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Calculated spin only magnetic moment of Cr^(x+) is 4.9 BM. Find the 'x' value. |
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Answer» Solution :Magnetic moment `mu= 4.9 BM = sqrt(n(n+2))` Number of unpaired ELECTRONS `= n = 4` with FOUR unpaired electrons, the configuration is `3d^(4)4s^(0)`. `Cr^(2+)` ion posses 4 unpaired electrons, THUS .X. value is 2. |
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