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The orbital configuration of `._(24)Cr` is `3d^(5) 4s^(1)`. The number of unpaired electrons in `Cr^(3+) (g)` isA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - C `Cr-4s^(2) 3d^(5), Cr^(+)-4s^(0) 3d^(3)` The correct answer is 3.
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