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Chemistry First Paper (Chem-111) 2015 Solved Question of NEHU: Part II

Solved Question

Solution to the First Year Final Examination Question Paper from NEHU, Shillong.

Paper - Chem - 111 (Part II: Descriptive Type Questions)

Group – A (Inorganic)

Click on Question No to view the solution:
[1 (a)] [1 (b)] [1 (c)]
[2 (a)] [2 (b)] [2 (c)]
[3 (a)] [3 (b)] [3 (c)] [3 (d)]

1 (a) State Pauli's exclusion principle. What is its importance? [1+2 = 3]

Answer: Pauli's Exclusion Principle states that, in an atom, no two electrons can have the same four electronic quantum numbers.

Paulis' Exclusion Principle helps explain a wide variety of physical phenomena. This principle gives the electron shell structure of atoms and the way atoms share electrons.

This principle is very useful in determining the maximum number of electrons that can occur in any quantum group. For K-shell, where n=1, l can have only one value, 0 and m can also have only one value 0. Hence, ms can be either \(\frac{1}{2}\) or \(-\frac{1}{2}\). Thus, there are two combinations of the quantum numbers. This shows that the K-shell, there is only one subshell l=0 and in this only two electrons of opposite spins can be accommodated.

1 (b) What is shielding effect? Explain the factors affecting it. [3]

Answer: Shielding effect is the effect exerted by inner shell electrons in an atom to the outer electrons. The electrons in lower shells shields the attractive nuclear force of the nucleus towards the outer shells electrons. It is also known as screening effect.

1 (c) Explain why cations are smaller and anions are larger in radii than their corresponding parent atom. [1.5]

Answer: Cations are formed by removal of one ore more electron from the outer most shell. This will lead to an increased effective nuclear charge and hence greater attraction from the nucleus. Thus, the radii decreases. Also, mostly formation of cations leads to complete removal of the outer most shell. Hence, the radii decreases. While for anions are formed by addition of one or more electrons to the outer shell. This will reduce the effective nuclear charge and then due to repulsion between electrons, the radii increases.

2 (a) Compare the relative stability of the following species and indicate their magnetic behaviours: \(\mathrm{O_2^-}\) and \(\mathrm{N_2^+}\) [2.5]

Answer: We know, electronic configuration of:
O : [He] 2s2 2p4
O- : [He] 2s2 2p4

N : [He] 2s2 2p3
N+: [He] 2s2 2p2
The MO diagram of \(\mathrm{O_2^-}\) and \(\mathrm{N_2^+}\) is as follows:

Molecular Orbital diagram of N2+ and O2-

\(BO = \frac{1}{2} (b - n)\) where BO is bond order, 'b' is numbers of electrons in bonding orbitals and 'n' is numbers of elctrons in anti-bonding orbitals.
Here the bond order of \(\mathrm{O_2^-}\) is 2.5 and of \(\mathrm{N_2^+}\) 1.5 therefore \(\mathrm{O_2^-}\) is more stable than \(\mathrm{N_2^+}\). And since both are having unpaired electrons hence they are paramagnetic.

2 (b) Discuss the use of radioactive isotopes in the treatment of diseases: [3]

2 (c) The uncertainties in position and velocity of a particle are 10-10 m and 5.27 x 10-24 m/s respectively. Calculate the mass of the particle (h = 6.625 x 10-34 J s). [2]

Answer: Given here, \[\Delta v = 5.27 \times 10^{-24} m/s\] \[\Delta x = 10^{-10} m \] \[h = 6.625 \times 10^{-34} Js\] We know, \[\Delta x \Delta p = \frac{h}{4\pi}\] \[\Delta p = m\Delta v\] Thus, \[\Delta x . m\Delta v = \frac{h}{4\pi}\] \[\text{therefore, } m = \frac{h}{4\pi\Delta v \Delta x}\] \[\text{or, } m = \frac{6.625 \times 10^{-34} Js}{4 \times 3.14 \times 5.27 \times 10^{-24} ms^{-1} \times 10^{-10} m}\] \[\text{or, } m = \frac{6.625 Kg m^2 s^{-2} s}{4 \times 5.27 m^2 s^{-1}}\] \[\text{thus, } m = 0.099 Kg\]

3(a) Describe the change in hybridization, if any, of the Al-atom in the following reaction: [2] \[\mathrm{AlCl_3} + \mathrm{Cl^-} \rightarrow \mathrm{AlCl_4^-}\]

Answer:

Reaction Scheme

In the above reaction, the hybridization of the central atom Al changes from sp2 to sp3.

3 (b) The hydrogen bonding in HF is stronger than that in H2O but the boiling point of water is 100°C while that of HF is only 19.5°C. Explain. [2]

Answer: HF has stronger hydrogen bonding than H2O because of higher electronegativity of Fluorine than Oxygen, hence higher dipole moment. But when we take a look at the structure of these two molecules, HF a linear molecule has possibility of forming only two hydrogen bonds while water has possibility to form four hydrogen bonds. More hydrogen bonds leads to stronger intermolecular interaction and that why water has higher boiling point than HF.

Hydrogen Bonding structure in HF and H2O

3 (c) Explain why dipole moment of hydrogen halides decreases from HF to HI. [2]

Answer:Dipole moment is the product of magnitude of charges and the distance of separation between the charges. Now magnitude of charges depends on electronegativity of the halides in hydrogen halides. Fluorine has highest electronegativity while electronegativity decreases down the group hence the magnitude of charges decreases. Also as we go down a group the atomic radii increases due to addition of more shell to the atom. Hence, bond length increases. But since the decrease in electronegativity is large, hence the dipole moment decreases as we go down the group.

3(d) state the Group Displacement Law. [1.5]

Answer: The Group Displacement law states that when an element emits α particles, then the new element formed will be two places to the left of the parent element, and when an element emits β particles, then the new element formed will be one place to the right of the parent element in the periodic table.

[next]
\(\nextSection\)

Group - B (Organic)

Click on Question No to view the solution:
[4 (a)] [4 (b)] [4 (c)] [4 (d)]
[5 (a)] [5 (b)] [5 (c)] [5 (d)]
[6 (a)]

4 (a) Draw the orbital structure of the following and mention their bond angles: [3]

\[\ce{CH_2=C=CH_2}\] \[\ce{H-C#C-H}\]

4 (b) Arrange the following carbocations in order of decreasing stability: C6H5+CH2 ; (C6H5)2+CH ; (C6H5)3+C Giver reason for your answer. [1.5]

4(c) What is resonance? On its basis, explain the acidic character of phenol. [2]

4(d) In CCl4, C-Cl bond has polarity but the molecule as a whole has no dipole moment. Explain. [1]

Answer: CCl4 is a symmetrical molecule with regular tetrahedral structure. Here the four C-Cl bonds are directed towards the corner of the tetrahedron as shown here: 

Structure of CCl4 Molecule

Thus, even when C-Cl bond is highly polar but due to the symmetrical structure of the molecule, the net dipole moment of each bond is cancelled by the resultant from the other three.

5 (a) Which of the following will have higher melting point and why? [1.5]

5(b) Explain why the staggered conformation of ethane is more stable than the eclipse conformation. [2]

5 (c) Convert the following Fischer projection formula to Sawhorse and Newman projection formula: [2]

5 (d) Indicate the type of isomerism exhibited by the following compounds. Write down the structures of the isomers separately in each case:
(i) CH3(Cl)C=C(Cl)CH3
(ii) C6H5CH(OH)CO2H [2]

6 (a) Complete the following reactions: [1x3=3]

[next]
\(\nextSection\)

Group - C (Physical)

Click on Question No to view the solution: [10 (a)] [10 (b)] [10 (c)]

No 10 (a): Define heat capacity at constant volume and constant pressure. Prove that: \(C_p - C_v = nR\) [4]

Answer: Heat capacity of a system is the amount of heat required to raise the temperature by 1 degree. Mathematically it can be written as follows: \begin{equation} \label{eq:heatCapa}C = \frac{q}{(T_2 - T_1)} \end{equation} Where, \(C\) is the Heat Capacity \(q\) is the amount of heat absorbed, \(T_1\) is the initial temperature \(T_2\) is the final temperature. \begin{equation} \label{eq:htCapa}\text{or, } C = \frac{dq}{dT}\end{equation} Now, at constant Volume, work done \(w = 0\), \begin{equation*}q = \Delta U \text{[From First Law]}\end{equation*} \begin{equation*}C_v = \left(\frac {q}{T_2 - T_1}\right)_v = \frac {\Delta U}{(T_2 - T_1)_v}\end{equation*} \begin{equation} \label{eq:htCv}\text{or, } C_v = \frac {dU}{dT}\end{equation} Where \(C_v\) is the heat capacity at constant volume.
Again, at constant pressure, \(q = \Delta U - w\) \begin{equation*}\text{or, } w = - P\Delta V\end{equation*} \begin{equation*} \text{or, }q = \Delta U + P\Delta V = \Delta H \end{equation*} Hence, \begin{equation} \label{eq:htCp} C_p = \frac {dH}{T_2 - T_1} = \frac {dH}{dT} \end{equation} Where \(C_p\) is the heat capacity at constant pressure. From first law of thermodynamics we know, \[U = H + PV\] Where U - internal energy, H - enthalpy, P is pressure and V is volume \[\text{or, } dU = dH + d(PV)\] \begin{equation} \label{eq:fullM} \text{or, }\frac{dU}{dT} = \frac{dH}{dT} + \frac{d(PV)}{dT} \end{equation} From ideal gas equation: \begin{equation} \label{eq:idealGas} PV = nRT \end{equation} Where n is no of moles, R - Gas constant, T is temperature From \ref{eq:fullM} and \ref{eq:idealGas} we get: \[ \frac{dU}{dT} = \frac{dH}{dT} + \frac{d(nRT)}{dT} \] \begin{equation} \label{eq:final} \text{or, } \frac{dU}{dT} = \frac{dH}{dT} + nR\frac{dT}{dT} \end{equation} From \ref{eq:htCv}, \ref{eq:htCp} and \ref{eq:final} we get: \[ C_v = C_p + nR \] \[ \text{or, } C_v - C_p = nR \]

10 (b) What is Joule-Thomson Effect? [1]

Answer: When a gas is allowed to expand freely in adiabatic condition, cooling takes place. This effect is known as Joule-Thomson Effect.

10 (c) Calculate \(q\), \(w\), \(\Delta U\) and \(\Delta H\) for the isothermal expansion of one mole of an ideal gas at 27℃ from a volume of 10 dm3 to a volume of 20 dm3 against a constant external pressure of 1 atm. [ 2.5]

Answer: Given Here,
Temparature, T = 27℃ = 300 °K
Initial Volume, V_1 = 10 dm3
Final Volume, \$V_2 = 20 dm3
Pressure, P = 1 atm
For isothermal process \(T\) is constant, therefore \(\Delta T = 0\) We know, \(\Delta U = C_v\Delta T = 0\) again, \begin{align*} \Delta H &= \Delta U + \Delta (PV) \\ &= 0 + \Delta (nRT) \text{ From Gas Law PV = nRT} \\ &;= R\Delta T = 0 \end{align*} Work done, \(w\) at constant pressure is given by: \begin{align*} w &= - \int_{V_1}^{V_2} PdV = - P(V_2 - V_1) \\ &= - 1 \text{ atm} (20 - 10) \text{ dm}^3 \\ &= - 10 \text{ dm}^3 \text{ atm} \end{align*} We know, \[R = 0.08206 dm^3 atm K^{-1} mol^{-1} = 8.314 J K^{-1} mol^{-1}\] \begin{align*} 1 atm &= \frac{8.314}{0.08206} \text{ J dm}^{-3} \\ &= 101.31 \text{ J dm}^{-3} \end{align*} Therefore, \begin{align*} w &= 10 \times 101.32 \text{ J dm}^{-3} \text{ dm}^3 \\ &= 1013.2 \text{ J} \end{align*} From first law of thermodynamics \[\Delta U = q + w \] \[q = - w \] \[q = - (-1013.2) \text{ J} \] \[q = 1013.2 \text{ J}\] For any queries comment out below.

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