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NH3 -- 107.5
PH3 -- 93.5
AsH3 -- 92
SbH3 -- 91.5
AsH3 -- 90
The H-E-H bond angle is expected to be a tetrahedral ideal of 109.5°, but since lone pairs repel more than bonding pairs, the actual angle would be expected to be slightly smaller. Two possible explanations are possible for the difference between NH3 and the other hydrides.
1. The N-H bond is short (1.015 Ã
) compared to the heavier analogs, and nitrogen is more electronegative than hydrogen, so the bonding pair will reside closer to the central atom and the bonding pairs will repel each other opening the H-N-H angle more than observed for PH3, etc.
2. The accessibility of the 2s and 2p orbitals on nitrogen allows for hybridization and the orbitals associated with N-H bonding in NH3 are therefore close to sp3 in character, resulting in a close to tetrahedral geometry. In contrast, hybridization of the ns and np orbitals for P, As, etc., is less accessible, and as a consequence the orbitals associated with P-H bonding in PH3 are closer to p in character resulting in almost 90° H-P-H angle. The lower down the Group the central atom the less hybridization that occurs and the closer to pure p-character the orbitals on E associated with the E-H bond.
This implies that as we go down the group, the %hydridisation decreases and mostly pure p orbitals are used
NF3 -- 102°9'
NCl3 -- 107.1° (unstable)
NBr3 -- highly unstable
NI3 -- highly unstable (explodes on touching by feather
http://www.chm.bris.ac.uk/motm/ni3/ni3j.htm)
PF3 -- 96.3°
PCl3 -- 100°6'
PBr3 -- 101°
PI3 -- 102°
AsF3 - 95.2°
Reason for Bond Angles increasing From -F to -I
The steric factor. Cl, Br and I are quite large in size. .So due to the increased size, the bond angle increases from -F to -I
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13 Answers
1asish, can u explain what does "accessibilty and less accessible" mean in above post??
11The accessibility of the 2s and 2p orbitals on nitrogen allows for hybridization and the orbitals associated with N-H bonding in NH3 are therefore close to sp3 in character, resulting in a close to tetrahedral geometry. In contrast, hybridization of the ns and np orbitals for P, As, etc., is less accessible, and as a consequence the orbitals associated with P-H bonding in PH3 are closer to p in character resulting in almost 90° H-P-H angle.
I think you are asking about this smsm.
The accessibility of 2s and 2p orbitals means that these orbitals can combine together to form
hybridized sp3 orbitals similar that are formed during the formation of methane. So
hybridization and the orbitals associated with N-H bonding in NH3 are therefore close to sp3 in
character, resulting in a close to tetrahedral geometry
But in the case of P, As these outer ns and np orbitals (n = 3 for P and n = 4 for As),
because of their larger size are not accessible,i.e, they cannot approach each other for the
formation of hybridized orbitals like Nitrogen ( the reason for this is since P and As are
larger in size, the outer orbitals are away from the nucleus and hence nuclear energy on
them is less). So here np and ns can't approach each other for hybridization. Here they
use their p orbital during the formation of any compound so they have more p character.
Hope am clear [1]
11Of course its from google, If I get the info in net that I know I directly copy and paste it.
