A) \[{{(C{{H}_{3}})}_{2}}SiC{{l}_{2}}\] is paramagnetic and bond order is greater than that for \[{{(C{{H}_{3}})}_{4}}Si\]
B) \[{{(C{{H}_{3}})}_{3}}SiCl\]is paramagnetic and bond order is lesser than that for\[C{{H}_{3}}SiC{{l}_{3}}\]
C) \[RN{{H}_{2}}+{{H}_{2}}OR-\overset{+}{\mathop{N{{H}_{3}}}}\,+O{{H}^{-}}\]is diamagnetic and bond order is lesser than that for \[K[{{H}_{2}}O]=\frac{[R-\overset{+}{\mathop{N{{H}_{3}}}}\,][O{{H}^{-}}]}{[RN{{H}_{2}}]}\]
D) \[{{K}_{b}}=\frac{[R-\overset{+}{\mathop{N{{H}_{3}}}}\,][O{{H}^{-}}]}{[R-N{{H}_{2}}]}\]is diamagnetic and bond order is more than that for \[p{{K}_{b}}=-\log {{K}_{b}}\]
Correct Answer: A
Solution :
Similarly \[CaOC{{l}_{2}}=0.\text{1}\times \text{63}.\text{5}=\text{ 6}.\text{35 g}\] has 16 electrons \[CaOC{{l}_{2}}=\frac{molar\,mass}{2}\] \[=\frac{127}{2}=63.5\] \[CaOC{{l}_{2}}=\frac{6.35}{10}\times 100=63.5%\] \[(AlC{{l}_{3}},B{{F}_{3}},FC{{l}_{3}}\,etc)\] bond order \[{{t}_{1/2}}=\frac{0.693}{k}or\,k=\frac{0.693}{{{t}_{1/2}}}=\frac{0.693}{6.93}=0.1\] \[\because \]has 15 electrons. \[k=\frac{2.303}{{{t}_{99%}}}\log \frac{a}{a-x}\therefore 0.1=\frac{2.303}{{{t}_{99%}}}\log \frac{100}{1}\]\[\therefore \] \[{{t}_{99%}}=\frac{2.303}{0.1}\times 2=46.06\min \] \[CaO\] bond order \[CaC{{O}_{3}}\xrightarrow{1073k}CaO+CO\] \[CaC+Si{{O}_{2}}\xrightarrow{{}}CaSi{{O}_{3}}\] bond order \[NaC{{O}_{3}}+2S{{O}_{2}}+{{H}_{2}}O\xrightarrow{{}}2NaHS{{O}_{3}}+C{{O}_{2}}(A)\] There is an unpaired electron in the anti bonding molecular orbital\[2NaHS{{O}_{3}}+N{{a}_{2}}S{{O}_{3}}\xrightarrow{{}}2N{{a}_{2}}S{{O}_{3}}+{{H}_{2}}O+C{{O}_{2}}(B)\]. Thus, the species is paramagnetic and the bond order is greater than that for \[N{{a}_{2}}S{{O}_{3}}+S\xrightarrow{{}}N{{a}_{2}}{{S}_{2}}{{O}_{3}}(C)\].
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