Before You Begin

Question #44866

Difficulty: 8/10 | 🟢 Free

The Question

Consider the homoleptic octahedral fluoro complexes of Group 9 metals in the +3 oxidation state: $\ce{[CoF6]^3-}$ , $\ce{[RhF6]^3-}$ , and $\ce{[IrF6]^3-}$ . Let $\mu_1, \mu_2,$ and $\mu_3$ be their respective spin-only magnetic moments (in BM).

Consider the homoleptic chloro complexes of Group 10 metals in the +2 oxidation state: $\ce{[NiCl4]^2-}$ , $\ce{[PdCl4]^2-}$ , and $\ce{[PtCl4]^2-}$ . Let $n_1, n_2,$ and $n_3$ be the number of unpaired electrons in these complexes, respectively.

Calculate the value of $Z = (\mu_1^2 + \mu_2^2 + \mu_3^2) + (n_1 + n_2 + n_3)$ . (Assume spin-only magnetic moments are calculated using the formula $\mu = \sqrt{n(n+2)}$ BM).

Options

A.
26
B.
52
C.
78
D.
32

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Lanthanoid Contraction cause|Spin-Only Magnetic Moment

Why This Question Is Tricky

Common Mistakes

The question tests the 'Hidden Variable' of the principal quantum number ( $n$ ) of the d-orbitals in determining Crystal Field Splitting Energy ( $\Delta$ ).

Students often memorize the spectrochemical series ( $\ce{F^-}$ and $\ce{Cl^-}$ as weak field ligands) and blindly apply it to all transition metals, incorrectly assuming that $\ce{[RhF6]^3-}$ , $\ce{[IrF6]^3-}$ , $\ce{[PdCl4]^2-}$ , and $\ce{[PtCl4]^2-}$ will be high-spin/tetrahedral like their 3d counterparts.

The 'Chemical Exception' is that 4d and 5d metals almost exclusively form low-spin complexes due to their inherently large $\Delta$ values, which overcome the pairing energy ( $P$ ) regardless of the ligand.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	D and F Block Elements - Class 12
Concept	Lanthanoid Contraction cause
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free