InterviewSolution
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InterviewSolution
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(a) Account for the following : (i) Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows the highest oxidation state of +4. (ii) Cr^(2+) is a strong reducing agent. (iii) Cu^(2+) salts are coloured while Zn^(2+) salts are white. (b) Complete the following equations : (i) 2MnO_(2)+4KOH+O_(2)overset(Delta)to (ii) Cr_(2)O_(7)^(2-)+14H^(+)+6I^(-)to |
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Answer» Solution :(a) (i) Mn can form double bonds with oxygen atoms and SHOWS an oxidation state of +7. FLUORINE is monovalent, only single bonds can be formed with fluorine and therefore the highest oxidation state shown by Mn with fluorine is +4. (ii) `Cr^(2+)` is a strong reducing agent. `Cr^(2+)` has the electronic configuration `d^(4)`. When it acts as a reducing agent, it is itself oxidised and changes to `Cr^(3+)`. The d-orbital configuration becomes `d^(3)`. It has stable half-filled `t_(2g)` configuration. (iii) The colour of transition metal IONS are due to d-d transitions of electrons i.e., from lower d-levels to higher d-levels `(t_(2g)" to "e_(G))`. `Cu^(2+)` ions contain an unpaired electron which on d-d transition gives blue colour. `Zn^(2+)` ions have no unpaired electron. There is no possibility of transition of electrons hence no colour. (B) (i) `2MnO_(2)+4KOH+O_(2)to2K_(2)MnO_(4)+2H_(2)O` (ii) `Cr_(2)O_(7)^(2-)+14H^(+)+6I^(-)to2Cr^(3+)+7H_(2)O+3I_(2)` |
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