MCQ
What is the bond angle in ammonia molecule?
- A106.5°
- B107°
- C108°
- D109.5°
Explanation:
The bond angle for a standard tetrahedral geometry is 109.5°. But the ammonia molecule contains a lone electron pair on nitrogen.
As it is closer to the N atom than the other orbitals, it creates a repulsive effect and pushes the Hydrogen atoms close to each other, thus reducing the bond angle to 107°.
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$Ag\left( s \right)\left| {AgBr\left( s \right)\,\left| {B{r^ - }\left( {0.01\,M} \right)} \right|\,\left| {{I^ - }\left( {0.02\,M} \right)} \right|\,AgI\left( s \right)} \right|Ag\left( s \right)$
the correct information is
[Given : $K_{sp}\,\left( {AgBr} \right) = 4 \times {10^{ - 13}}$ ,
$K_{sp}\,\left( {AgI} \right)$ $ = 8 \times {10^{ - 17}},\frac{{2.303\,RT}}{F} = 0.06\,V,\,\log \,2 = 0.3]$