A first row transition metal compound [M(H_2O)_6]^2+ has a magnetic moment of 2.

Question

A first row transition metal compound [M(H_2O)_6]^2+ has a magnetic moment of 2.83. Determine the electron configuration and number of unpaired electrons using the spin only case. Draw the electron configuration. Give the d electron configuration (in an octahedral ligand field) for the following compound for both high spin and low spin situations. Write out the Product c and Product e for each case. Assuming that Product c and Product e are constants regardless of configuration, how large does delta oct have to be in order to push it to a low spin compound? Remember LFSF lowers the over energy as does exchange energy, while columbic energy raises the energy. Which configuration do you predict and why?

Explanation / Answer

Magnetic moment given = 2.86

which imples suare root of [n(n+2) ]= 2.86 x 2.86

where n = number of unpaired electrons.

on solving the above quadratic we get two values 2 and -4.

retain the value 2, which means unpaired electrons in M +2= 2

so d electrons in M   = 8

electron configuration of M = 18 + 8 +2 = 28 lectons that is of nickel.

[Ar] 3d 8 4s2

this is an outer orbital ocmplex. , High spin complex.

RUTHENIUM COMPLEX

ruthenium : [Kr] 4d75s1

in its +3 state it will be [Kr] 4d55s0 (high spin configuration, 5 unpaired electrons.)

                                      [Kr] 4d55s0 (low spin configuration , 0nly one unpaired electron, 4 electrons are paired up )

CFSE will be very large since CO is strong ligand.

complex will be low spin complex. only one unpaired will be present.

please note that the symbols given in the question are not understood, clearly be given

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