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Chemistry Test - 21

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Chemistry Test - 21
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  • Question 1
    1 / -0
    The correct stability order of the following resonance structures is?
    (i) \(\mathrm{H}_{2} \mathrm{C}=\overrightarrow{\mathrm{N}}=\overline{\mathrm{N}}\)
    (ii) \(\mathrm{H}_{2} \mathrm{C}-\mathrm{N}=\overrightarrow{\mathrm{N}}\)
    (iii) \(\mathrm{H}_{2} \overline{\mathrm{C}}-\overrightarrow{\mathrm{N}} \equiv \mathrm{N}\)
    (iv) \(\mathrm{H}_{2} \overline{\mathrm{C}}-\mathrm{N}=\overrightarrow{\mathrm{N}}\)
    Solution

    No. of π π bonds ∝∝ Resonance Energy ∝∝ Stability

    +ve charge on electro positive element

    -ve charge on electro negative element

    unlike charges on adjacent atoms like charges to separate out far.

    (i) Two π π bonds unlike charges on adjacent atoms

    (iii) Two π π Bonds unlike charges on adjacent undesired atoms

    (ii) One π π bond unlike charges separated but on desired atoms.

    (iv) One π π bond unlike charges separated but on undesired atoms

    (i) > (iii) > (ii) > (iv)
    Hence, the correct option is (D)

  • Question 2
    1 / -0

    Nickel (Z = 28) combines with a uninegative mono dentate ligand X- to form a paramagnetic complex [NiX4]2-.The number of unpaired electron(s) in the nickel and geometry of this complex ion are, respectively

    Solution

    Number of unpaired electrons = 2

     

    Geometry = tetrahedral.
    Hence, the correct option is (C)

  • Question 3
    1 / -0

    Which of the following facts about the complex [Cr(NH3)6]Cl3 is wrong?

    Solution

    The complex [Cr(NH3)6]Cl3 involves d2sp3 hybridization as it involves (n - 1)d orbitals for hybridization. It is an inner orbital complex.
    Hence, the correct option is (C)

  • Question 4
    1 / -0

    The number of σ- and π-bonds in 5-oxohexanoic acid, respectively, is :

    Solution

    (18σ, 2π)

    Hence, the correct option is (A)

  • Question 5
    1 / -0

    The correct order of reducing ability for the four successive elements Cr, Mn, Fe and Co is: Their Eored values are given below.

    \(\mathrm{E}_{\mathrm{Mn}^{2+} / \mathrm{Mn}}^{\circ}=-1 \cdot 18 \mathrm{V}\)
    \(\mathrm{E}_{\mathrm{Cr}^{2+} / \mathrm{Cr}}^{\circ}=-0 \cdot 91 \mathrm{V}\)
    \(\mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\circ}=-0 \cdot 44 \mathrm{V}\)
    \(\mathrm{E}_{\mathrm{Co}^{2+} / \mathrm{Co}}^{\circ}=-0 \cdot 28 \mathrm{V}\)
    Solution

    The values of Eored gives us an idea of the ease with which metals get reduced. A higher negative value implies that it has greater tendency to get oxidised.

    For a metal to act as a reducing agent, it should itself get oxidised. Hence, the order of reducing ability is proportional to ease of oxidation which is inversly proportional to Eored.

    Hence, the answer will be:

    Mn > Cr > Fe > Co
    Hence, the correct option is (B)

  • Question 6
    1 / -0

    The boiling point elevation constant for toluene (b.pt = 110.7oC) is 3.32 K molality-1. The entropy of vaporization of toluene in JK-1 mole-1 is :

    Solution
    \(\begin{aligned} \Delta \mathrm{H}_{\mathrm{V}} &=\frac{\mathrm{M}_{\mathrm{C}_{7} \mathrm{H}_{8}} \times \mathrm{R} \times\left(\mathrm{T}_{\mathrm{b}}^{\circ}\right)^{2}}{1000 \times \mathrm{K}_{\mathrm{b}}} \\ &=\frac{92 \times 8.314 \times(383.7)^{2}}{1000 \times 3.32}=33.92 \mathrm{kJ} \mathrm{ml}^{-1} \\ \text {Now, } \Delta \mathrm{s} &=\frac{\Delta \mathrm{Hv}}{\mathrm{T}_{\mathrm{b}}}=\frac{33.92 \times 10^{3}}{383.7}=88.5 \mathrm{JK}^{-1} \mathrm{mol}^{-1} \end{aligned}\)
    Hence, the correct option is (B)
  • Question 7
    1 / -0

    Which one of the following reactions of Xenon compounds is not feasible?

    Solution

    The reaction XeO3+6HF⟶XeF6+3H2Ois not feasible because XeF6 formed will further produce XeO3 by getting hydrolysed.

    XeF6 + H2O → XeOF4 + 2HF

    XeOF4 + H2O → XeO2F2 + 2HF

    XeO2F2 + H2O → XeO3 + 2HF
    Hence, the correct option is (A)

  • Question 8
    1 / -0

    The shape of XeF4 is :

    Solution

    Due to presence of two lone pairs of electrons, it is square planar in shape. Lone pairs are arranged at pyramidal position as per Bent's Rule.
    Hence, the correct option is (B)

     

  • Question 9
    1 / -0

    How many moles of KMnO4 are needed to oxidise a mixture of 1 mole each of FeSO4, FeC2O4 and Fe2(C2O4)3 completely in acidic medium :

    Solution
    \(\begin{array}{cccc}{2+}{\mathrm{FeSO}_{4}} & \longrightarrow & \mathrm{Fe}^{3+}+\mathrm{SO}_{4}^{2-}+1 \overline{\mathrm{e}} & (\mathrm{n} \text { factor }=1) \\ \mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right) & \longrightarrow & \mathrm{Fe}^{3+}+2 \mathrm{CO}_{2}+3 \mathrm{e} & (\mathrm{n} \text { factor }=3) \\ \mathrm{Fe}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3} & \longrightarrow & 2 \mathrm{Fe}^{3+}+6 \mathrm{CO}_{2}+6 \mathrm{e} & (\mathrm{n} \text { factor }=6)\end{array}\)
    Equialents of \(\left\{\mathrm{FeSO}_{4}+\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)+\mathrm{Fe}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right\}\)
    \(=\{1 \times 1+1 \times 3+1 \times 6\}=10\) Equialents
    \(\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}\)
    \((\mathrm{n}\) factor \(=5)\)
    Equialents of \(\mathrm{KMnO}_{4}=10\) Equialents
    Moles of \(\mathrm{KMnO}_{4}=\frac{10}{5}=2 \mathrm{Moles}\)
    Hence, the correct option is (B)
  • Question 10
    1 / -0
    A real gas has equation of state \(P=\frac{R T}{V_{m}-b T}-\frac{a}{V_{m}^{3}}\left(V_{m}\right.\) is molar volume). Find a and \(b\) in terms of \(V_{c}\) and \(T_{c} .\) where \(V_{c}\) and \(T_{\mathrm{c}}\) are critical constants of gas.
    Solution
    At critical point (point of Inflection)
    \(\left(\frac{\partial \mathrm{P}}{\partial \mathrm{V}_{\mathrm{m}}}\right)_{\mathrm{T}_{\mathrm{C}}}=0=\frac{-\mathrm{RT}_{\mathrm{C}}}{\left(\mathrm{V}_{\mathrm{m}}-\mathrm{b} \mathrm{T}_{\mathrm{C}}\right)^{2}}+\frac{3 \mathrm{a}}{\mathrm{V}_{\mathrm{m}^{4}}}\)
    \(\left(\frac{\partial^{2} \mathrm{P}}{\partial \mathrm{V}_{\mathrm{m}^{2}}}\right)_{\mathrm{T}_{\mathrm{C}}}=0=\frac{2 \mathrm{RT}_{\mathrm{C}}}{\left(\mathrm{V}_{\mathrm{m}}-\mathrm{b} \mathrm{T}_{\mathrm{C}}\right)^{3}}-\frac{12 \mathrm{a}}{\mathrm{V}_{\mathrm{m}^{5}}}\)
    At critical point \(V_{m}=V_{C}\) rewriting
    \(\frac{3 \mathrm{a}}{\mathrm{V}_{\mathrm{C}}^{4}}==\frac{\mathrm{RT}_{\mathrm{C}}}{\left(\mathrm{V}_{\mathrm{C}}-\mathrm{bT}_{\mathrm{C}}\right)^{2}} \ldots\left(1^{\prime}\right)\)
    \(\frac{12 \mathrm{a}}{\mathrm{V}_{\mathrm{C}}^{5}}==\frac{2 \mathrm{RT}_{\mathrm{C}}}{\left(\mathrm{V}_{\mathrm{C}}-\mathrm{b} \mathrm{T}_{\mathrm{C}}\right)^{3}} \ldots\left(\mathrm{II}^{\prime}\right)\)
    \(\frac{I I^{\prime}}{I^{\prime}} \Rightarrow \frac{4}{V_{C}}=\frac{2}{V_{C}-b T_{C}} \Rightarrow 2 V_{C}=4 V_{C}-4 b T_{C}\)
    \(\frac{\mathrm{V}_{\mathrm{C}}}{2 \mathrm{T}_{\mathrm{C}}}=\mathrm{b}\)
    Substitute b in ( \("\) )
    \(\frac{3 \mathrm{a}}{\mathrm{V}_{\mathrm{C}^{4}}}=\frac{\mathrm{RT}_{\mathrm{C}}}{\left(\mathrm{V}_{\mathrm{C}}-\frac{\mathrm{v}_{\mathrm{C}}}{2 \mathrm{T}_{\mathrm{C}}} \mathrm{T}_{\mathrm{C}}\right)^{2}}=\frac{\mathrm{RT}_{\mathrm{C}}}{\mathrm{V}_{\mathrm{C}^{2}} / 4}=\frac{4 \mathrm{RT}_{\mathrm{C}}}{\mathrm{V}_{\mathrm{C}^{2}}}\)
    \(\mathbf{a}=\frac{4}{3} \frac{\mathrm{V}_{\mathrm{C}}^{4} \mathrm{RT}_{\mathrm{C}}}{\mathrm{V}_{\mathrm{C}}^{2}}\)
    \(\mathrm{a}=\frac{4}{3} \mathrm{RT}_{\mathrm{C}} \mathrm{V}_{\mathrm{C}}^{2}\)
    Hence, the correct option is (A)
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