MCQ
Any $p-$ orbital can accommodate up to:
- A$4$ electrons
- B$2$ electrons with parallel spins
- C$6$ electrons
- ✓$2$ electrons with opposite spins
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$(I)$ 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

|
${List-I }$ ${(Reactants) }$ |
${List-II }$ ${ Products }$ |
| ($A$)Phenol, $\mathrm{Zn} / \Delta$ |
$(I)$ Salicylaldehyde |
| ($B$)Phenol, $\mathrm{CHCl}_3, \mathrm{NaOH}, \mathrm{HCl}$ |
$(II)$ Salicylic acid |
| ($C$)Phenol, $\mathrm{CO}_2, \mathrm{NaOH}, \mathrm{HCl}$ |
$(III)$ Benzene |
| ($D$)Phenol, Conc. $\mathrm{HNO}_3$ |
$(IV)$ Picric acid |
(Molar Mass : Bromine $=80\; g / mol$ )