IR Spectroscopy MCQ Quiz in తెలుగు - Objective Question with Answer for IR Spectroscopy - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Explanation:

• In the case of the facial isomer of [Mo(CO)3(phosphite)3], each phosphine ligand is opposite to a CO ligand.
• The increased electron density over the M atom for a carbonyl compound increases the M-CO π -back bonding which results in a lowering of the vC–O stretching frequency of the complex.
• As the π -accepting abilities of phosphine increases, the electron density over the M atom decreases which results in in increase in the v_C–O stretching frequency of the complex.
• The C-O stretching bands as given below;

Phosphines:  PF3 (A); PCl3 (B); P(Cl)Ph2 (C); PMe3 (D)

ν(CO), cm^-1 ; 2090(i); 2040(ii); 1977(iii); 1945(iv)

• Now, the order of π-accepting abilities among the given phosphine is:

PF₃ (A) > PCl₃ (B)> PClPh₂ (C) > PMe₃ (D)

i.e. A > B > C > D

So, the correct combination for v_C–O stretching is: A →  2090; C → 1977; B → 2040; D → 1945 

Conclusion:-

Hence, The correct combination of the phosphine and the stretching frequency is (A–i), (B–ii), (C–iii), (D–iv)

Concept:

→ It has been observed that the effect on k(force constant) when an atom is replaced by an isotope is negligible but it does have an effect on ν due to changes in its new mass. 

→ This is because the reduced mass has an effect on the rotational and vibrational behaviour. The frequency of radiation ν that will bring about this change is identical to the classical vibrational frequency of the bond ν_m and it can be expressed as

E_rotation= hν = ΔE = hν_m = \(\frac{h}{2\pi}\sqrt{\frac{k}{μ}}\)

where, μ is reduced mass which is equal to \(\frac{m_{1}m_{2}}{m_{1}+m_{2}}\)

→ By increasing reduced mas frequency decreases.

Explanation:

→ In the IR spectrum of Co(CO)₄H, the three most intense bands are observed at 2121, 2062, and 2043 cm-¹, which are assigned to the stretching vibrations of the Co-CO bonds. The band at 1934 cm^-1 is assigned to the stretching vibration of the C-H bond.

→ When Co(CO)4H is replaced by Co(CO)₄D, the vibrational frequency of the Co-D bond is expected to appear in the IR spectrum. The stretching frequency of the Co-D bond is expected to be lower than that of the Co-H bond, due to the difference in atomic masses between hydrogen and deuterium.

→ Since the mass of deuterium (D) is greater than that of hydrogen (H), the stretching frequency of the Co-D bond is expected to be lower than that of the Co-H bond. 

\(\rm v \propto \frac{1}{\sqrt{μ}} \Rightarrow \quad \frac{v_{C o-H}}{v_{C o-D}}=\sqrt{\frac{μ_{C o-D}}{μ_{C o-H}}} \Rightarrow \frac{1934}{v_{C o-D}}=\sqrt{2} \Rightarrow v_{C o-D}=\frac{1934}{\sqrt{2}}=1367.7 \approx 1396\)

Conclusion: The correct answer is option 2.

Concept:

→ Infrared spectroscopy exploits the fact that molecules absorb frequencies that are characteristic of their structure. These absorptions occur at resonant frequencies, i.e. the frequency of the absorbed radiation matches the vibrational frequency.

→ The energies are affected by the shape of the molecular potential energy surfaces, the masses of the atoms, and the associated vibronic coupling.

→ The resonant frequencies are also related to the strength of the bond and the mass of the atoms at either end of it. Thus, the frequency of the vibrations are associated with a particular normal mode of motion and a particular bond type.

Explanation:

→ In order for a vibrational mode in a sample to be "IR active", it must be associated with changes in the dipole moment. A permanent dipole is not necessary, as the rule requires only a change in dipole moment

→ A molecule can vibrate in many ways, and each way is called a vibrational mode. For molecules with N number of atoms, geometrically linear molecules have 3N – 5 degrees of vibrational modes, whereas nonlinear molecules have 3N – 6 degrees of vibrational modes (also called vibrational degrees of freedom).

→ SCN− and -NCS both are monodentate ligand which can bond to a central atom through either of two or more donor atoms. Such ligands are also termed as ambidentate. Depending upon their structure binding modes can be determined through IR spectroscopy.

Conclusion: The correct answer is option 2.

Explanation:

A - I: N-H bonds of R-NH₂ have respective approximate symmetric and asymmetric stretching frequencies of 3300 and 3400 cm^-1. This is because N-H bonds are known to exhibit strong and broad absorption in this region due to their strong dipole moment.

B - III: N-O bonds of R-NO₂ have respective approximate symmetric and asymmetric stretching frequencies of 1350 and 1550 cm^-1. This is because N-O bonds are known to exhibit medium to strong absorption in this region due to their strong dipole moment.

C - II: C=O bonds of anhydride have respective approximate symmetric and asymmetric stretching frequencies of 1790 and 1810 cm^-1. This is because C=O bonds are known to exhibit strong absorption in this region due to their strong dipole moment.

Therefore, the correct matching between the functional groups in List I and their respective approximate symmetric and asymmetric stretching frequencies in List II is:

A - II: 3300 and 3400 cm^-1

B - III: 1350 and 1550 cm^-1

C - I: 1790 and 1810 cm^-1

Conclusion: The correct answer is option 4.

Concept:

• The number of CO bands in the IR spectrum can be helpful in determining the isomers.
• Group theory is usually applied to predict the number of CO-stretching bands
• A popular application of it is the distinction of isomers of metal carbonyl complexes.

Explanation:

Set (i): trigonal bipyramidal isomers

(A) axial‐Fe(CO)4L belongs to the point group of C_3V, it shows 3 CO bands in IR spectroscopy.

(B) equitorial‐Fe(CO)4L belongs to C_2v point group. So, it gives 4 CO bands in IR.

Set (ii): Octahedral isomers

(C) fac‐Mo(CO)3L₃ gives 2 CO bands in IR.

(D) mer‐Mo(CO)3L₃ gives 3 CO bands in IR.

Conclusion:

number of CO bands in IR for given isomers is:

(A) 3

(B) 4

(C) 2

(D) 3

Explanation:

   Its frequency is 1748 ^cm-1.

    Its frequency is 1740 cm-1. But due to the double bond, the frequency is less due to 20 so,  Its frequency is about 1720 cm-1.

    Its frequency is 1715 ^cm-1.  Due to resonance frequency is less due to 30, So,  Its frequency is about 1685 ^cm-1. So less than 1700 ^cm-1.

 Its frequency is about 1735 ^cm-1.

    due aromatic ring and ester so their frequency is less than 1700 cm-1.

  Its frequency is about 1750 cm-1.

  Its frequency is >1700 cm-1.

So there are 3 structures that are below 1700 ^cm-1.