MCQ
Which of the following is violating Hund's rule
- A
- B
- C
- ✓All of the above
✓
Answer
Correct option: D.
All of the above
d
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→$(1)\,CH_3C \equiv C - CH_3$ $(2)\,CH_3CH_2 -CH_2 -CH_3$
$(3)\, CH_3CH_2C \equiv CH$ $(4)\, CH_3CH = CH_2$
$\begin{matrix}
C{{H}_{3\,\,\,\,\,\,\,\,\,\,\,}} \\
|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
C{{H}_{3}}-C=CH-C{{H}_{3}} \\
\end{matrix}\xrightarrow{Hg{{(OAc)}_{2}}/EtOH}\underset{(90\%)}{\mathop{(A)}}\,$
→C{{H}_{3\,\,\,\,\,\,\,\,\,\,\,}} \\
|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
C{{H}_{3}}-C=CH-C{{H}_{3}} \\
\end{matrix}\xrightarrow{Hg{{(OAc)}_{2}}/EtOH}\underset{(90\%)}{\mathop{(A)}}\,$
Product $(A)$ of the above reaction is
For which of the following complexes, identify the complex which ${\Delta _0}$ for $II > {\Delta _0}$ for $I$
→| $I$ | $II$ | |
| $[Cr(H_2O)_6]^{+2}$ | $[Cr(H_2O)_6]^{+3}$ | $(a)$ |
| $[Fe(H_2O)_6]^{+3}$ | $[Fe(CN)_6]^{-3}$ | $(b)$ |
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The reaction is called $\begin{array}{*{20}{c}}
{2C{H_3} - C - O{C_2}{H_5}} \\
{||\,\,\,} \\
{O\,\,\,}
\end{array}$ ${\xrightarrow{{{C_2}{H_5}ONa}}}$ $\begin{array}{*{20}{c}}
{C{H_3} - C - C{H_2} - C - O{C_2}{H_5}} \\
{||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||\,\,\,\,\,\,\,} \\
{O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O\,\,\,\,\,}
\end{array}$ $ + \,{C_2}{H_5}OH$
→{2C{H_3} - C - O{C_2}{H_5}} \\
{||\,\,\,} \\
{O\,\,\,}
\end{array}$ ${\xrightarrow{{{C_2}{H_5}ONa}}}$ $\begin{array}{*{20}{c}}
{C{H_3} - C - C{H_2} - C - O{C_2}{H_5}} \\
{||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||\,\,\,\,\,\,\,} \\
{O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O\,\,\,\,\,}
\end{array}$ $ + \,{C_2}{H_5}OH$