8.2 Organic chemistry isomerism — Chemistry STD 11 Science — Question

$(I)$ It is easier to remove $2 \mathrm{p}$ electron than $2 \mathrm{s}$ electron

$(II)$ $2 \mathrm{p}$ electron of $\mathrm{B}$ is more shielded from the nucleus by the inner core of electrons than the $2 s$ electrons of $Be.$

$(III)\; 2 s$ electron has more penetration power than $2 \mathrm{p}$ electron.

$(IV)$ atomic radius of $\mathrm{B}$ is more than $\mathrm{Be}$

(Atomic number $\mathrm{B}=5, \mathrm{Be}=4$ )

The correct statements are

. . . . . (Given that the value of solubility product of $A B \left( K _{ sp }\right)=2 \times 10^{-10}$ and the value of ionization constant of $H B \left( K _{ a }\right)=1 \times 10^{-8}$ )

$\begin{array}{*{20}{c}}
  {C{H_3} - C - C - C{H_3}} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,|} \\ 
  {O\,\,\,\,\,\,\,\,\,O} 
\end{array}$  $X$ in the above reaction is