Characterisation of Inorganic Compounds MCQ Quiz in தமிழ் - Objective Question with Answer for Characterisation of Inorganic Compounds - இலவச PDF ஐப் பதிவிறக்கவும்

Last updated on Apr 1, 2025

பெறு Characterisation of Inorganic Compounds பதில்கள் மற்றும் விரிவான தீர்வுகளுடன் கூடிய பல தேர்வு கேள்விகள் (MCQ வினாடிவினா). இவற்றை இலவசமாகப் பதிவிறக்கவும் Characterisation of Inorganic Compounds MCQ வினாடி வினா Pdf மற்றும் வங்கி, SSC, ரயில்வே, UPSC, மாநில PSC போன்ற உங்களின் வரவிருக்கும் தேர்வுகளுக்குத் தயாராகுங்கள்.

Latest Characterisation of Inorganic Compounds MCQ Objective Questions

Top Characterisation of Inorganic Compounds MCQ Objective Questions

Characterisation of Inorganic Compounds Question 1:

The compound which gives a fragment at 𝑚/𝑧 = 124[M+H]+ is

  1. F1 Teaching Savita 12-1-24 D10
  2. F1 Teaching Savita 12-1-24 D11
  3. F1 Teaching Savita 12-1-24 D12
  4. F1 Teaching Savita 12-1-24 D13

Answer (Detailed Solution Below)

Option 2 : F1 Teaching Savita 12-1-24 D11

Characterisation of Inorganic Compounds Question 1 Detailed Solution

Concept:-

  • Mass spectrometry is an analytical tool useful for measuring the mass-to-charge ratio (m/z) of one or more molecules present in a sample.
  • A mass spectrum is a histogram plot of intensity vs. mass-to-charge ratio (m/z) in a chemical sample, usually acquired using an instrument called a mass spectrometer.
  • In a typical MS procedure, a sample, which may be solid, liquid, or gaseous, is ionized, for example by bombarding it with a beam of electrons. This may cause some of the sample's molecules to break up into positively charged fragments or simply become positively charged without fragmenting.
  • This process is known as electron ionization (EI, formerly known as electron impact ionization and electron bombardment ionization).
  • These ions (fragments) are then separated according to their mass-to-charge ratio, for example by accelerating them and subjecting them to an electric or magnetic field: ions of the same mass-to-charge ratio will undergo the same amount of deflection.​
  • The mass value of any ion is its true mass, i.e., the sum of the masses of each atom (most common isotope) in that single ion (accurate), and not its molecular weight calculated from chemical atomic weights (integral atomic mass, weighted averages of weights of all of the isotopes).
  • The precise mass of some elements are given below on the C-12 scale

qImage641c229912e2397d6f0d0c01

Explanation:-

Option B

First, the electron bombardment which lead to the formation of molecular ion (O+)
Followed by the homolytic cleavage 

F1 Teaching Savita 12-1-24 D14

Conclusion:-

So [M+H]+ =124 in option 2

Characterisation of Inorganic Compounds Question 2:

For a transition metal M, the correct order of 13C NMR spectral shift [in ppm relative to [Si(CH3)4] for the moieties M-CH3, M-CO and M-C6H5, is

  1. M-CH3 < M-C6H5 < M-CO
  2. M-CO < M-CH3 < M-C6H5
  3. M-C6H5 < M-CH3 < M-CO
  4. M-CO < M-C6H5 < M-CH3

Answer (Detailed Solution Below)

Option 1 : M-CH3 < M-C6H5 < M-CO

Characterisation of Inorganic Compounds Question 2 Detailed Solution

Explanation:-

13C NMR:

  • The Carbon-13 nuclear magnetic resonance or 13C NMR spectroscopy is the application of nuclear magnetic resonance spectroscopy to carbon (C). 13C NMR spectroscopy helps to identify the nature or environment of C atoms in organic molecules just as 1H NMR spectroscopy helps to detect the H atoms.
  • The nature or environment of C atoms in organic molecules can be identified by chemical shift (δ) values.
  • The chemical shift values of some common functional groups are as follows:
Functional Group Chemical Shift (\(\delta\))
RCH3 10 to 30 ppm
RCH=CHR 100 to 140 ppm
Amide carbonyl 150 to 180 ppm
Aldehyde or Ketone 180 to 220 ppm
  • From the above table, we can conclude that, the chemical shift value will be highest for M-CO (180 to 220 ppm), followed by M-C6H(100 to 140 ppm) and M-CH3 (10 to 30 ppm).
  • Thus, the correct order of 13C NMR spectral shift [in ppm relative to [Si(CH3)4] for the moieties M-CH3, M-CO and M-C6H5, is M-CH3 < M-C6H5 < M-CO

Conclusion:-

Hence, for a transition metal M, the correct order of 13C NMR spectral shift [in ppm relative to [Si(CH3)4] for the moieties M-CH3, M-CO, and M-C6H5, isM-CH3 < M-C6H5 < M-CO

Characterisation of Inorganic Compounds Question 3:

(R3Ge)2 on photolysis gives a radical which shows ESR spectrum. The ESR signals carrying the signature of 73Ge (I = 9/2) are in terms of

  1. Nine lines 
  2. Ten lines
  3. Two lines
  4. One line 

Answer (Detailed Solution Below)

Option 2 : Ten lines

Characterisation of Inorganic Compounds Question 3 Detailed Solution

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.

F1VinantiTeaching10.04.23D3
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.

Characterisation of Inorganic Compounds Question 4:

The compound which shows IR frequencies at both 3314 and 2126 cm-1 is

  1. CH3(CH2)4CH2SH
  2. CH3(CH2)4CH2C≡N
  3. CH3(CH2)4CH2C≡C-H
  4. CH3(CH2)2C≡C(CH2)2CH3

Answer (Detailed Solution Below)

Option 3 : CH3(CH2)4CH2C≡C-H

Characterisation of Inorganic Compounds Question 4 Detailed Solution

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:

F1 Priya Teaching 28-5-24 D2

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 

qImage1988

Bond Functional Group Frequency in cm-1 (Instensity*)
C-H Alkane 3000-2850 (s)
  -CH3 1450 and 1375 (m)
  -CH2 1465 (m)
  Alkene 3100-3000 (m), 1000-650 (s)
  Aromatic 3100-3050 (s), 900-690 (s)
  Alkyne (terminal) 3300 (s)
  Aldehyde 2850 and 2750 (w)
C=C Alkene 1630 (w-m)
  Aromatic 1600-1475 (w-m)
C-C Alkyne 2150-2100 (w-m)
C=O Aldehyde 1740-1720 (s)
  Ketone 1725-1705 (s)
  Carboxylic Acid 1730-1700 (s)
  Ester 1750-1730 (s)
  Amide 1670-1640 (s)
  Anhydride 1810 and 1760 (s)
C-O Alcohol, Either, Ester, Carboxylic Acid, Anhydride 1300-1000 (s)
O-H Alcohol, Phenol  
  Free 3650-3600 (m)
  Hydrogen-bonded 3500-3200 (m)
  Carboxylic Acids 3300-2400 (m)
N-H Primary and Secondary Amide and Amide 3550-3060 (m-s); 1640-1550 (m-s)
C-N Amine 1250-1000 (m-s)
C=N Imine and Oxime 1690-1640 (w-s)
C=N Nitrile 2260-2240 (m)
N=O Nitro (R-NO2) 1550 and 1350 (s)
C-Cl Chloride 800-600 (s)

 

qImage1989

Characterisation of Inorganic Compounds Question 5:

For an octahedral Cu2+ complex depicting axial EPR spectrum (g|| > g⊥), the geometry of Cu2+ and the orbital containing the unpaired electron are, respectively

  1. Tetragonally elongated, \(\rm d_x^2 − y^2\)
  2. Tetragonally compressed, \(\rm d_z^2\)
  3. Tetragonally elongated, \(\rm d_z^2\)
  4. Tetragonally compressed, \(\rm d_x^2 − y^2\)

Answer (Detailed Solution Below)

Option 1 : Tetragonally elongated, \(\rm d_x^2 − y^2\)

Characterisation of Inorganic Compounds Question 5 Detailed Solution

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 dx2-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 dxy, dxz, and dyz orbitals are closer to the ligands, while the dx2-y2 and dz2 orbitals are farther away.

Screenshot 2024-02-27 182145

→ This results in a splitting of the d orbitals into two energy levels: a lower energy level consisting of the dxy, dxz, and dyz orbitals, and a higher energy level consisting of the dx2-y2 and dz2 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 dxy 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 gvalue, indicating that the electron has a greater magnetic moment along the axis of the complex.

Conclusion:

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

Characterisation of Inorganic Compounds Question 6:

The correct order of carbonyl stretching frequency for the given compounds is

F2 Madhuri Teaching 27.03.2023 D157

  1. A > B > C
  2. B > C > A
  3. C > A > B
  4. B > A > C

Answer (Detailed Solution Below)

Option 4 : B > A > C

Characterisation of Inorganic Compounds Question 6 Detailed Solution

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.

Characterisation of Inorganic Compounds Question 7:

The correct order of C=O stretching frequency in IR spectrum for the following compounds is

F1 Madhuri Teaching 06.02.2023 D54F1 Madhuri Teaching 06.02.2023 D55F1 Madhuri Teaching 06.02.2023 D56

  1. A > C > B
  2. B > C > A
  3. C > B > A
  4. B > A > C

Answer (Detailed Solution Below)

Option 3 : C > B > A

Characterisation of Inorganic Compounds Question 7 Detailed Solution

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

F1 Madhuri Teaching 06.02.2023 D57F1 Madhuri Teaching 06.02.2023 D58F1 Madhuri Teaching 06.02.2023 D59

Hence, the correct option is C. 

Characterisation of Inorganic Compounds Question 8:

In a 400 MHz 1H NMR spectrometer, a proton resonates at 1560 Hz higher than that of tetramethylsilane. The chemical shift value of this proton is _______ ppm. (Round off to one decimal place)

(Chemical shift of tetramethylsilane is fixed at zero ppm)

Answer (Detailed Solution Below) 3.9

Characterisation of Inorganic Compounds Question 8 Detailed Solution

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. 

Characterisation of Inorganic Compounds Question 9:

The selection rule in mossbauer spectroscopy is

  1. ΔMI = 0
  2. ΔMI = ±1
  3. ΔMI = 0, ±1
  4. ΔMI = 0, ±1, ±2,...

Answer (Detailed Solution Below)

Option 3 : ΔMI = 0, ±1

Characterisation of Inorganic Compounds Question 9 Detailed Solution

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 (ΔMI) during the absorption or emission process.
  • The selection rule for magnetic dipole transitions in Mössbauer spectroscopy is given as:
    • ΔMI = 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 ΔMI = 0, no change occurs in the orientation of the magnetic moment of the nucleus.
  • If ΔMI = ±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 ΔMI = 0, ±1, which means the magnetic quantum number can change by 0 or ±1 during a transition. This is correctly given by option 3.

Characterisation of Inorganic Compounds Question 10:

The EPR spectrum of p-benzosemiquinone radical anion consists of :

  1. a quintet with intensity 1 : 2 : 3 : 2 : 1
  2. a quintet with intensity 1 : 1 : 1 : 1 : 1
  3. a quartet with intensity 1 : 3 : 3 : 1
  4. a quintet with intensity 1 : 4 : 6 : 4 : 1

Answer (Detailed Solution Below)

Option 4 : a quintet with intensity 1 : 4 : 6 : 4 : 1

Characterisation of Inorganic Compounds Question 10 Detailed Solution

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.

    • F1 Sourav NET 09 01 25 ANKIT  125

  • 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.

Get Free Access Now
Hot Links: teen patti boss teen patti real cash teen patti comfun card online