Q. Bond angle of $109{^\circ}28'$ is found in
A) $NH_{3}$
B) $H_{2}O$
C) [image:4_image58.emf]
D) [image:4_image59.emf]
Explanation
(d)
In ${NH}_{4}^{+}$ ion, N is $sp^{3}$ hybridised therefore, bond angle in ${NH}_{4}^{+}$ (tetrahedral shape) is $109{^\circ}28'$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.