The behaviour of an electron in an atom is described mathematically by a wave function, or orbital.Spin of the electron produce angular momentum equal to S= sqrt(s(s+1)) (h)/(2pi)where S= +1/2. Total spin of an atom=+ n/2orh/2 Where n is the number of unpaired electron. The substance which contain species with unpaired electrons in their orbitals behave as paramagnetic substances. The paramagnetism is expressed in terms of magnetic moment the magnetic moment of an atom mu_s sqrt(s(s+1)) (eh)/(2pi mc) = sqrt((n/2)(n/2+1)) (eh)/(2pi mc) s = n/2 impliesmu_s = sqrt(n(n+2)) B.M n = number of unpaired electrons 1 B.M. (Bohr magneton) = (eh)/(4pi mc) If magnetic moment is zero the substances is di-magnetic. Which of the following ion has lowest magnetic moement.

The behaviour of an electron in an atom is described mathematically by

1.	The behaviour of an electron in an atom is described mathematically by a wave function, or orbital.Spin of the electron produce angular momentum equal to S= sqrt(s(s+1)) (h)/(2pi)where S= +1/2. Total spin of an atom=+ n/2orh/2 Where n is the number of unpaired electron. The substance which contain species with unpaired electrons in their orbitals behave as paramagnetic substances. The paramagnetism is expressed in terms of magnetic moment the magnetic moment of an atom mu_s sqrt(s(s+1)) (eh)/(2pi mc) = sqrt((n/2)(n/2+1)) (eh)/(2pi mc) s = n/2 impliesmu_s = sqrt(n(n+2)) B.M n = number of unpaired electrons 1 B.M. (Bohr magneton) = (eh)/(4pi mc) If magnetic moment is zero the substances is di-magnetic. Which of the following ion has lowest magnetic moement.
Answer» `Fe^(2+)` `MN^(2+)` `Cr^(3+)` `V^(3+)` Solution :Lowest NUMBER of UNPAIRED `e^-` have lower magnetic MOMENT