$(iii)$ Size $(iv)$ Energy level
electronegativity depends on $s$ - character, more is the $s$ - character, more is the electronegativity. So, the order for electronegativity should be $sp \,> \,sp ^2\,>\,sp ^3$
bond angle between same hybrid orbitals is in the following order
$sp \left(180^{\circ}\right)\,>\, sp ^2\left(120^{\circ}\right)\,>\,sp ^3\left(109.5^{\circ}\right)$
Size $\rightarrow$ more $s$-character $\rightarrow$ more compact $\rightarrow$ smaller size.
So, the given order is correct, $sp \,<\, sp ^2\,<\,sp ^3$.
Energy level $\rightarrow$ since $s$-orbitals have lower energy closer than $p$ - as they allow closer proximity of the $e^{-}$to the nuclear. Thus, more $p$ - character more energy.
So, the order will be $sp \,<\,sp ^2\,<\,sp ^3$
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$CO\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \rightleftharpoons C{O_2}\left( g \right)$ is
Reason $R:$ The increasing nuclear charge outweighs the shielding across the period.
In the light of the above statements, choose the most appropriate from the options given below:
