[Technicalhuman]

Since nitrogen as 3 singly filled electron in each of the outermost p orbitals, why can't the hydrogen form a sigma bond with those three and the lone pair of electrons in the s orbital just remain as it is? I think it has something to do with the positioning of the lone pair if it remains in the s orbital.

I am thinking that it has to hybridize to form an 4 sp3 orbitals so that the NH3 would have a electron pair geometry of tetrahedral. But I am basing this on the VSEPR model. But it doesn't really explain the need for the s orbitals electrons to be hybridized if we ignore the known VSEPR model. So why must it undergo hybridization?

[AWK]

VSEPR model consists a set of rules coming from quantum chemical calculation and in majority of cases gives a correct explanation. Why you disregard it?

[Technicalhuman]

VSEPR model consists a set of rules coming from quantum chemical calculation and in majority of cases gives a correct explanation. Why you disregard it?

Hmm i was thinking that since we have 3 singly filled p orbitals why don't the hydrogens bond to it without any need to hybridize. the same way PH3 does not hybridize. Why is that so?

[sjb]

VSEPR model consists a set of rules coming from quantum chemical calculation and in majority of cases gives a correct explanation. Why you disregard it?

Hmm i was thinking that since we have 3 singly filled p orbitals why don't the hydrogens bond to it without any need to hybridize. the same way PH₃ does not hybridize. Why is that so?

Why do you feel that the phosphorus in phosphine is not hybridised?

[Technicalhuman]

VSEPR model consists a set of rules coming from quantum chemical calculation and in majority of cases gives a correct explanation. Why you disregard it?

Hmm i was thinking that since we have 3 singly filled p orbitals why don't the hydrogens bond to it without any need to hybridize. the same way PH₃ does not hybridize. Why is that so?

Why do you feel that the phosphorus in phosphine is not hybridised?

Hi i thought it isn't as the bond angle is closer to 90 than 109.5 degrees. So wouldn't that indicate that no hybridization occurs and instead its just regular overlap between the p orbitals of P and s orbital of H?

[AWK]

Non-hybridized orbitals should show exactly 90 degrees (in this case).

[Technicalhuman]

Non-hybridized orbitals should show exactly 90 degrees (in this case).

I guess that there is a little distortion by the two lone pairs in the p and s orbitals? Because even in NH3 where there is hybridization the bond angle should be 109.5 degrees yet it shows that the bond angle is less than that. So i thought that the lone pair does not just affect unhybridized orbitals and instead it affects all orbitals. So it wouldn't be surprising that the angle is not exactly 90 degrees but it is close to it. So i thought that since it was closer to 90 degrees it indicates that no hybridization occurs.

Which part of that was wrong?

Thanks