A magnetic moment of 1.73 BM will be shown by one among the following : (a) [CoCl₆]⁴⁻ (b) TiCl₄ (c) [Cu(NH₃)₄]

Question 58 of 67

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

(a)

[CoCl_6]^{4-}

(b)

TiCl_4

(c)

[Cu(NH_3)_4]^{2+}

(d)

[Ni(CN)_4]^{2-}

Puducherry TnboardTamil Nadu HSC (DGE) Board 2023MCQ· 1mImportance★★★★★

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Using the spin-only formula $\mu=\sqrt{n(n+2)}$ BM, $\mu=1.73$ BM requires exactly $n=1$ unpaired electron; only $[Cu(NH_3)_4]^{2+}$ (Cu $^{2+}$ , $d^9$ , always 1 unpaired electron) matches.

Check each complex: (a) $[CoCl_6]^{4-}$ — overall charge $-4$ with 6 $Cl^-$ ( $-6$ ) gives Co oxidation state $+2$ , so $Co^{2+}$ is $3d^7$ ; $Cl^-$ is a weak-field ligand so this is high-spin, giving 3 unpaired electrons ( $\mu=\sqrt{3\times5}=3.87$ BM).

(b) $TiCl_4$ — Ti is $+4$ , so $Ti^{4+}$ is $3d^0$ , no unpaired electrons ( $\mu=0$ , diamagnetic).

(c) $[Cu(NH_3)_4]^{2+}$ — $Cu^{2+}$ is $3d^9$ ; irrespective of ligand field, a $d^9$ configuration always has exactly one unpaired electron, giving $\mu=\sqrt{1\times3}=\sqrt3=1.73$ BM. …

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

Concept understanding — Valence Bond Theory of Coordination Compounds

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

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