8.8   Calculate the ‘spin only’ magnetic moment of  M^{2+}\; _{(aq)} ion (Z=27).

Answers (1)

Atomic number (Z)= 27
So the electronic configuration cobalt (Co) is 3d^{7}, 4s^{2}
M^{2+}\; _{(aq)} ion means, it loses its two electrons and become d^{7} configuration. And it has 3 unpaired electrons
So, \mu = \sqrt{n(n+2)} , where n = no. of unpaired electron
by putting the value of n= 3
we get, \mu = \sqrt{15}
                 \approx 4\ BM

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