Question

A transition metal 'M' among Sc, Ti, V , Cr, Mn and Fe has the highest second ionisation enthalpy. The spin only magnetic moment value of M+
ion is ______ BM (Near integer) (Given atomic number Sc : 21, Ti : 22, V : 23, Cr : 24, Mn : 25, Fe : 26)

Answer 1

The second ionisation enthalpy is highest for Mn, because the second ionisation removes an electron from a stable [Ar]3$d^5$ configuration, making it highly stable. The spin-only magnetic moment is calculated using the formula:
$\mu = \sqrt{n(n+2)}{ BM}$
For Mn, with $n = 5$ (because 3$d^5$ has 5 unpaired electrons), the magnetic moment is:
$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 6 { BM}.$
Thus, the correct answer is (6).

Answer 2

The second ionisation enthalpy is highest for Mn, because the second ionisation removes an electron from a stable [Ar]3$d^5$ configuration, making it highly stable. The spin-only magnetic moment is calculated using the formula:
$\mu = \sqrt{n(n+2)}{ BM}$
For Mn, with $n = 5$ (because 3$d^5$ has 5 unpaired electrons), the magnetic moment is:
$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 6 { BM}.$
Thus, the correct answer is (6).