Bond angle of 109° 28' is found in :

Question

Bond angle of 109° 28' is found in : (A) NH 3 (B) H2O (C)  overset⊕CH5 (D)  overset⊕NH4

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/9a58aa358f1169c3930defa5424731c5-.png" /> 109° 28', it means pure regular tetrahedral. (a) NH3 sp3 pyramidal (b) H2O sp3 bent (c) CH5+ (d)NH4+ sp3 tetrahedral

Q. Bond angle of $10 9^{\circ} 2 8^{'}$ is found in :

$N H_{3}$

$H_{2} O$

$C \oplus H_{5}$

$N \oplus H_{4}$

$10 9^{\circ} 2 8^{'}$ , it means pure regular tetrahedral.
(a) $N H_{3} s p^{3}$ pyramidal
(b) $H_{2} O s p^{3}$ bent
(c) $C H_{5}^{+}$
(d) $N H_{4}^{+} s p^{3}$ tetrahedral

Q. Bond angle of 109∘28′ is found in :

NH3​

H2​O

C⊕H5​

N⊕H4​