MCQ
Decreasing order of bond angle is
- ✓$BeCl_2 > NO_2 > SO_2$
- B$BeCl_2 > SO_2 > NO_2$
- C$SO_2 > BeCl_2 > NO_2$
- D$SO_2 > NO_2 > BeCl_2$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| Column $-I$ | Column $-II$ |
| (Atomic number) | (Position of element in Periodic table) |
| $(A)$ $Z = 37$ | $(P)$ $p-$ block |
| $(B)$ $Z = 42$ | $(Q)$ $f-$ block |
| $(C)$ $Z = 34$ | $(R)$ $d-$ block |
| $(D)$ $Z = 92$ | $(S)$ $s-$ block |
$\begin{array}{*{20}{c}}
{Cl\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,} \\
{C{H_3} - CH - COOH}
\end{array}\mathop {\xrightarrow{{(i)\,N{H_3}}}}\limits_{1\,mole\,\,(ii)\,{H_2}O} [X]$
Product $[X]$ will be :
