For phosphorous, we can initially assume that the bond angle of $\ce{PH3}$ is close enough to $90^\circ$ for no hybridisation to be necessary (semi-proven by this answer of Martin) and likewise for $\ce{PF3}$. The bond lengths are $142~\mathrm{pm}$ and $156~\mathrm{pm}$, respectively. The difference in bond lengths is only half of that of the ...

2014年7月3日 · The $\ce{H-N-H}$ bond angle in ammonia is around 107 degrees. Therefore, the nitrogen atom in ammonia is roughly $\ce{sp^3}$ hybridized and the 4 orbitals emanating from nitrogen (the orbitals used for the 3 bonds to hydrogen and for the lone pair of electrons to reside in) point generally towards the corners of a tetrahedron.

Nitrogen is more electro-negative than phosphorous and hence attracts electron more towards itself from bond pair with hydrogen, as compared to attraction of electrons by phosphorous in P H 3. The electrons in N H 3 come closer to nitrogen and …

As we move down the group, radius of elements increases and electronegativity decreases so bond angle decreases. so bond order is N H 3 > P H 3 > A s H 3 > S b H 3 107 93 92 91

As electronegativity of P is much lower than N, so in N H 3 nitrogen attracts all bond electron towards the centre and electron-electron repulsion is higher so it has a higher bond angle. Bond angle in ammonia is 107 & that in phosphene is 93.5.

Hence repulsion between bond pairs in N F 3, is less than N H 3. As a result the bond angle decreases to 102.4 ° , whereas in N H 3 it decreased to 107.3 ° only. P H 3 and P F 3 are also pyramidal in shape with one lone pair on P .

Therefore the bond angle in P H 3 molecule lesser than in N H 3 bond angle in P H 3 is 93.5 and in N H 3 is 107.8. Option A is correct. Was this answer helpful?

2017年10月10日 · But in the case of PH3 the electron cloud is farther away from the central atom(P) than in the case of NH3. So the bond pair - bond pair repulsion is comparatively lesser, causing the 3 H atoms to move closer together to an angle of almost 90°, resembling the px, py, and pz orbitals, as a consequence it's lone pairs are present in almost pure ...

2015年7月10日 · I noticed the fact that all the hydrides of the elements belonging to group IV has bond angle 109.5° while in group V it varies from 107.3° for $\ce{NH3}$ to 91.3° for $\ce{SbH3}$. Similarly we see in group VI that water has highest bond angle then it significantly decreases to 89.5° for $\ce{H2Te}$.

2014年7月31日 · In case of $\ce{NH3}$ due to higher bond pair bond pair repulsion (since electronegativity of $\ce{N}$ atom is very high hence it attracts bonded electrons of $\ce{N-H}$ bond towards itself) bond pair moves away from each other and hence shows greater bond angle. This may be considered one of the major factors for less hybridization.)