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 :
[c] 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. |
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