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Q.

${Yb}^{2 +}$ and ${Lu}^{3 +}$ are diamagnetic due to

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a

Fully ‘f’ sub shells

b

Parlty filled ‘f’ sub shells

c

Vacant ‘f’ sub shells

d

Partly filled ‘d’ sub shells

answer is B.

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Detailed Solution

${Yb}^{2 +} = 4 f^{14} 5 d^{0} 6 s^{0} {Lu}^{3 +} = 4 f^{14} 5 d^{0} 6 s^{0}$
${Yb}^{2 +}, {Lu}^{3 +}$ have no unpaired electrons hence, diamagnetic.

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