A) \[{{H}_{2}}O\]
B) \[{{H}_{2}}S\]
C) \[{{H}_{2}}Se\]
D) \[{{H}_{2}}Te\]
Correct Answer: D
Solution :
\[{{H}_{2}}O\] \[{{H}_{2}}S\] \[CC{{l}_{4}}+{{I}_{2}}\,\to \] \[{{H}_{2}}Te\] \[{{104.5}^{o}}\] \[2NaI+C{{l}_{2}}\,\to \,2KCl+{{I}_{2}}\] \[{{91}^{o}}\] \[{{90}^{o}}\] As we go down the group electronegativity decreases due to which repulsion between bonded pairs of electron also decreases. Hence, bond angle decreases.
You need to login to perform this action.
You will be redirected in
3 sec
