Select the Correct statements about the magnetic nature of the following compounds:
(A) [Cr(NH3)6]3+ is paramagnetic
(B) [Ni(Cl)4]2- is diamagnetic
(C) [Ni(CN)4]2- is diamagnetic
(D) [Co(NH3)6]3+is paramagnetic
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (C) only
A complex is paramagnetic if it has at least one unpaired electron, and diamagnetic if all its electrons are paired. To decide, find the d electron count of the metal ion and check whether the ligands are strong field (they pair the electrons) or weak field (they do not):
weak field: \( \ce{Cl^{-}} \), \( \ce{F^{-}} \), \( \ce{H2O} \)
strong field: \( \ce{CN^{-}} \), CO, and \( \ce{NH3} \) with \( \ce{Co^{3+}} \)
(A) \( \ce{[Cr(NH3)6]^{3+}} \) is paramagnetic — Correct. Cr (Z = 24) is [Ar]3d54s1; losing the 4s electron and two 3d electrons gives Cr3+: 3d3. The three electrons occupy the t2g orbitals singly whatever the ligand, so there are three unpaired electrons.
(B) \( \ce{[Ni(Cl)4]^{2-}} \) is diamagnetic — Incorrect. Ni (Z = 28) is [Ar]3d84s2; losing the two 4s electrons gives Ni2+: 3d8, and the weak field Cl− cannot pair its electrons, so the complex is tetrahedral with two unpaired electrons and is paramagnetic.
(C) \( \ce{[Ni(CN)4]^{2-}} \) is diamagnetic — Correct. The strong field CN− pairs the eight 3d electrons of \( \ce{Ni^{2+}} \) (Ni, [Ar]3d84s2, minus the two 4s electrons); the complex is square planar (dsp2) with no unpaired electron.
(D) [Co(NH3)6]3+is paramagnetic — Incorrect. Co (Z = 27) is [Ar]3d74s2; losing the two 4s electrons and one 3d electron gives Co3+: 3d6, and NH3 is strong field for Co3+, so all six electrons pair in t2g (d2sp3); the complex is diamagnetic.