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Choose the Best Answer · Q7

Q.A magnetic moment of 1.73 BM will be shown by one among the following (NEET)

(a) TiCl₄
(b) [CoCl₆]⁴⁻
(c) [Cu(NH₃)₄]²⁺
(d) [Ni(CN)₄]²⁻
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Step 1. μs = √[n(n+2)] BM = 1.73 BM corresponds to n = 1 unpaired electron (√[1×3] = 1.73).

Step 2. TiCl₄: Ti is +4 here, a d⁰ configuration -- zero unpaired electrons, diamagnetic (μs = 0), ruled out.

Step 3. [CoCl₆]⁴⁻: Co²⁺ is d⁷; Cl⁻ is a weak-field ligand, so this is high spin, t2g⁵eg², giving 3 unpaired electrons (μs = √[3×5] = 3.87 BM), ruled out. …

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