Question

In $H_{2}O$, the bond angle HOH is ${104}^{\circ }{28}^{\prime }$ but in $H_{2}S,H_{2}Se$ and $H_{2}Te$ the bond angles are pretty close to ${90}^{\circ }$. This suggests that:

• A. oxygen uses $s{p}^2$-hybrid orbitals while $S,Se$ and $Te$ use $s{p}^3$-hybrid orbitals for bonding with the hydrogen atoms
• B. oxygen uses $s{p}^3$-hybrid orbitals to bond with the two hydrogen atoms while $S,Se$ and $Te$ use almost pure p orbitals
• C. oxygen uses $s{p}^3$-hybrid orbitals while $S,Se$ and $Te$ utilize d orbitals for bonding with the hydrogen atoms
• D. all the atoms use pure p orbitals to bond with the hydrogen atoms

Answer 1

The correct option is B oxygen uses $s{p}^3$-hybrid orbitals to bond with the two hydrogen atoms while $S,Se$ and $Te$ use almost pure p orbitals

$s{p}^3$ orbitals have an angle of ${109}^0{28}^{\prime }$ which hints that $O$ is using this hybridisation to bond whereas the pure p orbitals are at $90$ degress with each other hinting that $S,Se$ and $Te$ use almost pure p orbitals