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Chemistry Test - 13

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Chemistry Test - 13
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  • Question 1
    1 / -0

    Which of the following can act as reducing agent

    Solution

    Stability of metal carbonyl is deduced from EAN rule.

    EAN rules states – those complex’s in which effective atomic number is numerically equal to atomic number of noble gas element found in same period in which metal is situated, are most stable.

    EAN (effective atomic number) =

    (Atomic monber ofcentral metal ion)( Oxidation state)+(Number of e'sgiven by ligands)

    A. [Co(CO)4]

    EAN = 27 – (–1) + 2 × 4

    = 36 (inert gas atomic number)

    Since, it already has EAN = 36 so it won’t give up or take in any electron.

    B. EAN = 25 – (0) + 2 × 6 = 37

    It has EAN = 37 so to attain inert gas electronic configuration it will donate one electron

    [Mn(CO)6] -e- [Mn(CO)6]+

    EAN = 25 – (+1) + 2 × 6 = 36

    It will act as reducing agent.

    C. [Mn(CO)5]

    EAN = 25 – (0) + 2 × 5 = 35

    To gain inert gas electronic configuration, it will accept one electron.

    [Mn(CO)5) +e- [Mn(CO)5]

    EAN = 25 – (1) + 2 × 5 = 36

    It will act as oxidizing agent.

    D. [Cr(CO)6]

    EAN = 24 – (0) + 2 × 6 = 36

    It already has inert gas configuration so it won’t exchange electrons.

    Hence option (B) is correct.

  • Question 2
    1 / -0
    Consider the following galvanic cell at \({ }_{25^{\circ} \mathrm{C}}\) Pt, \(H_{2_{(q)}} 1\) atm \(/ \mathrm{HCl}(100 \mathrm{ml}, 0.1 \mathrm{M}) \| \mathrm{H}_{2} \mathrm{SO}_{4}(0.05 \mathrm{M}) \mid \mathrm{Pt}, \mathrm{H}_{2(\mathrm{g})} 1 \mathrm{atm}\)

    On adding 50 ml of 0.1 M NaOH into the Anode compartment, the emf. Of cell
    Solution
    \(P t, H_{2_{(q)}} 1 a t m / H A(100 m l, 0.1 M) \| H_{2} S O_{4}(0.05 M) \mid P t, H_{2(g)} 1 a t m\)
    Here \(E=-0.059 \log \frac{\left[\mathrm{H}^{+}\right] \text {anode }}{\left[\mathrm{H}^{+}\right] \text {cathode }}\)
    \(\therefore \mathrm{E}=0.059\left(\mathrm{pH}_{\text {anode }}-\mathrm{pH}_{\text {cathode }}\right)\)
    Initially; \(\mathrm{pH}_{\text {anode }}=1 \mathrm{pH}_{\text {cathode }}=\) remain constant in both cases = 1 \(\therefore \mathrm{E}_{\text {initial }}=0.059(1-1)=0 \mathrm{V}\)
    \(\therefore \mathrm{HCl}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}\)
    5 mole remaining mole of \(\mathrm{HCl}=5\) and total volume become \(150 \mathrm{ml}\). \(\therefore\left[\mathrm{H}^{+}\right]\) after reaction \(=\frac{5}{150}=\frac{1}{30} \mathrm{M} \Rightarrow \mathrm{pH}=1.477\)
    \(\therefore \mathrm{E}_{\text {final }}=0.059(1.477-1)=0.028\)
    \(\therefore E\) increase by \(0.028 \mathrm{v}\)
  • Question 3
    1 / -0

    In which of the following transformations, the bond order has increased and magnetic behavior has decreased?

    Solution

    If there are some unpaired electrons then the molecule is paramagnetic, if it has no unpaired electrons then the molecule is diamagnetic.

    A. \(C_{2}^{\oplus}=11 \mathrm{e}^{\prime} \mathrm{s}=\sigma_{1 s} 2 \cdot \sigma_{1 s^{*} 2}, \sigma_{2 s^{2}}, \quad \sigma_{2 s^{* 2}}, \pi_{2 p y}^{2}, \pi_{2 p x}^{1}\)
    There is one unpaired electron thus paramagnetic \(\mathrm{C}_{2}=12 \mathrm{e}^{\prime} \mathrm{S} \sigma_{1 s} 2, \sigma_{1 s^{*} 2}, \sigma_{2 s^{2}}, \sigma_{2 s^{* 2}}, \pi_{2 p y}^{2}, \pi_{2 p x}^{2}\)
    There are no unpaired electrons thus diamagnetic \(C_{2}^{\oplus} \Rightarrow B . O .=\frac{7-4}{2}=1.5\) (Paramagnetic)
    \(\mathrm{C}_{2}=\mathrm{B.O}=\frac{8-4}{2}=2\) (Diamagnetic)
    \(\mathrm{B} \cdot \mathrm{NO}^{\oplus}=14 \mathrm{e}^{\prime} \mathrm{s}=\sigma_{1 s^{2}}, \quad \sigma_{1 s^{2}}, \quad \sigma_{2 s^{2}}, \quad \sigma_{2 s^{* 2}}, \quad \sigma_{2 P_{z}^{2}}, \pi_{2 p x}^{2}, \pi_{2 p y}^{2}\)
    B.0. \(=\frac{10-4}{2}=3\) Diamagnetic
    \(B .0=\frac{10-5}{2}=2.5\) Paramagneti
    B. \(0=10-6 / 2=2\)
    Unpaired e's so paramagnetic \(O_{2}^{\oplus}=15 \mathrm{e}^{\prime} \mathrm{s} \sigma_{1 s^{2}} \quad \sigma_{1 s^{2}} \quad \sigma_{2 s^{2}} \quad \sigma_{2 s^{2}} \quad \sigma_{2 P_{z}^{2}}\left(\pi_{2 p x}^{2}, \pi_{2 p y}^{2}\right)\left(\pi_{2 p x}^{*_{1}}, \pi_{2 p y}^{* 0}\right)\)
    \(0=\frac{10-5}{2}=2.5\)
    Unpaired e's so paramagnetic \(\mathrm{D} . \mathrm{N}_{2}=14 \mathrm{e}^{\prime} \mathrm{s}\)
    \(\sigma_{1 s^{2}} \quad \sigma_{1 s^{2}} \quad \sigma_{2 s^{2}} \quad \sigma_{2 s^{2}}\left(\pi_{2 p x}^{2}, \pi_{2 p y}^{2}\right) \sigma_{2 p z}^{2}\)
    \(B .0=\frac{10-4}{2}=3\)
    No unpaired electron so diamagnetic \(N_{2}^{\oplus}=13 \mathrm{e}^{\prime} \mathrm{s}\)
    \(\sigma_{1 s^{2}} \quad \sigma_{1 s^{2}} \quad \sigma_{2 s^{2}} \quad \sigma_{2 s^{* 2}}\left(\pi_{2 p x}^{2}, \pi_{2 p y}^{2}\right) \sigma_{2 p z^{1}}\)
    B.0 \(=\frac{9-4}{2}=\frac{5}{2}=2.5\)
    Unpaired electrons so paramagnetic.
    Hence, option (A) is correct.
  • Question 4
    1 / -0

    x% Oleum in water is treated with 0.5lof 2.75 M Ca(OH)2 solution. The resulting solution required 15.7 gm of H3PO3 (Assume strong acid) solution for complete neutralization. Calculate the amount of free SO3 in 100 gms of oleum

    Solution
    Moles of \(\mathrm{H}_{2} \mathrm{SO}_{4}=\frac{100+\mathrm{x}}{98}\)
    Moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) taken \(=2.75 \times 0.5=1.375\)
    Moles of \(\mathrm{H}_{3} \mathrm{PO}_{3}\) used \(=\frac{15.7}{82}\)
    since \(\mathrm{H}_{3} \mathrm{PO}_{3}\) is dibasic, \(\therefore\) moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) left \(=\frac{15.7}{82}\)
    \(\therefore\) Moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) used with \(\mathrm{H}_{2} \mathrm{SO}_{4}=1.375-\frac{15.7}{82}\)
    \(\therefore \frac{100+x}{98}=1.375-\frac{15.7}{82}\)
    Solving, \(x=15.64\) \(\mathrm{SO}_{3}=\frac{15.64}{18} \times 80=71.1 \mathrm{g}\)
  • Question 5
    1 / -0

    If P,Q,R and S are elements of 3rd period of p–block in modern periodic table and among these one element is metal and rest are non-metal and their order of electronegativity is also given asP < Q < R < S .Then in which of the following release of is relatively easier.

    Solution

    In any X – O – H, we take into account electronegative difference between X and O, and O and H. If electronegative difference between X and O is greater than O and H, X – O bond will be more polar and will break easily giving OH. If electronegative difference between O and H is larger than X and O, O – H bond will be more polar and easier to break.
    Electronegativity of S is largest, so electronegative difference between S and O will be least in S – O – H. It will be easier to break O – H bond, givingH.
    Hence option (B) is correct.

  • Question 6
    1 / -0

    28 g of N2 and 6.0 g of H2 are heated over a catalyst in a Closed 1L flask at 4500C. The entire equilibrium mixture Required 500 ml of 1.0 M H2SO4 for neutralization. The Value of KC  (in standard unit) for the following reaction is:

    Solution

  • Question 7
    1 / -0

    Bakelite is a

    Solution

    Structure of bakelite

    Hence option (b) is correct.

  • Question 8
    1 / -0

    If the speed of an electron in first Bohr orbit of hydrogen be 'x', then speed of the electron in the second orbit of He+ is:

    Solution
    Bohr velocity of an electron in nth orbit \((\mathrm{Vn})=\) Vozn where, \(\mathrm{V}_{\mathrm{O}=2.18 \times 106 \mathrm{m} / \mathrm{s}}\)
    Given \(V_{1}(H)=x=V_{0}\)
    Now \(V_{2}(H e+)=V \circ z z=V o\)
    Therefore the speed of electron in the second orbit of \(\mathrm{He}+\) is \(\mathrm{x}\) which is equal to the speed of an electron in the first Bohr orbit of hydrogen.
  • Question 9
    1 / -0

    Fruity smell is given by:

    Solution

    Esters are pleasant smelling liquids having fruity smell.

  • Question 10
    1 / -0

    2-Phenylethanol may be prepared by the reaction of phenylmagnesium bromide with:

    Solution

    2-Phenylethanol may be prepared by the reaction of phenylmagnesium bromide with oxirane. It is an example of nucleophilic attack on epoxide ring which opens the epoxide ring.

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