Valence Bond Theory and CFT Bonding-Valence bond approach and basic ideas of Crystal field theory, colour and magnetic properties;: NEET MCQs with Answers

Explanation: Each hybridisation scheme fixes a characteristic arrangement of the hybrid orbitals in space. A → III: four sp³ hybrid orbitals point to the corners of a tetrahedron, e.g. [NiCl₄]²⁻ and Ni(CO)₄. B → IV: four dsp² hybrid orbitals lie in one plane at 90° to each other, giving square planar, e.g. [Ni(CN)₄]²⁻. C → I: five sp³d hybrid orbitals give a trigonal bipyramid, e.g. Fe(CO)₅. D → II: six d²sp³ hybrid orbitals point to the corners of an octahedron, e.g. [Co(NH₃)₆]³⁺. Hence A-III, B-IV, C-I, D-II.

Explanation: Cr²⁺ is 3d⁴, so n = 4 and μ = √[4(4 + 2)] = √24 = 4.90 BM.

Explanation: The spectrochemical series runs SCN⁻ (S-bonded) < F⁻ < C₂O₄²⁻ < H₂O < NH₃ < en < CN⁻ < CO. Picking out the four ligands gives SCN⁻ < F⁻ < C₂O₄²⁻ < CN⁻ in order of increasing field strength.

Explanation: Iron is in the +2 state (3d⁶) and cyanide is a strong-field ligand, so ∆₀ > P. All six electrons therefore occupy the lower t₂g set, giving the low-spin diamagnetic configuration t₂g⁶ e₇⁰.

Explanation: Carbon monoxide is a strong-field ligand, so in Ni(0) the 4s electrons shift into 3d giving 3d¹⁰ 4s⁰. With no unpaired electrons the complex is diamagnetic, and sp³ hybridisation of the empty 4s and 4p orbitals gives a tetrahedron.

Explanation: Mn³⁺ is 3d⁴ and the strong-field CN⁻ forces the low-spin t₂g⁴ arrangement, freeing two inner 3d orbitals. Hybridisation is therefore d²sp³, giving an octahedral inner-orbital complex.

Explanation: Ligand field strength increases H₂O < NH₃ < en, so ∆₀ increases in that order. Since λ is inversely proportional to ∆₀, the absorbed wavelength increases in the reverse order: en < NH₃ < H₂O.

Explanation: The greater the negative charge on the metal, the more electron density is back-donated into the π* orbitals of CO, weakening and lengthening the C–O bond. [Fe(CO)₄]²⁻ carries the largest negative charge and therefore has the longest C–O bond.

Explanation: Co³⁺ is d⁶ and cyanide is a strong-field ligand, so ∆₀ > P and all six electrons pair up in the t₂g set. The complex is therefore low spin and diamagnetic.

Explanation: Fe³⁺ is d⁵ and with the weak-field water ligand it is high spin, t₂g³ e₇². CFSE = 3(–0.4∆₀) + 2(+0.6∆₀) = –1.2∆₀ + 1.2∆₀ = 0.

Explanation: Six hybrid orbitals point to the corners of an octahedron, so a six-coordinate complex with sp³d² or d²sp³ hybridisation is octahedral. As per NCERT, [Co(NH₃)₆]³⁺ actually uses d²sp³ (it is an inner-orbital, diamagnetic complex), but the shape is the same.

Explanation: In Ni(CO)₄ nickel is Ni(0), and strong-field CO forces the 4s electrons into 3d to give 3d¹⁰ with every electron paired. The sp³ complex is therefore tetrahedral and diamagnetic, so describing it as paramagnetic is incorrect.

Explanation: Paramagnetism requires unpaired electrons. [Cr(NH₃)₆]³⁺: Cr³⁺ = [Ar]3d³. Even with a strong field the three electrons occupy the three t₂g orbitals singly, so three unpaired electrons remain → paramagnetic ✔ [Cr(CO)₆]: Cr(0) = 3d⁵4s¹; CO is a very strong-field ligand and all six electrons pair up as t₂g⁶ → diamagnetic [Fe(CO)₅]: Fe(0) with strong-field CO → all electrons paired → diamagnetic [Fe(CN)₆]⁴⁻: Fe²⁺ = 3d⁶ with strong-field CN⁻ → low spin t₂g⁶ → diamagnetic Only the chromium ammine complex is paramagnetic, and note that a d³ ion is unavoidably paramagnetic whatever the ligand.

Explanation: The cyanide ion is classed as a pseudohalide — an ion that resembles the halides in charge and in many of its reactions, but is polyatomic. Pseudohalides such as CN⁻, SCN⁻ and N₃⁻ are strong coordinating ligands because they can bond to a metal in two complementary ways at once: a σ-bond donated from the ligand to the metal, and a π-bond accepted back from the filled metal d-orbitals. This combination of σ-donation and π-acceptance produces an unusually large crystal-field splitting, which is why CN⁻ sits near the top of the spectrochemical series and forms low-spin complexes. Option (4) is plainly wrong — strong-field ligands give low-spin, not high-spin, complexes — and a negative charge alone does not make a ligand strong-field, as the weak-field halides show.

Explanation: Manganese is +2 here, since water is neutral and the complex ion carries a charge of +2. Mn (Z = 25) = [Ar]3d⁵4s², so Mn²⁺ = [Ar]3d⁵. Water is specified as a weak-field ligand, which means Δ₀ is smaller than the pairing energy. Under those conditions the electrons occupy the orbitals singly as far as possible before pairing, giving the high-spin arrangement t₂g³e₉². All five electrons therefore remain unpaired, n = 5, and μ = √35 = 5.92 BM. This is the maximum spin-only moment attainable in the 3d series.