As the $s-$ character of a hybrid orbital decreases
$(I)$ The bond angle decreases $(II)$ The bond strength increases
$(III)$ The bond length increases $(IV)$ Size of orbitals increases
$s$-character, $75\, \%\,p$-character $sp ^2=33 \,\%\, s$-character, $66 \,\% \,p$-character $sp =50\, \%\, p -$ character. The more $s$-character a bond has, the stronger and shorter the bond is. Hence the bond length decrease with increase in $s$ character. An $sp-sp$ bond is strongest and $s p ^3- sp ^3$ bond is weakest.
The bond angle of $sp ^3$ is $109.5, sp ^2$ is $120$ and $sp$ is $180$ . An $sp$ orbital is half $s$ character, $sp ^2$ is $1 / 3 s$ character and $sp ^3$ is $1 / 4 s$ character, so increasing the s character corresponds to increasing the bond angle.
The size of the orbital depends upon the value of principal quantum number$(n)$. Greater the value of $n$, larger is the size of the orbital and lesser the $s$-character
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$C{O_{(g)}} + \frac{1}{2}{O_{2(g)}} \to C{O_{2(g)}}$
How are $\Delta E$ and $\Delta H$ releated for the reaction?
| List $I$ (Spectral Series for Hyddrogen) | List $II$ (Spectral Region/Higher Energy State) |
| $A$. Lyman | $I$. Infrared region |
| $B$. Balmer | $II$. $UV$ region |
| $C$. Paschen | $III$. Infrared region |
| $D$. Pfund | $IV$. Visible region |
Choose the correct answer from the options given below :-
select the correct statement
$Ph - C \equiv C - Ph \to $ 