Decreasing order of bond angle of (NH(3...

Decreasing order of bond angle of `(NH_(3), PH_(3), AsH_(3))` is

`NH_(3) gt PH_(3) gt AsH_(3)`

`NH_(3) gt AsH_(3) gt PH_(3)`

`PH_(3) gt NH_(3) gt AsH_(3)`

`AsH_(3) gt PH_(3) gt NH_(3)`

Text Solution

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The correct Answer is:

To determine the decreasing order of bond angles for the compounds \( NH_3 \) (ammonia), \( PH_3 \) (phosphine), and \( AsH_3 \) (arsine), we can follow these steps: ### Step 1: Understand the molecular geometry All three compounds have a trigonal pyramidal shape due to the presence of a lone pair of electrons on the central atom (N, P, or As). ### Step 2: Consider the electronegativity of the central atoms - **Nitrogen (N)** has the highest electronegativity among the three. - **Phosphorus (P)** has a lower electronegativity than nitrogen. - **Arsenic (As)** has the lowest electronegativity of the three. ### Step 3: Analyze the effect of electronegativity on bond angles - In \( NH_3 \), the higher electronegativity of nitrogen causes the bonding electrons to be closer to the nitrogen atom, leading to greater lone pair-bond pair repulsion. This results in a larger bond angle, approximately \( 107^\circ \). - In \( PH_3 \), the lower electronegativity of phosphorus means the bonding electrons are less attracted to the central atom, which reduces the repulsion between bond pairs. Thus, the bond angle decreases to around \( 93.5^\circ \). - In \( AsH_3 \), the even lower electronegativity of arsenic leads to further reduction in the bond angle, resulting in a bond angle of approximately \( 91.8^\circ \). ### Step 4: Arrange the bond angles in decreasing order Based on the analysis: - \( NH_3 \) has the largest bond angle. - \( PH_3 \) has a smaller bond angle than \( NH_3 \). - \( AsH_3 \) has the smallest bond angle. Thus, the decreasing order of bond angles is: \[ NH_3 > PH_3 > AsH_3 \] ### Final Answer The decreasing order of bond angles is \( NH_3 > PH_3 > AsH_3 \). ---

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Decreasing order of bond angle of (NH_(3), overset(o+)(NH_(4)), PCl_(5), SCl_(2)) is

`NH_(3) gt SCl_(2) gt PCl_(5) gt overset(o+)(NH_(4))`

`NH_(3) gt SCl_(2) gt overset(o+)(NH_(4)) gt PCl_(5)`

`overset(o+)(NH_(4)) gt NH_(3) gt SCl_(2) gt PCl_(5)`

`overset(o+)(NH_(4)) gt NH_(3) gt PCl_(5) gt SCl_(2)`

The decreasing order oof bond angles in BF_(2), NH_(3), PF_(3) and I_(3)^(-) is:

`I_(3)^(-) gtNH_(3) gtPF_(3) gtBF_(3)`

`BF_(3) gt I_(3)^(-)gt PF_(3) gt NH_(3)`

`BF_(3) gt NH_(3) gtPF_(3) gt I_(3)^(-)`

`I_(3)^(-) gt BF_(3) gtNH_(3) gt PF_(3)`

The decreasing order of bond angles from NH_(3) (107^(@)) to SbH_(3)(91.3^(@)) down the group 15 elements is due to :

increasing bond pair : bond pair repulsion

increasing p-orbital character on `sp^(3)` orbital

decreasing lone pair- bond pair repulsion

decreasing electronegativity.