question_answer

Out of \[A+B\to \]and \[=k{{[A]}^{2}}[B]\] (Z of \[=k[A]{{[B]}^{2}}\])        the colourless species are

A) \[=k{{[A]}^{2}}{{[B]}^{2}}\]and \[=k[A][B]\]

B) \[N_{2}^{-}<{{N}_{2}}<N_{2}^{2-}\] and \[N_{2}^{2-}<N_{2}^{-}<{{N}_{2}}\]

C) \[{{N}_{2}}<N_{2}^{2-}<N_{2}^{-}\]and\[N_{2}^{-}<N_{2}^{2-}<{{N}_{2}}\]

D) \[{{C}_{2}},{{C}_{3}},{{C}_{5}}\] and \[{{C}_{6}}\]

Correct Answer: C

Solution :

Key Idea Species without unpaired electrons are colourless because transition of electron is not possible. In \[4l+2\] is present as \[2l+1\] \[4l-2\] Hence, \[2{{n}^{2}}\] is colourless. In \[\Delta H\]Co is present as \[\Delta S\] Due to presence of unpaired electrons, \[{{C}_{(graphite)}}+C{{O}_{2}}(g)\xrightarrow[{}]{{}}2CO(g)\]is coloured. In \[170J{{K}^{-1}},\]is present as \[{{H}_{2}}COH.C{{H}_{2}}OH\] Due to absence of unpaired electron, \[2C{{O}_{2}}\] is colourless. In \[\underset{\begin{smallmatrix}  \text{ }\!\!|\!\!\text{ } \\  \text{CHO} \end{smallmatrix}}{\mathop{\text{CHO}}}\,\]Ni is present as \[1.5\times {{10}^{-s}}\]. Since, unpaired electrons are present, \[4.5\times {{10}^{-10}},\] is coloured. Hence, \[C{{N}^{-}}+C{{H}_{3}}COOH\]and \[HCN+C{{H}_{3}}CO{{O}^{-}}\]are colourless species.