons,
μ = √[n (n + 2)]
μ = √[5 (5 + 2)]
∴ μ = 5.91 BM
IV. Ni2+
Ni2+ has 2 lone pair of electrons,
μ = √[n (n + 2)]
μ = √[2 (2 + 2)]
∴ μ = 2.82 BM
The arrangement the given ions in the decreasing order of their magnetic moment is Fe3+ > Mn3+ > Ni2+ > V4+