Calculate the spin-only magnetic moment of the high-spin Fe^2+ ion (atomic number 26) in Bohr magnetons using mu=sqrtn(n+2) (two decimals).

JEE Main levelNumerical answerDifficulty 2 of 3ss209

Calculate the spin-only magnetic moment of the high-spin $Fe X^{2 +}$ ion (atomic number 26) in Bohr magnetons using $μ = n (n + 2)$ (two decimals).

Numerical answer type, answer in BM, tolerance 0.01.

Fe X^{2 +}

3 d^{6}

; in the high-spin state 4 electrons are unpaired, so

n = 4

μ = 4 \times 6 = 24 = 4.899 \approx 4.90

BM.