Chemistry MCQs for NEET — Practice Questions with Answers

(b) SO₂ gas can readily reduces acidified KMnO₄ solution because KMnO₄ is an oxidising agent that easily oxidise SO₂.

$2{\mathrm{MnO}}_4^{-}+ 5{\mathrm{SO}}_2 + 2{\mathrm{H}}_2\mathrm{O} \to 2{\mathrm{Mn}}^{2+} + 5{\mathrm{SO}}_4^{2-} + 4{\mathrm{H}}^{+}$

While other options such as NO₂ (strong oxidising agent), CO₂ (neither oxidising agent nor reducing agent cannot decolourise acidified KMnO₄ solution.

(c) HgCl₂ and I₂ both when dissolved in water containing I^- ions, the pair of species formed is  ${\mathrm{HgI}}_4^{2-} \mathrm{and} {\mathrm{I}}_3^{-}$

(c) Eu(63) = 4f⁷.5d°, 6S², Eu²⁺ = Af⁷

In lanthanoids series, ionic radius decreases and covalent character increases, thus basicity decreases. Lanthanons are less reactive than aluminium due to high ionisation potential. The reason for this high ionisation potential is lanthanoid contraction. Ce⁴⁺ is a good oxidising agent, it is easily converted to Ce³⁺.

(C) When SO₂ is passed through acidified K₂Cr₂O₇ solution, green chromium sulphate is formed. In this reaction, oxidation state of Cr changes from +6 to +3.

K₂Cr₂O₇ + H₂SO₄ + 3SO_2  $\to$K₂SO₄ + Cr₂(SO₄)₃ + H₂O

The appearance of green colour is due to the reduction of chromium metal.

Because of the lanthanoid contraction Zr (atomic radii 160 pm) and Hf (atomic radii 158 pm) have nearly same atomic radii.

Lanthanoids include the elements from lanthanum La (Z = 57) to lutetium Lu(Z = 71). zirconium Zr(40) belong to the second transition series (4d) and Hf (72) belongs to third transition series (5d). Lanthanoid contraction is associated with the intervention of the 4f orbitals which are filled before the 5d-series of elements starts. The filling of 4f if orbitals before 5d-orbitals results in regular decrease in atomic radii which compensates the expected increase in atomic size with increasing atomic number. As a result of this lanthanoid contraction, the elements of second and third transition series have almost similar atomic radii.

       0                      III

(a) Fe +H₂O+O₂→Fe₂O₃. x [H₂O]
          (From air)
    
       0                  II
(b) Fe+CuSO₄→FeSO₄+Cu

      0              0
(c) Fe+5CO_^→Fe(CO₅)

      0              III
(d) Fe+H₂O→Fe₂O₃+H₂



∴ Formation of Fe(CO)₅ from Fe does not involve oxidation of iron because there 18 no change in oxidation state.

(c) ₆₄Gd = [Xe]4f⁷5d¹6s²

FeSO₄ will require the least amount of acidified KMnO₄ for complete oxidation.

(a) Lanthanoid contraction is the regular decrease in atomic and ionic radii of lanthanides. This is due to the imperfect shielding or poor screening effect of orbitals due to their diffused shape, which unable to counterbalance the effect of the increased nuclear charge. Hence, the net result is a contraction in size of lanthanoids.

Since OH^- are generated from weak acid (H₂O), a weak acid (like CO₂) should be used to remove it because of strong acid (HCl) reverse the reaction. KOH increases the concentration of OH^-, thus again shifts the reaction in backward side. CO₂ combines with OH^- to give carbonate which is easily removed. SO₂ reacts with water to give strong acid, so it cannot be used.