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Class 12 Tuition > Learn Intext Exercise 8.7 > Calculate the ‘spin only’ magnetic moment...
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Calculate the ‘spin only’ magnetic moment of M2+(aq) ion (Z = 27).

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Magnetic moment=√n(n+2), here n- number of unpaired electrons M2+ contain 5 unpaded electron in it's valance sheel i.e. in 3d Therefore magnetic moment=√n(n+2)=√3*(5)=√15=3.87BM
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Magnetic moment=√n(n+2), here n- number of unpaired electrons M2+ contain 5 unpaded electron in it's valance sheel i.e. in 3d Therefore magnetic moment=√n(n+2)=√3*(5)=√15=3.87BM
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An experienced physics professional for iit-jee & neet

The atomic number Z =27 corresponds to Cobalt and Cobalt has 3 unpaired electrons(n=3). The spin magnetic moment mu(S) =sqrt =sqrt =sqrt(15) =3.87 BM (approx)
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Z = 27 3d7 4s2 M2+ = 3d7 3d7 = i.e., 3 unpaired electrons n = 3 μ ≈ 4 BM
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Z = 27 [Ar] 3d7 4s2 M2+ = [Ar] 3d7 3d7 = i.e., 3 unpaired electrons n = 3 μ ≈ 4 BM read less
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μs=√(n(n+2)), n= number of unpaired electrons. hence, 3.870 B.M.
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Z=27 is Cobalt No of unpaired electron in Colbalt +2 is three Spin on magnetic moment √15= 4(almost)
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Sr. CHEMISTRY FACULTY ( UPTO 12 TH CLASS , JEE , NEET PREPARATION)

Magnetic moment have formula is √n(n+2). In this question unpaired electrons is 3. So magnetic moment is √3(3+2). Magnetic moment is √15 BM. =3.8729 BM. Beacuse when M goes to M2+ then 2 electrons loss by 4s and 7 electrons present in 3d so M2+ have outermost configuration is 3d7 so...
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Magnetic moment have formula is√n(n+2). In this question unpaired electrons is 3. So magnetic moment is√3(3+2). Magnetic moment is√15 BM. =3.8729 BM. Beacuse when M goes to M2+ then 2 electrons loss by 4s and 7 electrons present in 3d so M2+ have outermost configuration is 3d7 so having 3 unpaired electrons. read less
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PhD in Physical Chemistry with an experience of 8+ years in teaching.

3.87 BM
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