The spin-only magnetic moment value of Mn+ ion formed among Ni, Zn, Mn and Cu that has the least enthalpy of atomisation is . (in nearest integer) Here n is equal to the number of diamagnetic complexes among K2[NiCl4], [Zn(H2O)6]Cl2, K3[Mn(CN)6] and [Cu(PPh3)3I]
Answer
0
Solution
Enthalpy of atomisation ( kJ mol-1): Mn 281, Ni 430, Cu 339, Zn 126.
Zn ( 3d104s2) has no unpaired d electron for metallic bonding, its enthalpy of atomisation is the least M = Zn.
Value of n:
K2[NiCl4]: Ni2+ (3d8), Cl- weak field, tetrahedral (sp3) — 2 unpaired electrons, paramagnetic.
[Zn(H2O)6]Cl2: Zn2+ ( 3d10) — diamagnetic.
K3[Mn(CN)6]: Mn3+ (3d4), CN- strong field, t2g4eg0 — 2 unpaired electrons, paramagnetic.
[Cu(PPh3)3I]: Cu+ ( 3d10) — diamagnetic.
n = 2, and the ion is Zn2+ ( 3d10, no unpaired electron).
= nu(nu + 2) = 0(0 + 2) = 0 BM
Answer: 0