A) \[\underset{\begin{smallmatrix} \text{ }\!\!|\!\!\text{ } \\ \text{H} \end{smallmatrix}}{\overset{\begin{smallmatrix} \text{H} \\ \text{ }\!\!|\!\!\text{ } \end{smallmatrix}}{\mathop{\text{C}}}}\,=\underset{\begin{smallmatrix} \text{ }\!\!|\!\!\text{ } \\ \text{H} \end{smallmatrix}}{\overset{\begin{smallmatrix} \text{H} \\ \text{ }\!\!|\!\!\text{ } \end{smallmatrix}}{\mathop{\text{C}}}}\,+\text{H}-\xrightarrow[{}]{{}}\text{H}-\underset{\begin{smallmatrix} \text{ }\!\!|\!\!\text{ } \\ \text{H} \end{smallmatrix}}{\overset{\begin{smallmatrix} \text{H} \\ \text{ }\!\!|\!\!\text{ } \end{smallmatrix}}{\mathop{\text{C}}}}\,-\underset{\begin{smallmatrix} \text{ }\!\!|\!\!\text{ } \\ \text{H} \end{smallmatrix}}{\overset{\begin{smallmatrix} \text{H} \\ \text{ }\!\!|\!\!\text{ } \end{smallmatrix}}{\mathop{\text{C}}}}\,-\text{H}\]
B) \[\text{mo}{{\text{l}}^{\text{-1}}}\]
C) \[\text{mo}{{\text{l}}^{\text{-1}}}\]
D) \[\text{mo}{{\text{l}}^{\text{-1}}}\]
Correct Answer: B
Solution :
Key Idea Bond order \[{{B}_{2}}{{O}_{3}}\]where, \[CaO\] number of electrons in bonding MO \[Si{{O}_{2}}\] number of electrons in anti bonding MO \[BeO\] \[{{[Co{{(en)}_{2}}C{{l}_{2}}]}^{+}}\] \[{{[Co{{(N{{H}_{3}})}_{3}}C{{l}_{3}}]}^{0}}\] \[{{[Co(en)C{{l}_{2}}{{(N{{H}_{3}})}_{2}}]}^{+}}\] \[{{[Co{{(en)}_{3}}]}^{3+}}\] \[A{{l}_{2}}{{O}_{3}}\] \[4.5\times {{10}^{4}}\] \[A{{l}_{2}}{{O}_{3}}\] \[\text{A}l=27\text{g}\,\text{mo}{{\text{l}}^{-1}}\] \[9.0\times {{10}^{3}}g\]\[8.1\times {{10}^{4}}g\] Hence, the increasing order of B.O is, \[2.4\times {{10}^{5}}g\]
You need to login to perform this action.
You will be redirected in
3 sec
