Some species are given
Ni2+, Fe2+, Co2+, V3+ and Ti2+
How many species has magnetic moment (spin only) less than 3 BM.
Correct Answer :
Solution :
The correct answer is 3.
To determine how many of the given species have a spin-only magnetic moment less than 3 BM, we need to calculate the number of unpaired electrons for each transition metal ion.
The spin-only magnetic moment () is given by the formula:
where is the number of unpaired electrons.
For to be less than 3 BM, the value of must be less than 3 (since for , ). Therefore, we are looking for species with , i.e., or .
Let us analyze the valence shell electron configuration and count the number of unpaired electrons () for each ion:
1. Ni2+ (Titanium to Nickel series, Z = 28):
Electronic configuration of Ni: [Ar] 3d8 4s2
Electronic configuration of Ni2+: [Ar] 3d8
Number of unpaired electrons () = 2
(which is < 3 BM).
2. Fe2+ (Z = 26):
Electronic configuration of Fe: [Ar] 3d6 4s2
Electronic configuration of Fe2+: [Ar] 3d6
Number of unpaired electrons () = 4
(which is > 3 BM).
3. Co2+ (Z = 27):
Electronic configuration of Co: [Ar] 3d7 4s2
Electronic configuration of Co2+: [Ar] 3d7
Number of unpaired electrons () = 3
(which is > 3 BM).
4. V3+ (Z = 23):
Electronic configuration of V: [Ar] 3d3 4s2
Electronic configuration of V3+: [Ar] 3d2
Number of unpaired electrons () = 2
(which is < 3 BM).
5. Ti2+ (Z = 22):
Electronic configuration of Ti: [Ar] 3d2 4s2
Electronic configuration of Ti2+: [Ar] 3d2
Number of unpaired electrons () = 2
(which is < 3 BM).
Thus, the species having a magnetic moment less than 3 BM are Ni2+, V3+, and Ti2+.
Therefore, there are 3 such species.
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