Consider the following ions:
(i) Ni2+ (ii) Co2+ (iii) Cr2+ (iv) Fe3+
(Given, at. no. Cr = 24, Fe = 26, Co = 27, Ni = 28)
The correct order of magnetic moment of these ions is
(i) < (ii) < (iii) < (iv)
(iv) < (ii) < (iii) < (i)
(i) < (iii) < (ii) < (iv)
(iii) < (iv) < (ii) < (i)
A.
(i) < (ii) < (iii) < (iv)
Magnetic moment,
where, n = number of unpaired electrons
Higher the number of unpaired electrons, more the value of μ.
The electronic configuration of given species are-
Ni2+ = [Ar]3d8 (two unpaired electrons)
Co2+ = [Ar]3d7 (three unpaired electrons)
Cr2+ = [Ar]3d4 (four unpaired electrons)
Fe3+ = [Ar]3d5 (five unpaired electrons)
Thus, the order of magnetic moment is-
The electron configuration of the oxide ion is much most similar to the electron configuration of the
sulphide ion
nitride ion
oxygen atom
nitrogen atom