MCQ
In a sigma bond
- ASidewise as well as end to end overlap of orbitals take place
- BSidewise overlap of orbitals takes place
- ✓End to end overlap of orbitals takes place
- DNone of the above
A covalent bond resulting from the formation of a molecular orbital by the end-to-end overlap of atomic orbitals, denoted by the symbol $\sigma$.
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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
$NO_2^+ \to NO_2 \to NO_2^-$