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JEE Main2013Chemistryd and f Block ElementsActual

Which of the following arrangements does not represent the correct order of the property stated against it?

Options

  1. ACo 3 + < Fe 3 + < Cr 3 + < Sc 3 + :​ Stability in aqueous solution
  2. BSc < Ti < Cr < Mn :​ Number of oxidation states
  3. CV 2 + < Cr 2 + < Mn 2 + < Fe 2 + :​​ Paramagnetic behaviour
  4. DNi 2 + < Co 2 + < Fe 2 + < Mn 2 + :​ Ionic size

Correct answer

C. V 2 + < Cr 2 + < Mn 2 + < Fe 2 + :​​ Paramagnetic behaviour

Step-by-step solution

No. of unpaired electrons 2 3 V 2 + 1 8 Ar &#160;4 S 0 &#160;3 d 3 &#160; = &#160; 3 2 4 Cr 2 + 1 8 Ar &#160;4 S 0 &#160;3 d 4 &#160; = &#160; 4 2 5 Mn 2 + 1 8 Ar &#160;4 S 0 &#160;3 d 5 &#160; = &#160; 5 2 6 Fe 2 + 1 8 Ar &#160;4 S 0 &#160;3 d 6 &#160; = &#160; 4 On moving left to right in the periodic table, ionic size decreases. Sc shows only one oxidation state, and it is +3 but Mn shows variable oxidation state from +2 to +7.

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