The correct match of spin-only magnetic moment for the complexes cis-[Fe(phen)₂(NCS-N)₂] (A) and [Fe(phen)₃]Cl₂ (B) at 300 K is (phen = 1, 10-phenanthroline)

Concept:

Spin-only magnetic moment:

The magnetic behaviour based only on the contribution of unpaired electrons can be approximated by the spin-only magnetic moment.

μ = √[n(n+2)] BM

where,

• μ = spin-only magnetic moment in Bohr magnetons (BM).
• n= No. of unpaired electrons.
• BM is the Bohr magneton (9.274 x 10^-24 J/T).

Explanation:

cis-[Fe(phen)2(NCS-N)2] =High spin complex

Fe+2(\(t_{2g}^{4}\:e_{g}^{2}\)), No. of unpaired electrons(n)=4

\(\mu = { \sqrt{n(n+1)} }\)

or, \(\mu= {\sqrt{4(4+1)} }\)

or,\(\mu= {\sqrt{4(5)} }\)

or, \(\mu= 4.89\) BM

[Fe(phen)3]Cl2 :

1,10-Phenanthroline ligand complexes act as a low spin complexes if the temperature is \(\leq 300\)k.

Fe²⁺(\(t_{2g}^{6}\:e_{g}^{0}\)), No. of unpaired electrons=0

So spin only magnetic moment (n)=0BM

Conclusion:

Hence the correct answer is  option (3) .