Question Details

Some species are given

Ni2+, Fe2+, Co2+, V3+ and Ti2+

How many species has magnetic moment (spin only) less than 3 BM.

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Correct Answer :

3

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 (μs) is given by the formula:

μs=n(n+2) BM

where n is the number of unpaired electrons.
For μs to be less than 3 BM, the value of n must be less than 3 (since for n=3, μs=3(3+2)=153.87 BM). Therefore, we are looking for species with n<3, i.e., n=1 or n=2.

Let us analyze the valence shell electron configuration and count the number of unpaired electrons (n) 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 (n) = 2
μs=2(2+2)=82.83 BM (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 (n) = 4
μs=4(4+2)=244.90 BM (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 (n) = 3
μs=3(3+2)=153.87 BM (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 (n) = 2
μs=2(2+2)=82.83 BM (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 (n) = 2
μs=2(2+2)=82.83 BM (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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