The value of spin only magnetic moment for the complex, [Pt(CN)4]2− is..........
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0 BM
The spin only magnetic moment depends only on the number of unpaired electrons, n:
\( \displaystyle \mu=\sqrt{n(n+2)}\ \mathrm{BM} \)
Platinum is +2 here (x + 4(−1) = −2). Pt (Z = 78) is [Xe]4f145d96s1; losing the 6s electron and one 5d electron gives \( \ce{Pt^{2+}} \): 5d8. For 4d and 5d metals the crystal field splitting is large, and \( \ce{CN^{-}} \) is a strong field ligand. So \( \ce{[Pt(CN)4]^{2-}} \) is a square planar complex (dsp2). All eight d electrons pair up in the four lower-energy d orbitals, leaving the highest one empty.
\( \displaystyle n=0 \)
\( \displaystyle \mu=\sqrt{0\times(0+2)}=0\ \mathrm{BM} \)
The complex is diamagnetic.