Chemistry GS TIFR- Tata Institute of Fundamental Research_ Questions 1-5_ Solved
1. The structure of nitrous oxide (N2O) can best be described by a resonance of two structures. The hybridization of nitrogen atoms in the two resonance structures are:
(i) both sp
(ii) one sp another sp3
(iii) both sp2
(iv) one sp another sp2
Answer: The two resonance structures for nitrous oxide (N2O) are:
The first structure shows the nitrogen atoms in a triple bond with a positive charge on one nitrogen and a negative charge on the oxygen. The second structure shows the nitrogen atoms in a double bond with a positive charge on one nitrogen and a double bond with oxygen.
The hybridization of the nitrogen atoms can be determined by looking at the number of regions of electron density around each nitrogen atom.
In the first structure, the nitrogen atom has two regions of electron density from the triple bond and one lone pair of electrons. This gives a total of three regions of electron density, which corresponds to sp hybridization.
In the second structure, the nitrogen atom has three regions of electron density from the double bond and one lone pair of electrons. This gives a total of four regions of electron density, which corresponds to sp³ hybridization.
Therefore, the hybridization of the nitrogen atoms in the two resonance structures of N2O are sp and sp³, respectively.
2. Which statement is true for the following chemical reactions?
CuCO3 + heat → CuO + CO2
CuO + SnO → Cu + SnO2
(i) CO2 is oxidised and SnO2 is reduced
(ii) CuCO3 is oxidized and CuO is reduced
(iii) CuO is oxidized and SnO is reduced
(iv) SnO is oxidized and CuO is reduced
Answer: The first statement is a thermal decomposition reaction where copper(II) carbonate (CuCO3) decomposes into copper(II) oxide (CuO) and carbon dioxide (CO2) upon heating.
The second statement is a redox reaction where copper(II) oxide (CuO) reacts with tin(IV) oxide (SnO2) to form copper (Cu) and tin(IV) oxide (SnO2). In this reaction, copper(II) oxide is reduced to copper while tin(IV) oxide is oxidized to tin(IV) oxide.
Therefore, both statements are true and describe different types of chemical reactions.
3. Among the following, the one that gives a positive Iodoform test upon reaction with I2 and NaOH is:
The correct Answer is B
Answer: The organic compounds that give a positive Iodoform test are those that contain a methyl ketone (CH3-CO-) or a methyl alcohol (CH3-OH) group adjacent to carbon that has a single bond to two other carbons (i.e., a methyl group).
Therefore, among the following options, the compound that gives a positive Iodoform test upon reaction with I2 and NaOH is the one that contains a methyl ketone or a methyl alcohol group adjacent to such a carbon.
Without knowing the options provided, it is not possible to determine the correct answer. Please provide the options for a more accurate response.
4. Dinitrogen tetroxide (N2O4) breaks down into nitrogen dioxide (NO2). If the reaction is reversible and endothermic, which condition will give the largest yield of NO2?
(i) High temperature and high pressure.
(ii) High temperature and low pressure.
(iii) Low temperature and high pressure.
(iv) Low temperature and low pressure.
Answer: The breakdown of dinitrogen tetroxide (N2O4) into nitrogen dioxide (NO2) is an example of an endothermic reaction, meaning that it requires energy to proceed. The forward reaction can be represented as follows:
N2O4 ⇌ 2NO2 (ΔH > 0)
The equilibrium constant (K) for this reaction can be expressed as follows:
K = [NO2]²/[N2O4]
where [N2O4] and [NO2] represent the molar concentrations of the respective species at equilibrium.
To maximize the yield of NO2, we want to shift the equilibrium towards the product side of the reaction. This can be achieved by either increasing the temperature or decreasing the pressure. According to Le Chatelier's principle, a system at equilibrium will respond to a change in conditions in a way that counteracts the change.
Therefore, if we increase the temperature, the reaction will shift towards the endothermic direction, i.e., towards the product side, to absorb the additional heat. Similarly, if we decrease the pressure, the reaction will shift toward the side with fewer moles of gas, which in this case is the product side.
Thus, the condition that will give the largest yield of NO2 is high temperature and low pressure.
3. The melting temperature of a human protein is 33 °C in 100 mM sodium phosphate buffer at pH 6.5. What should the predominant conformation of the protein in the same buffer at physiological temperature conditions be?
(ii) α helical
(iii) β sheet
(iv) Mixture of alpha helix and beta sheet.
Answer: The melting temperature (Tm) of a protein is the temperature at which it undergoes a transition from its folded to the unfolded state. In this case, the Tm of the human protein is 33 °C in a 100 mM sodium phosphate buffer at pH 6.5. This means that at temperatures below 33 °C, the protein should be mostly folded, while at temperatures above 33 °C, it should be mostly unfolded.
Physiological temperature conditions for humans are typically around 37 °C. Since this is above the Tm of the protein, the predominant conformation of the protein in the same buffer at physiological temperature conditions should be mostly unfolded. However, it's important to note that the actual conformation of the protein will depend on a variety of factors, including the specific sequence and structure of the protein, as well as the presence of any stabilizing factors such as ligands or other molecules.
4. 20 μl of a 10 mg/ml solution of protein A (molecular weight 10 kDa) is added to 20 μl of a 10 mg/ml solution of protein B of molecular weight 20 kDa. What is the molar ratio of the two proteins (A:B) in the mixture?
(i) 1 :1
Answer: First, we need to calculate the number of moles of each protein in 20 μl of a 10 mg/ml solution. We can use the formula:
moles = mass (mg) / molecular weight (MW)
For protein A:
moles of A = 10 mg / 10 kDa = 1 μmol
For protein B:
moles of B = 10 mg / 20 kDa = 0.5 μmol
The total moles of protein in the mixture is:
total moles = moles of A + moles of B = 1 μmol + 0.5 μmol = 1.5 μmol
The molar ratio of A to B can be calculated as:
A:B = moles of A / moles of B
A:B = 1 μmol / 0.5 μmol = 2:1
Therefore, the molar ratio of protein A to protein B in the mixture is 2:1.
5. When the velocity of enzyme activity is plotted against substrate concentration, which of the following is obtained?
(i) Hyperbolic curve
(iii) Straight line with positive slope
(iv) Straight line with negative slope
Answer : When the velocity of enzyme activity is plotted against substrate concentration, a hyperbolic curve known as the Michaelis-Menten curve is obtained. The Michaelis-Menten equation describes the relationship between enzyme velocity and substrate concentration and is given by:
V = Vmax [S]/(Km + [S]),
where V is the velocity of the enzyme-catalyzed reaction, Vmax is the maximum velocity of the reaction, [S] is the substrate concentration, and Km is the Michaelis constant, which represents the substrate concentration at which the reaction velocity is half of Vmax. The Michaelis-Menten curve shows that the reaction velocity increases with increasing substrate concentration, but eventually levels off as the enzyme becomes saturated with substrate.
5. Phosphoric acid is tribasic with three pKa values of 2.14, 6.86, and 12.4, respectively. The ionic form of the acid that predominates at pH 3.2 is:
Answer: At pH 3.2, the hydrogen ion concentration is relatively high, and the acid will be mostly dissociated into its ionic form. To determine which ionic form predominates at this pH, we need to compare the pH to the pKa values of the acid.
The first pKa value of phosphoric acid is 2.14, which means that at pH values lower than 2.14, most of the acid will be in its fully protonated form H3PO4. At pH values between 2.14 and 6.86, the acid will be partially dissociated, and a mixture of H3PO4, H2PO4-, and HPO42- ions will be present in the solution. At pH values between 6.86 and 12.4, the acid will be mostly dissociated, and a mixture of H2PO4-, HPO42-, and PO43- ions will be present in the solution. At pH values higher than 12.4, most of the acid will be in its fully deprotonated form PO43-.
Since the pH of 3.2 is between the first and second pKa values of phosphoric acid, the acid will be partially dissociated, and a mixture of H3PO4, H2PO4-, and HPO42- ions will be present in the solution. However, to determine which of these ionic forms predominates, we need to compare their relative concentrations. At pH values close to the second pKa, the ratio of the concentrations of the H2PO4- and HPO42- ions is approximately 1:1, so we can assume that these two forms are present in roughly equal amounts. The concentration of the H3PO4 form will be relatively small, as it will have mostly dissociated at pH 3.2.
Therefore, the ionic form of phosphoric acid that predominates at pH 3.2 is H2PO4-.