# Calculate the ‘spin only’ magnetic moment of M2+(aq) ion (Z = 27).

sumit kumar
8 years ago
For, Mn2+ ion
Z= 27
Electronic Configuration of Mn2+= [Ar] 3d7
Number of unpaired electrons = 3
Spin only magnetic moment =$\sqrt{n(n+2))} =\sqrt{3(3+2)}$= 3.87 B.M
Thanks & Regards
Sumit Kumar
Riya Nath
35 Points
5 years ago
The atomic number of Mn is 25. This implies that no. of electrons present is 25.Since, it is given Mn2+ , so number of electrons present now is 23 with the electronic configuration= 1s2 2s2 2p6 3s2 3p6 3d5.No. Of unpaired electrons, n= 5So magnetic moment= √n(n+2)=√5(5+2)= 5.92 BM
Ujjwal Rana
26 Points
5 years ago
Z=27 implies element is Co.Co²+ carries 7 electrons in d orbital. For aq solution H2O is the ligand which is weak field ligand. So metal will tend to form tetrahedral structure, carrying 3 unpaired electrons in t2g.So spin only magnetic moment would be √(3(3+2)) = √15 = 3.87
mani sharma
24 Points
5 years ago
Z=27 means the element is cobalt when 2 electrons are sabstracted then the number of in paired electrons in d orbital is 5 so under root5(5+2)=root 35 =4.87
Ujjwal Rana
26 Points
5 years ago
Plz dont post such stupid answers. While removing 2 electrons. Theyll be removed from s orbital not d. That leaves us with 7 d orbitals then how on Earth can you have 5 unpaired electrons. Plz clear your concepts before posting such stupid answers.
Ujjwal Rana
26 Points
5 years ago
More over root 35 is 5.9 approx. Such persons should focus on studies rather than wasting time by posting such answers
Riya Nath
35 Points
5 years ago
And how on earth the atomic number of manganese becomes 27? People, please go and study the periodic table.
Keshav Negi
24 Points
5 years ago
First of all
its Written M2+ not Mn2+..
Z= 27 i.e Cobalt.
Co2+ the 2 electron will be removed from s orbital not d orbital.
7 electrons in the outermost shell.
i.e 3 valence electrons.
Since for an aqueous solution. H2O is a weak field ligand.
magnetic moment  =√(3(3+2)) = √15 = 3.87
Piyal Mukherjee
24 Points
5 years ago
See Mn is 25 and Mn2+ is 23With the configuration 3d5 at the end.Thus half filled condition, hence 5 unpaired eletrons......as d orbital can have at the max 10 eletrons.So magnetic moment, mew=under root n (n+2) B.M.(Bohr Megneton)[Where n is no. of unpaired electrons]Ans. Mew=root 35 thus , Paramagnetic{If mew was 0 then dimagnetic...no unpaired eletrons}.
samruddhi A Bogar
11 Points
5 years ago
Mn2+ =23 with 3d^5 in its end .Number of unpaired electrons is 5 .Hence magnetic moment is calculated with n= 5 which will give correct ans as 5.92
Abhay
11 Points
5 years ago
There are 2 methods to calculate magnetic moment.First: √n(n+2) where n= number of unpaired electronsSecond: √4s(s+1) where s= 1/2 × n (total spin of an atom..)
Akansha
19 Points
4 years ago
The electronic configuration of manganese is 3d5 4 s2 and for mn+2 is 3d5 so 5 unpaired electrons are present in mn+2 Formula for magnetic moment is √ n(n+2)Where n = no. Of unpaired electronsFor mn+2=√ n(n+2), = √5(5+2) = √5(7) =√35 =5.91BM approx.Where BM stands for Bohr magneton it is a unit of magnetic momentAkansha panwar Student of b.sc bio 2 nd year
13 Points
3 years ago
Mn has atomic number 25 its confrogation is 3d5 4s2  but if it is Mn2+ it will have 23 electrons it will hav 3d5 confrigation . no of unpaired electron will be 5 so the value will be 5.7 Bm
2 years ago
Hello student,

The given atomic number Z = 27 denotes Co
Now, Electronic Configuration of Co2+ = [Ar] 3d7
Hence, number of unpaired electrons in Co2+ = 3
The spin only magnetic moment is = $\sqrt{n(n+2))} =\sqrt{3(3+2)}$ =  3.87 B.M

Hope it helps.