Question

Which is the correct decreasing order of bond angle in the following series?

• A. $N{{H}_{3}}>{{H}_{2}}O>{{H}_{2}}S$
• B. ${{H}_{2}}S>N{{H}_{3}}>{{H}_{2}}O$
• C. ${{H}_{2}}S>{{H}_{2}}O>N{{H}_{3}}$
• D. ${{H}_{2}}O>N{{H}_{3}}>{{H}_{2}}S$

Answer 1

Answer: (A)

$N{{H}_{3}}>{{H}_{2}}O>{{H}_{2}}S$

Answer 2

: For $N{{H}_{3}}$ bond angle $(H\overset{\wedge }{\mathop{N}}\,H)$ is $107{}^\circ$ . For ${{H}_{2}}O$ bond angle $(H\overset{\wedge }{\mathop{O}}\,H)$ is $104.5{}^\circ$ . For ${{H}_{2}}S$ bond angle $(H\overset{\wedge }{\mathop{S}}\,H)$ is $92{}^\circ$ .