Test on Inorganic Chemistry: Spectra and Complexes

Question 1: A Tanabe-Sugano diagram is provided in the study materials for the d1 electron configuration.

A. Ano

B. Ne

Explanation: The study materials provide Tanabe-Sugano diagrams for d2, d3, d4, d5, d6, d7, and d8 configurations. No Tanabe-Sugano diagram for d1 is presented in the provided materials.

Question 2: According to the study materials, which statement(s) accurately describe(s) the process of electron excitation and relaxation in quantized energy states?

A. An electron can be excited from a ground state to an excited state by absorbing energy.

B. From an excited state, an electron relaxes back to its original ground state, releasing energy as photons.

C. Energy states are non-quantized, allowing electrons to occupy any energy level.

D. Electrons in an excited state will spontaneously move to a lower energy state without releasing any energy.

Explanation: The study materials state that 'Energy states are quantized; Electron can be excited from ground state to excited state by absorbing energy; From excited state, it will relax back to original ground state; this process releases energy as photons of a specific wavelength'. Therefore, options 0 and 1 are correct, while options 2 and 3 contradict the provided information.

Question 3: The spectrochemical series indicates that carbon monoxide (CO) is a weaker field ligand than fluoride (F-), resulting in a smaller d-orbital splitting energy.

A. Ano

B. Ne

Explanation: According to the ligand spectrochemical series in the study materials, CO is positioned to the far right (CO), indicating it is a strong field ligand, while F- is positioned earlier (F-), indicating it is a weaker field ligand. Therefore, CO would result in a larger d-orbital splitting energy, not smaller than F-.

Question 4: Which statement correctly describes the relationship between metal-carbonyl bonding and its infrared stretching frequency?

A. A stronger metal-carbon bond leads to a higher C-O stretching frequency.

B. Increased π-backbonding from the metal to the carbonyl ligand results in a lower C-O stretching frequency.

C. Introducing a stronger π-acceptor ligand into a complex will typically decrease the C-O stretching frequency.

D. A negative charge on the metal-carbonyl complex tends to increase the C-O stretching frequency.

Explanation: According to the study materials, the stronger the M-C bond, which is due to stronger π-backbonding from the metal, the weaker the C-O bond. A weaker C-O bond means a decreasing bond order in C-O, which in turn decreases the IR frequency (wavenumber). Therefore, increased π-backbonding leads to a lower C-O stretching frequency. Introducing another π-acceptor ligand introduces competition for metal electron density, leading to less M-C backbonding and a higher wavenumber for CO. A negative charge on the molecule causes metal orbitals to expand, leading to better overlap and a stronger M-C bond, which weakens the C-O bond and thus lowers its IR frequency.

Question 5: In the MO approach to ligand classification, for a σ-donor, π-acceptor ligand, the metal's t2g orbitals are described as bonding orbitals.

A. Ano

B. Ne

Explanation: The study materials state under 'Ligand classification – MO approach' for 'σ -donor, π -acceptor' ligands that the 'Frontier orbitals: ... t2g are bonding orbitals (t2g)'.