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Which of the following species is not paramagnetic?

Answer:

Explanation:

To identify the magnetic nature we need to check the molecular orbital configuration. If all orbitals are fully occupied, species is diamagnetic while when one or  more molecular orbitals is/are singly occupied , species is paramagnetic

(a) $NO (7+8=15)-\sigma1s^{2},\sigma^{*}1s^{2},\sigma2s^{2},\sigma^{*}2s^{2},\pi2p_x^2$

           $=\pi2p_y^2,\pi2p_z^2,\pi^{*}2p_x^1$ $=\pi^{*}2p_y^0$

   One unpaired electron is present.

              Hence , it is paramagnetic

(b)  $CO(6+8=14)-\sigma1s^{2},\sigma^{*}1s^{2},\sigma2s^{2},\sigma^{*}2s^{2},\pi2p_x^2$

      $=\pi2p_y^2,\sigma2p_z^2$     

    No unpaired electron is present

   Hence , it is diamagenetic

(c)    $O_{2}(8+8=16)-\sigma1s^{2},\sigma^{*}1s^{2},\sigma2s^{2},\sigma^{*}2s^{2},\sigma2p_z^2,\pi2p_x^2$

        $=\pi2p_x^2,\pi^{*}2p_x^1$ $=\pi^{*}2p_y^1$

  Two unpaired electrons are present. Hence it is paramagnetic

(d)      $B_{2}(5+5)-\sigma1s^{2},\sigma^{*}1s^{2},\sigma2s^{2},\sigma^{*}2s^{2},\pi2p_x^1$

          $=\pi2p_x^2,\pi^{*}2p_x^1$

           $=\pi2p_y^1$

     Two unpaired electrond are present. Hence it is paramagnetic

The sodium salt of an organic acid 'X' produces effervescence with conc. H₂SO₄ . 'X' reacts with the acidified aqueous CaCl₂ solution to give a white precipitate which decolorises acidic solution of KMnO₄. 'X' is

Answer:

Explanation:

The reaction talkes place as follows 

   $Na_{2}C_{2}O_{4}+H_{2}SO_{4}\rightarrow Na_{2}SO_{4}+H_{2}O$

      {x}                           (conc.}                                                             

                                                          $+CO\uparrow +CO{2}\uparrow$

                                                              effervescence

    $Na_{2}C_{2}O_{4}+CaCl_{2}\rightarrow CaC_{2}O_{4}+2NaCl$

           {X}                                        white ppt.

$5 Ca C_{2}O_{4}+2KMnO_{4}+8H_{2}SO_{4}\rightarrow K_{2}SO_{4}+5CaSO_{4}+2MnSO_{4}$

                                purple                                                                                          colourless

                                                                                          $+ 10CO_{2}+8H_{2}O$

     hence , X is Na₂C₂O₄.

The major products obtained in the following reaction is 

Answer:

Explanation:

An alkyl halide in the presence of a bulkier base removes a proton from a carbon adjacent to the carbon bonded to the halogen. This reaction is called E2 ( $\beta$ -elimination reaction)

The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are

Answer:

Explanation:

Cl_{2 ,} Br₂  and I₂  form a mixture of halide and hypohalites when react with cold dilute alkalies while a mixture of halides and haloate when react with concentrated cold alkalies.

    $Cl_{2}+2NaOH \rightarrow NaCl+NaClO+H_{2}O$

                          cold and dilute

$\therefore$     Cl^- and ClO^- are obtained as products when chlorine gas reacts with cold and dilute aqueous NaOH.

The Tyndall effect is observed only when the following conditions are satisfied

  1. The diameter of the dispersed particles is much smaller than the wavelength of the light used.

   2. The diameter of the dispersed particle is not much smaller than the wavelength of the light used.

   3. The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude.

   4. The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.

Answer:

Explanation:

Colloidal solutions show Tyndall effect due to the scattering of light by colloidal particles in all directions in space. It is observed only under the following conditions.

   (i) The diameter of the colloids should not be much smaller than the wavelength of light used.

   (ii)  The refractive indices of the dispersed phase and dispersion medium should differ greatly in magnitude.

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