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Chemistry Test - 14

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Chemistry Test - 14
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  • Question 1
    1 / -0

    If the valence shell electronic configuration for an element is \(\mathrm{ns}^{2} \mathrm{np}^{5}\), this element belongs to the group of:

    Solution

    If the valence shell electronic configuration for an element is \(\mathrm{ns}^{2} \mathrm{np}^{5}\), this element belongs to the group ofhalogens.

    The electronic configuration \(\mathrm{ns}^{2}, \mathrm{np}^{5}\) shows that the elements has 7 electrons in its valence shell and it is a halogen.The group of halogens is the only periodic table group that contains elements in three of the main states of matter at standard temperature and pressure. All of the halogens form acids when bonded to hydrogen. Most halogens are typically produced from minerals or salts. The middle halogens—chlorine, bromine, and iodine—are often used as disinfectants. Organobromides are the most important class of flame retardants, while elemental halogens are dangerous and can be lethally toxic.

  • Question 2
    1 / -0

    Consider the Newman projection formulas shown below:

    Which of the following statements is correct?

    Solution

    The Newmann's projection is the most stable conformation about C1 - C2 bond of 1-chloro 2-methyl propane. The Cl atom and the methyl groups are trans to each other and the steric interactions are minimized. As we can see both A and B are the mirror image of each other. Hence, the 3 statement is correct Compounds B and C are enantiomers.

  • Question 3
    1 / -0

    In a CCP lattice of X and Y, X atoms are present at the corners while Y atoms are at face centres. Then, the formula of the compound would be if one of the X atoms from a corner is replaced by Z atoms (also monovalent)?

    Solution

    Effective number of atoms of \(X, Y\) and \(Z\) are:

    According to question:

    For X:

    Number of atoms \(=\frac{1}{8} \times 7\) (out of 8 corners, \(X\) is occupying 7)

    For Y:

    Number of atoms\(=\frac{1}{2} \times 6=3 (\mathrm{Y}\) is occupying all face centers)

    For Z:

    Number of atoms\(=\frac{1}{8}({Z}\) is occupying one corner out of 8 )

    Therefore, the formula of the compound will be:

    \(X_{\frac{7}{8}} Y_{3} Z_{\frac{1}{8}}\)

    \(=X_{7} Y_{24} Z\)

     

  • Question 4
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    The state of a thermodynamic system is described by its measurable or macroscopic (bulk) properties. These are:

    Solution

    The state of a thermodynamic system is described by its measurable or macroscopic (bulk) properties. These arepressure, volume, temperature, and amount.

    State of the System: The state of a thermodynamic system is described by its measurable (or macroscopic) or bulk properties. Thus, the state of a gas can be described by quoting its pressure \( (p)\), volume \( (v)\), temperature \( (T)\), and amount \( (n)\). These variables are called state variables or state functions because their values depend only on the state of the system and not on how it is reached.

  • Question 5
    1 / -0

    In the phenomenon of electric discharge through gases at low pressure, the coloured glow in the tube appears as a result of:

    Solution

    In the phenomenon of electric discharge through gases at low pressure, the coloured glow in the tube appears as a result of collisions between charged particles emitted from cathode and atoms of gas.

    The discharge of electricity through liquefied gases is an interesting phenomenon which can be systematically studied with the help of a discharge tube. In discharge tube collision between charged particles emitted from cathode and atoms of the gas results to colorless glow in tube.

  • Question 6
    1 / -0

    Which of the following ligands form a chelate?

    Solution

    In chelation, ring formation occurs because two atoms from the same ligand coordinate with the metal atom. This cannot happen if the ligand is monodentate.

    Oxalate is a bidentate ligand so it forms a chelate. It can coordinate with both of its negatively charged O atoms.

    Acetate, cyanide and ammonia are monodentate ligands. They do not form chelates.

  • Question 7
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    From which of the following tertiary butyl alcohol is obtained by the action of methyl magnesium bromide?

    Solution

    Tertiary butyl alcohol is obtained by the action of methyl magnesium bromide on acetone CH3COCH3.

    Methyl magnesium bromide is CH3MgX and falls in the category of Grignard Reagents. The methyl group bears a partial negative charge and the metal bears a partial positive charge. Ketones react with a Grignard reagent to give tertiary alcohols as addition products.

    Aldehydes give secondary alcohol and formaldehyde gives primary alcohol. The reaction is a step-up reaction as the number of carbon atoms increases in the product.

    The reaction of acetone with methyl magnesium halide is given as follows:

    The product of the reaction is tertiary butyl alcohol.

  • Question 8
    1 / -0

    \(\mathrm{Fe}^{3+}\) compounds are more stable than \(\mathrm{Fe}^{2+}\) compounds because:

    Solution

    Stability depends on the electronic configurations of respective ions/elements.

    \(F e^{+3}\) has an electronic configuration of \([A r] 3 d^{5} 4 s^{0}\)

    \(F e^{+2}\) has an electronic configuration of \([A r] 3 d^{6} 4 s^{0}\).

    \(\mathrm{Fe}^{+3}\) being half filled is more stable than \(\mathrm{Fe}^{+2}\) which is incompletely filled.

  • Question 9
    1 / -0

    A buffer solution contains \(100~ mL\) of \(0.01 M ~CH _{3} COOH\) and \(200 ~mL\) of \(0.02 M~CH _{3} COONa\). \(700 ~mL\) of water is added to this solution. \(pH\) before and after dilution are: \(\left(p K_{a}=4.74, \log 4 = 0.602\right)\)

    Solution

    \(pH\) (before dilution)

    \(=p K_{a}+\log \frac{[\text { salt }]}{[\text { acid }]}\)

    \(=4.74+\log \frac{200 \times 0.02}{100 \times 0.01}\)

    \(=4.74+\log 4\)

    \(=5.34\)

    Dilution has no effect on \(pH\) of buffer solution hence \(pH\) (after dilution) \(= 5.34\)

  • Question 10
    1 / -0

    In solid state \(\mathrm{N}_{2} \mathrm{O}_{5}\) exists as:

    Solution

    In solid state \(\mathrm{N}_{2} \mathrm{O}_{5}\) exists as \(\left[\mathrm{NO}_{3}\right]^{-}\left[\mathrm{NO}_{2}\right]^{+}\).

    • Solid \(\mathrm{N}_{2} \mathrm{O}_{5}\) is a salt, consisting of separated anions and cations.
    • The cation is the linear nitronium ion \(\mathrm{NO}_{2}^{+}\)and the anion is the planar nitrate \(\mathrm{NO}_{3}^{-}\)ion.
    • Thus, the solid could be called nitronium nitrate.
    • Both nitrogen centers have oxidation state \(+5\).
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