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

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

Paper - Chem - 111 (Part I: Objective Type Questions)

Group - A (Inorganic)

SECTION - I

1. Put a Tick (\(\checkmark\)) mark against the correct answer in the brackets provided for it: [1 x 6 = 6]

NB: Answers are [ \(\checkmark\) ] checkmarked

  1. Which of the following represents the correct set of four quantum numbers of a 4d1 electron?
    1. 4, 3, 2, +1/2 [ ]
    2. 4, 2, 1, 0 [ ]
    3. 4, 3, -2, +1/2 [ ]
    4. 4, 2, +1, +1/2 [ \(\checkmark\) ]
  2. Which of the following electronic configuration belongs to d-block elements?
    1. [Rn] 6s27s2 [ ]
    2. [Xe] 4f15d16s2 [ ]
    3. [Xe] 4f146d16s2 [ ]
    4. [Xe] 5d16s2[ \(\checkmark\) ]
  3. \(^{234}_{90}\text{Th} \) disintegrates giving the end-product \(^{206}_{82}\text{Pb}\). The total number of particles emitted are
    1. 7 alpha and 4 beta particles [ ]
    2. 7 alpha and 6 beta particles[ \(\checkmark\) ]
    3. 4 alpha and 4 beta particles [ ]
    4. 4 alpha and 6 beta particles [ ]
  4. Bond order of 1.5 is shown by
    1. O2+ [ ]
    2. O2-[ \(\checkmark\) ]
    3. O22- [ ]
    4. O2 [ ]
  5. The type of hybridization of Be in BeCl2 is
    1. sp [ \(\checkmark\) ]
    2. sp2 [ ]
    3. sp3 [ ]
    4. dsp2 [ ]
  6. Among the following, the element with the highest ionization enthalpy is
    1. oxygen [ ]
    2. nitrogen[ \(\checkmark\) ]
    3. carbon [ ]
    4. boron [ ]

Section - II

Answer the following in not more than 5 sentences each:

1. Explain why half-filled and fully filled subshells have extra stability. [2]

Answer:There are two main reason for the extra stability of half-filled and fully filled shubshells.

  1. Exchange Energy: The more the exchange energy of electrons, more stable is the configuration. While pairing of electrons in an orbital leads to instability therefore, half-filled orbitals are very stable because there is no pairing but there is maximum exchange energy. Similarly there is very high exchange energy in fully filled orbitals but due to pairing, the overall stabilization is less.
  2. Symmetrical distribution if charges: In both cases of half-filled and fully filled subshells, there is a symmetrical distribution of charges in all its orbitals. Consider the p subshell, each p orbital is containing either one electron in half-filled or two electron in fully filled subshell, therefore the charge distribution is very symmetrical.

2. Distinguish between bonding and anti-bonding molecular orbitals. [2]

Answer:

Bonding Molecular OrbitalAnti-Bonding Molecular Orbital
(a) Bonding Molecular Orbital is formed by the positive addition of the Atomic Orbital wave functions \(\Psi_{BO} = \Psi_A + \Psi_B\) (1) Anti Bonding Molecular Orbital is formed by the negative addition of the Atomic Orbital wave functions \(\Psi_{BO} = \Psi_A - \Psi_B\)
(b) Bonding MO has lower energy than anti-bonding MO. (2) Anti-bonding MO has higher energy than Bonding MO
(c) Electron charge density is high between nuclei, hence repulsion between nuclei is low. This lead to favourable stable bonds. (3) Electron charge density is less between nuclei, hence repulsion between nuclei is high. This lead to unfavourable stable bonds.
[next]

Group - B (Organic)

SECTION - I

1. Put a Tick (\(\checkmark\)) mark against the correct answer in the brackets provided for it: [1 x 6 = 6]

NB: Answers are [ \(\checkmark\) ] checkmarked

  1. CH3CH=CH2 is more stable than CH2=CH2 because of
    1. resonance effect [ ]
    2. inductive effect [ ]
    3. electromeric effect [ ]
    4. hyperconjugation [ \(\checkmark\) ]
  2. Find the wrong statement
    1. A nucleophile is a Lewis base [ ]
    2. All nucleophiles are negatively charged species [ \(\checkmark\) ]
    3. A positively charged species is not a nucleophile [ ]
    4. Some nucleophiles are neutral molecules [ ]
  3. If a compound contains two dissimilar asymmetric carbon atoms, it can give rise to
    1. two optical isomers [ ]
    2. three optical isomers [ ]
    3. four optical isomers [ \(\checkmark\) ]
    4. six optical isomers [ ]
  4. Which of the following is not a meta directing group?
    1. -+N(CH3)3 [ ]
    2. -NO2 [ ]
    3. -COOH [ ]
    4. -CH3 [ \(\checkmark\) ]
  5. Which of the following compounds does not react with ammooniacal AgNO3 solution?
    1. Acetylene [ ]
    2. 1-Butyne [ ]
    3. Propyne [ ]
    4. 2-Butyne [ \(\checkmark\) ]
  6. Markovnikoff's addition of HBr is not applicable to
    1. propene [ ]
    2. 1-butene [ ]
    3. 1-pentene [ ]
    4. 2-butene [ \(\checkmark\) ]

SECTION - II

Answer the following questions

1. Draw all the stereoisomeric forms of 2-bromo-3-chlorobutane and indicate the enantiomeric pairs [2]

Answer: 2-bromo-3-chlorobutane has four (4) stereoisomers as follows (drawn in Fischer projection):

Structure 1 & 2 and 3 & 4 are enantiomeric pairs

2. Cyclopropane undergoes ring opening reactions to give open chain addition compounds but cyclopentante does not. Explain. [2]

Answer: According to Bayer's strain theory, cycloalkanes are planar and hence lower cycloalkanes suffers strain from the normal tetrahedral angle of 109.5°. The more the strain, more unstable is the compound. Among cyclopropane and cyclopentane, cyclopropane bond angle is only 60° with angle strain of about 24° while cyclopentante bond angle is about 108° hence the strain it suffers is only 0.75°. Thus, cyclopropane is very unstable and hence undergoes ring opening reaction.

[next]

Group - C (Physical)

SECTION - I

Put a Tick (\(\checkmark\)) mark against the correct answer in the brackets provided for it: [1 x 6 = 6]

NB: Answers are [ \(\checkmark\) ] checkmarked

  1. The real gases show nearly ideal behaviour at
    1. low pressure and low temperature [ ]
    2. high pressure and high temperature [ ]
    3. low pressure and high temperature [ \(\checkmark\) ]
    4. high pressure and low temperature [ ]
  2. The expression derived form kinetic theory of gases is given as
    1. \(PV = \frac{1}{3}mNC^2\) [ \(\checkmark\) ]
    2. \(PV = \frac{3}{2}mNC^2\) [ ]
    3. \(PV = \frac{1}{2}mNC^2\) [ ]
    4. None of the above [ ]
  3. Which of the following properties is not associated with a crystalline solid?
    1. Anisotropy [ ]
    2. Isotropy [ \(\checkmark\) ]
    3. Sharp melting point [ ]
    4. Definite geometry [ ]
  4. The unit of surface tension is
    1. kg m-1 s-1 [ ]
    2. kg m2 s-1 [ ]
    3. N m2 [ ]
    4. N m-1 [ \(\checkmark\) ]
  5. $\Delta H\degree$ represents the enthalpy change
    1. at 0 °C and 1 atm pressure [ ]
    2. at 0 K and 1 atm pressure [ ]
    3. at 25 °C and 1 atm pressure [ \(\checkmark\) ]
    4. None of the above [ ]
  6. Which of the following laws deals with the entropy change for an irreversible process?
    1. First law of thermodynamics [ ]
    2. Second law of thermodynamics [ \(\checkmark\) ]
    3. Third law of thermodynamics [ ]
    4. Zeroth law of thermodynamics [ ]

SECTION - II

Answer the following Questions

1. Define the term coefficient of viscosity. What is its unit in CGS system?

Answer: The ratio of the shearing stress to the velocity gradient is a measure of the viscosity of the fluid and is called the coefficient of viscosity. Mathematically it can be represented as follows: \[ \eta = \frac{F\partial z}{A \partial v} \] where \(\eta\) is the coefficient of viscosity, F is the force, A is the area \(\partial v \over \partial z\) is the shear velocity. The unit of \(\eta\) in CGS is dynes sec cm-2 and is also known as poises.

2. Give the mathematical formulation of the first law of thermodynamics. What form will it assume when applied to an adiabatic process?

Answer: The mathematical formulation of the first law of thermodynamics is given as follows: \begin{equation} \label{eq:firstlaw} dU = q + w \end{equation} Where, \(dU\) is the change in internal energy of the system \(q\) is the heat transfer in the system \(w\) is the work done. Now \(w = -P\Delta V\) where, \(P\) is pressure and \(\Delta V\) is change in volume. Hence equation \ref{eq:firstlaw} can be written as: \begin{equation} \label{eq:firstlawP} dU = q - P\Delta V \end{equation} When first law is applied to an adiabatic process (no heat transfer) \(q = 0\) the first law equation \ref{eq:firstlawP} takes the following form: \[ dU = w = -P\Delta V \]

Continue to the Part II (Descriptive Type) Question year 2015

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F Diengdoh.com: Chemistry First Paper (Chem-111) 2015 Solved Question of NEHU: Part I
Chemistry First Paper (Chem-111) 2015 Solved Question of NEHU: Part I
Solution to the First Year Final Examination Question Paper from NEHU, Shillong. Paper - Chem - 111 (Part I: Objective Type Questions)
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