The correct order of bond angle in the given | vedclass

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(A) $1$. $SiO_4^{4-}$ has $sp^3$ hybridization with a tetrahedral geometry,so the bond angle is $109^{\circ} 28'$.$2$. In $NCl_3$ and $PCl_3$,both hav

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$SiO_4^{4-} > NCl_3 > PCl_3 > SbH_3 > H_2Te$

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(A) $1$. $SiO_4^{4-}$ has $sp^3$ hybridization with a tetrahedral geometry,so the bond angle is $109^{\circ} 28'$.$2$. In $NCl_3$ and $PCl_3$,both have a lone pair. The bond angle in $NCl_3$ $(107^{\circ})$ is greater than in $PCl_3$ $(100^{\circ})$ because the electronegativity of $N$ is higher,pulling bonding electrons closer to the central atom,and the $s$-character in the hybrid orbitals decreases down the group.$3$. In $SbH_3$,the bond angle is close to $90^{\circ}$ because the bonding involves almost pure $p$-orbitals.$4$. In $H_2Te$,the bond angle is approximately $90^{\circ}$ (slightly higher due to lone pair-lone pair repulsions). Comparing $SbH_3$ and $H_2Te$,$H_2Te$ has a slightly larger bond angle due to the presence of two lone pairs on $Te$ compared to one on $Sb$.$5$. Thus,the correct order is $SiO_4^{4-} > NCl_3 > PCl_3 > SbH_3 > H_2Te$.

List-$I$ (Molecule Species)	List-$II$ (Property/Shape)
$A$. $SO_2Cl_2$	$I$. Paramagnetic
$B$. $NO$	$II$. Diamagnetic
$C$. $NO_2^-$	$III$. Tetrahedral
$D$. $I_3^-$	$IV$. Linear

The correct order of bond angle in the given species is:

Similar Questions

The number of $S-S$ bonds in $SO_3, S_2O_3^{2-}, S_2O_6^{2-}$ and $S_2O_8^{2-}$ respectively are

Match List-$I$ with List-$II$ :

List-$I$ (Molecule Species) List-$II$ (Property/Shape)

$A$. $SO_2Cl_2$ $I$. Paramagnetic

$B$. $NO$ $II$. Diamagnetic

$C$. $NO_2^-$ $III$. Tetrahedral

$D$. $I_3^-$ $IV$. Linear

Choose the correct answer from the options given below :

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The formal charges of $C$ and $O$ atoms in $CO_2$ $(\ddot{O}=C=\ddot{O})$ are,respectively:

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