Characterisation of Inorganic Compounds MCQ Quiz in मल्याळम - Objective Question with Answer for Characterisation of Inorganic Compounds - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:-

• EPR or EPR spectroscopy is used to study a paramagnetic species with one or more unpaired electrons. While diamagnetic species are always EPR silent.
• Paramagnetic species such as free radicals, diradicals, and metal complexes containing paramagnetic metal centers are always EPR active and exhibit an EPR spectrum.
• For a paramagnetic metal ion with one unpaired electron, the total spin quantum number is

• S=. There are two possible spin states with Ms= and Ms=

• By applying a magnetic field Bo, the interaction between the unpaired electrons and the magnetic field leads to the splitting of the energy levels.
• Therefore, by supplying appropriate microwave radiation to the sample, electron spin transitions between the two energy states occur. The system is then in resonance, and the recording of these transitions represents the EPR spectrum.

​Explanation:-

• On photolysis (R3Ge)2 will undergo homolytic cleavage to give two R3Ge° radicals.

The one unpaired electron (or radical) is responsible for the paramagnetism and  EPR spectrum of the compound.

• The number of lines observed in the EPR spectrum is given by

2nI+1,

where n is the number of equivalent nuclei with the magnetic spin number I.

• For each R3Ge° radical, the value of I for 73Ge is 9/2.
• For R3Ge°, the number of lines observed in the EPR spectrum for the 73Ge nucleus (73Ge, 100% I= 9/2) for n=1 is,

​= 2 × 1 ×  + 1

= 10 lines.

Conclusion:-

• Hence, the ESR signals carrying the signature of 73Ge ( I = 9/2) are in terms of 10 lines.​

Concept:

→ IR Spectroscopy measures the vibrations of atoms, and based on this it is possible to determine the functional groups.

→ Generally, stronger bonds and light atoms will vibrate at a high stretching frequency (wavenumber).

→ The greater the masses of attached atoms, the lower the IR frequency at which the bond will absorb.

Explanation: 

In CH3(CH2)4CH2C≡C-H stretching frequencies are:

Conclusion: Out of four groups IR frequencies at both 3314 and 2126 cm-1 is shown by CH3(CH2)4CH2C≡C-H, hence the correct answer is option 3.

Additional Information 

Concept:

Mössbauer spectroscopy:

This technique is used in solid-state physics and chemistry to study the characteristics of atomic nuclei in a material. It is a type of gamma-ray spectroscopy that offers comprehensive details on the electrical and magnetic environment around particular atomic nuclei.

A sample, radioactive source like iron-57 is often exposed to a source of gamma rays in Mössbauer spectroscopy. The sample's nuclei absorb the gamma rays, which causes the nuclei to get excited to higher energy states. In mineralogy, this method is frequently used to determine the valence state of iron whether it's Fe(0), Fe(II) or (III).

Explanation:

The magnetic dipole selection rule for Mössbauer spectroscopy is,

\(\Delta m_{I}=0,\pm1\)  ,where m_I= magnetic quantum no.

for quadruple splitting, \(I> \frac{1}{2} (G.S./E.S.)\) , where I= nuclear spin

\(57Fe=(I_{G.S}=\frac{1}{2}) \:and(I_{E.S}=\frac{3}{2})\)

\(Fe(spin G.S.)=\frac{1}{2} \), degeneracy for G.S=\((2nI+1)\)

=\((2\times1\times\frac{1}{2})+1=2\)

\(Fe(spin E.S.)=\frac{3}{2} \), degeneracy for E.S=\((2\times1\times\frac{3}{2})+1=4\)

Mössbauer lines in ⁵⁷Fe in the presence of static magnetic field=\(2+4=6\)

Conclusion:

Hence the correct answer is Six.

Concept:

→ In a Cu²⁺ ion, there are nine electrons in the d orbital, with two electrons occupying each of the d orbitals and one electron occupying the d_x²-y² orbital. When Cu²⁺ forms a complex, it can coordinate with various ligands to form a complex with a specific geometry.

→ In an octahedral complex, Cu²⁺ is surrounded by six ligands arranged at the corners of an octahedron. These ligands can be either of two types: axial or equatorial. The axial ligands lie along the axis of the octahedron, while the equatorial ligands lie in the plane perpendicular to the axis.

→ When the octahedral Cu²⁺ complex is tetragonally elongated, it means that the axial ligands are further away from the Cu²⁺ ion than the equatorial ligands. This results in a distortion of the octahedral geometry, which affects the electronic structure of the complex.

Explanation:

→ In a tetragonally elongated octahedral complex, the d_xy, d_xz, and d_yz orbitals are closer to the ligands, while the d_x²-y² and d_z² orbitals are farther away.

→ This results in a splitting of the d orbitals into two energy levels: a lower energy level consisting of the d_xy, d_xz, and d_yz orbitals, and a higher energy level consisting of the d_x²-y² and d_z² orbitals.

→ The unpaired electron in a Cu²⁺ complex contributes to its magnetic properties and can be studied using Electron Paramagnetic Resonance (EPR) spectroscopy.

→ In the case of a Cu²⁺ complex with an axial EPR spectrum (g_|| > g_⊥), it suggests that the unpaired electron is in the d_xy orbital, which is oriented perpendicular to the axial ligands and parallel to the magnetic field. This results in a larger g_|| value compared to the g_⊥ value, indicating that the electron has a greater magnetic moment along the axis of the complex.

Conclusion:

The correct answer is Tetragonally elongated, \(\rm d_x^2 − y^2\).

Concept:

Infrared Spectroscopy:

• Infrared spectroscopy (IR) is an absorption method widely used in both qualitative and quantitative analyses. IR spectroscopy is the measurement of the interaction of infrared radiation with the matter by absorption, emission, or reflection.  IR spectroscopy is used for the identification of organic compounds. 
• An IR spectrum is a graph plotted with the infrared light absorbed on the Y-axis against and frequency or wavelength on the X-axis.
• The value of the IR stretching frequency increases with increases in bond strength and decreases with reduced mass of the system.
• The IR stretching frequency of a bond can be expressed as ;

 \(\vartheta = {1 \over {2\pi }}\sqrt {{k \over \mu }} \), where k is the force constant and µ is reduced mass. 

Explanation:

• The carbonyl stretching frequency of a molecule increases with increases in the %S character in the carbonyl bond.
• In compounds A, B, and C the carbon atom of the carbonyl group (C=O) is sp², sp, and sp² hybridized respectively. The %S character in the carbon atom in compounds A, B, and C are 33%, 50%, and 33% respectively.
• Compound B has the highest carbonyl stretching frequency as the %S character is highest.
• Because of more strain in compound A, there is more %S character present in compound A. Thus compound A has more carbonyl stretching frequency than compound C.
• The correct order of carbonyl stretching frequency for the given compounds is B > A > C.

​Conclusion:

Hence, the correct order of carbonyl stretching frequency for the given compounds is  B > A > C.

Concept: -

 IR Spectroscopy

• Also known as Infrared Spectroscopy.
• It refers to the analysis of the interaction of a molecule with infrared light.
• The major use of infrared spectroscopy is to determine the functional groups of molecules, relevant to both organic and inorganic chemistry.
• The IR spectroscopy theory utilizes the concept that molecules tend to absorb specific frequencies of light that are characteristic of the corresponding structure of the molecules. The energies are reliant on the shape of the molecular surfaces, the associated vibronic coupling, and the mass corresponding to the atoms.
• For simple aldehydes and ketones, the stretching vibration of the carbonyl group has a strong infrared absorption between 1710 and 1740 cm-1.

We know by Hooke's law

\(\nu = \frac{1}{2\pi}\sqrt{\frac{k}{m_1m_2/(m_1 + m_2)}} = \frac{1}{2\pi}\sqrt{\frac{k}{μ}}\\ \\ and, \bar{\nu} = c/v\\ \\ \therefore \bar{\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{m_1m_2/(m_1 + m_2)}} = \frac{1}{2\pi c}\sqrt{\frac{k}{μ}}\\ \)

Where m₁ and m₂ are the masses of the atom, μ is reduced mass, and k is force constant i.e. measure of bond strength.

Thus, we can say: -

The vibrational frequency of a bond is directly proportional to bond strength and inversely to the reduced mass.

So, any factor which will increase the bond strength with an increase in the C=O stretching frequency.

• Hence, any electronegative group present in the vicinity of the carbonyl group will increase its C=O stretching frequency.

​Explanation: -

Field effect: -

• It has been found that two groups often influence each other's vibrational frequencies by a through-space interaction.
• These interactions can be electrostatic as well as in nature.
• The most common example of field effect is the interaction between carbonyl and halogen groups.
• For example, in α-chloroketone the C=O stretching frequency is high when the chloro group is equatorial than the axial form.
• The reason behind this is the electronic repulsive interaction between the lone pair electrons of chlorine with the oxygen of the carbonyl group, which changes the hybridization of oxygen and hence increases the C=O stretching frequency.

Conclusion:

Thus, by taking the electronegativity of -Cl and the field effect, the order of frequency will be

​

Hence, the correct option is C. 

Explanation:-

The chemical shift of Nuclear Magnetic Resonance Spectroscopy is the relative resonant frequency of the nucleus with respect to the magnetic field. 

Calculation:-

δ(ppm) = ѴHz / ѴMHz , δ(ppm) 

δ(ppm) = 1560Hz / 400 MHz 

δ(ppm) = 3.9 ppm. 

Concept:

• Mössbauer spectroscopy is a technique that relies on the Mössbauer effect, which is the recoil-free, resonant absorption and emission of gamma rays by atomic nuclei bound in a solid.
• It provides detailed information on the hyperfine interactions within a material, such as magnetic and electric quadrupole interactions, by observing the splitting of nuclear energy levels.
• Mössbauer spectroscopy is highly sensitive and can detect small changes in the oxidation state, electronic environment, and magnetic properties of the nuclei in a sample.
• It is named after Rudolf Mössbauer, who discovered the effect in 1958 and was awarded the Nobel Prize in Physics in 1961.
• The technique is widely used in various fields, including solid-state physics, chemistry, materials science, geology, and biology, to study both crystalline and amorphous materials.
• The selection rule in Mössbauer spectroscopy refers to the allowed changes in the magnetic quantum number (ΔM_I) during the absorption or emission process.
• The selection rule for magnetic dipole transitions in Mössbauer spectroscopy is given as:
  • ΔM_I = 0, ±1
  • This means that the magnetic quantum number MI can change by 0 or ±1 between the initial and final states of the nucleus during the transition.

Explanation:

• In Mössbauer spectroscopy, the transitions are typically between different nuclear energy levels that are split by hyperfine interactions such as magnetic or electric quadrupole interactions.
• When a nucleus absorbs or emits a gamma-ray photon, the magnetic quantum number (MI) of the nucleus changes according to the selection rule ΔMI = 0, ±1.
• This selection rule results from the conservation of angular momentum during the transition.
• If ΔM_I = 0, no change occurs in the orientation of the magnetic moment of the nucleus.
• If ΔM_I = ±1, there is a change in the orientation by one quantum unit, either increasing or decreasing the magnetic quantum number by 1.

Conclusion:

The correct selection rule in Mössbauer spectroscopy is ΔM_I = 0, ±1, which means the magnetic quantum number can change by 0 or ±1 during a transition. This is correctly given by option 3.

Concept:

The Electron Paramagnetic Resonance (EPR) spectrum provides insights into the electronic structure of radicals through hyperfine splitting. Key points related to the EPR spectrum and intensity patterns are:

• EPR spectroscopy detects unpaired electrons and is useful for studying radicals and transition metal complexes.
• The hyperfine splitting in EPR spectra occurs due to interactions between unpaired electrons and nearby nuclear spins, leading to multiple lines.
• The number of lines observed in the spectrum can be predicted using the formula (2nI + 1), where n is the number of equivalent nuclei and I is the nuclear spin.

Explanation: 

• EPR Spectrum of p-Benzoquinone Radical Anion: The p-benzosemiquinone radical anion has an unpaired electron that interacts with equivalent protons in the molecule.

  • ​

• This interaction leads to a hyperfine splitting pattern in the EPR spectrum:

  • ​Number of prtons = 4

  • Number of lines observed = 2nI + 1 = 2× 4× 1/2 + 1 = 5

• Intensity pattern for a quintet is 1 : 4 : 6 : 4 : 1.

Conclusion:

The correct answer is a quintet with intensity 1 : 4 : 6 : 4 : 1, as this matches the EPR spectrum pattern for the p-benzosemiquinone radical anion.

The correct answer is 

Concept:-

Chemical Shift (δ): The chemical shift is a measure of the resonance frequency of a nucleus relative to a standard in a magnetic field. It is measured in parts per million (ppm) and provides information about the electronic environment of the nucleus.
Tetramethylsilane (TMS): TMS is used as an internal standard in 1H NMR spectroscopy. Its chemical shift is defined as 0 ppm. The frequency difference between the sample and TMS is used to calculate the chemical shift.
Instrument Frequency: The instrument frequency refers to the operating frequency of the NMR spectrometer, which is typically given in MHz (megahertz). For this problem, the instrument frequency is 400 MHz.
Frequency Difference: The frequency difference (ν _sample −ν _TMS ) is the difference in resonance frequency between the sample proton and the TMS proton, measured in Hz.

Explanation:-

To determine the chemical shift value in ppm (parts per million) for a proton in ^1H NMR, we need to use the provided information and apply the formula for chemical shift:

\(\delta ({ppm}) = \frac{\nu_{{sample}} - \nu_{{TMS}}}{{Instrument Frequency (in MHz)}}\)

Here:
\(- \nu_{{sample}} - \nu_{{TMS}} = 1560 { Hz} \\ - Instrument frequency = 400 { MHz}\)

Using the formula:

\(\delta ({ppm}) = \frac{1560 { Hz}}{400 { MHz}} = \frac{1560}{400000000} \times 10^6 = \frac{1560}{400} = 3.9 { ppm}\)

Therefore, the chemical shift value of this proton is approximately 3.9 ppm .

Conclusion:-

So,  the chemical shift value of this proton is approximately 4.0 ppm