(a) What type of isomerism is shown by the complex [Co(NH)(SCN)]?
(b) Why is [NiCl] paramagnetic while [Ni(CN)] is diamagnetic? (Atomic number of Ni = 28)
(c) Why are low-spin tetrahedral complexes rarely observed?
(b) Why is [NiCl] paramagnetic while [Ni(CN)] is diamagnetic? (Atomic number of Ni = 28)
(c) Why are low-spin tetrahedral complexes rarely observed?
Solution
(a) SCN is ambidentate (S- or N-bonded) → the complex shows \textbf{linkage isomerism}.
(b) Ni = 3d.
[NiCl]: Cl is weak-field — no pairing, sp tetrahedral, 2 unpaired electrons → paramagnetic.
[Ni(CN)]: CN is strong-field — the electrons pair, dsp square planar, 0 unpaired → diamagnetic.
(c) In a tetrahedral field the splitting is small, Δ = (4/9)Δ; it never exceeds the pairing energy P, so electrons always occupy the upper set rather than pair — low-spin tetrahedral complexes are therefore rarely observed.
(b) Ni = 3d.
[NiCl]: Cl is weak-field — no pairing, sp tetrahedral, 2 unpaired electrons → paramagnetic.
[Ni(CN)]: CN is strong-field — the electrons pair, dsp square planar, 0 unpaired → diamagnetic.
(c) In a tetrahedral field the splitting is small, Δ = (4/9)Δ; it never exceeds the pairing energy P, so electrons always occupy the upper set rather than pair — low-spin tetrahedral complexes are therefore rarely observed.
CBSE Class 12 Board · 2017 · Subjective