Read the passage carefully and answer the Questions.
According to Valence Bond theory, the metal atom or ion under the influence of ligands can use its (n-1)d, ns, np or ns, np, nd orbitals for hybridisation to yield a set of equivalent orbitals of definite geometry such as octahedral, tetrahedral, square planar and so on. These hybridised orbitals are allowed to overlap with ligand orbitals that can donate electron pairs for bonding. While the VB theory, to a larger extent, explains the formation, structures and magnetic behaviour of coordination compounds, it suffers from the many shortcomings. The crystal field theory (CFT) is an electrostatic model which considers the metal-ligand bond to be ionic arising purely from electrostatic interactions between the metal ion and the ligand. The five d orbitals in an isolated gaseous metal atom/ion have same energy, i.e., they are degenerate. This degeneracy is maintained if a spherically symmetrical field of negative charges surrounds the metal atom/ion. However, when this negative field is due to ligands in a complex, it becomes asymmetrical and the degeneracy of the d orbitals is lifted. It results in splitting of the d orbitals. The pattern of splitting depends upon the nature of the crystal field.
Which among the following are diamagnetic complexes?
\( [Ni(CN)_4]^{2-}, [Co(C_2O_4)_3]^{3-}, [Fe(CN)_6]^{4-}, [CoF_6]^{3-}, [Mn(CN)_6]^{3-} \)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \([Ni(CN)_4]^{2-}, [Co(C_2O_4)_3]^{3-}, [Fe(CN)_6]^{4-}\)
A complex is diamagnetic when it has no unpaired electrons. To decide this, find the oxidation state of the metal from the charges, write the d configuration of the ion from the atomic number, and then see whether the ligands pair the electrons. A strong field ligand such as \( \ce{CN-} \) forces pairing (low spin); a weak field ligand such as \( \ce{F-} \) does not (high spin).
\( \ce{[Ni(CN)4]^2-} \): x + 4(−1) = −2, so Ni is +2. Ni (Z = 28) is \( \mathrm{[Ar]\,3d^{8}\,4s^{2}} \); losing the two 4s electrons gives \( \ce{Ni^2+} \): \( \mathrm{3d^{8}} \). \( \ce{CN-} \) pairs all eight electrons into four 3d orbitals, and the free 3d orbital is used in \( \mathrm{dsp^{2}} \) (square planar) hybridisation: 0 unpaired, diamagnetic.
\( \ce{[Co(C2O4)3]^3-} \): x + 3(−2) = −3, so Co is +3. Co (Z = 27) is \( \mathrm{[Ar]\,3d^{7}\,4s^{2}} \); losing the two 4s electrons and one 3d electron gives \( \ce{Co^3+} \): \( \mathrm{3d^{6}} \). In this complex the six electrons are paired in the \( t_{2g} \) set (\( t_{2g}^{6} \)): 0 unpaired, diamagnetic.
\( \ce{[Fe(CN)6]^4-} \): x + 6(−1) = −4, so Fe is +2. Fe (Z = 26) is \( \mathrm{[Ar]\,3d^{6}\,4s^{2}} \); losing the two 4s electrons gives \( \ce{Fe^2+} \): \( \mathrm{3d^{6}} \). \( \ce{CN-} \) pairs them as \( t_{2g}^{6} \): 0 unpaired, diamagnetic.
\( \ce{[CoF6]^3-} \): Co is again +3, \( \mathrm{3d^{6}} \). \( \ce{F-} \) is a weak field ligand, so the electrons spread out as \( t_{2g}^{4}e_g^{2} \): 4 unpaired, paramagnetic.
\( \ce{[Mn(CN)6]^3-} \): x + 6(−1) = −3, so Mn is +3. Mn (Z = 25) is \( \mathrm{[Ar]\,3d^{5}\,4s^{2}} \); losing the two 4s electrons and one 3d electron gives \( \ce{Mn^3+} \): \( \mathrm{3d^{4}} \). Even with \( \ce{CN-} \), four electrons in three \( t_{2g} \) orbitals give \( t_{2g}^{4} \): 2 unpaired, paramagnetic.
So the diamagnetic complexes are \( \ce{[Ni(CN)4]^2-} \), \( \ce{[Co(C2O4)3]^3-} \) and \( \ce{[Fe(CN)6]^4-} \).
Shortcut: \( \ce{[CoF6]^3-} \) is a high spin complex of a weak field ligand, which rules out Options (1) and (3). An ion with \( \mathrm{d^{4}} \), like \( \ce{Mn^3+} \), can never have all its electrons paired in an octahedral field, which rules out Option (4).