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JEE Advanced levelNumerical answerDifficulty 3 of 3df208

Terbium has atomic number 65 and ground-state configuration $[Xe] 4 f^{9} 6 s^{2}$ . Calculate the spin-only magnetic moment of $Tb X^{4 +}$ in BM, using $μ = n (n + 2)$ (answer to two decimals).

Numerical answer type, answer in BM, tolerance 0.01.

Answer: 7.94 BM

Removing

6 s^{2}

and two

4 f

electrons gives

Tb X^{4 +}

with

4 f^{7}

, so n = 7 unpaired electrons. Spin-only

μ = 7 \times 9 = 63 = 7.94

BM.

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