Question

Choose the correct code of characteristics for the given order of hybrid orbitals of same atom,
$sp<{sp}^2<{sp}^3$
$\left(I\right)$ Electronegativity
$\left(II\right)$ Bond angle between same hybrid orbitals
$\left(III\right)$ Size
$\left(IV\right)$ Energy level

• A. $\left(II\right),\left(III\right),\left(IV\right)$
• B. $\left(III\right),\left(IV\right)$
• C. $\left(II\right),\left(IV\right)$
• D. $\left(I\right),\left(II\right),\left(III\right),\left(IV\right)$

Answer 1

Answer: (B)

$\left(III\right),\left(IV\right)$

Given $\to$ $Sp<{Sp}^2<{Sp}^3$

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 $\to$ more $S-$character$\to$more compact$\to$smaller size.

So, the given order is correct, $Sp<{Sp}^2<{Sp}^3$.

Energy level$\to$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 $Sp<{Sp}^2<{Sp}^3$

Answer:$B$ $iii,iv$