# Why the formation of PH4+ is difficult compared to NH4+?

My textbook asks this as an objective question. What I had in mind was that the lone pair on phosporous was more distributed as compared to nitrogen(since nitrogen is more electronegative and smaller in size ) and hence the statement. But the option given in the textbook says the answer is lone pair of phosphorous resides in almost pure s orbital. How do you know if the lone pair is placed in pure s orbital or pure p orbital

The bond angle in $$\ce{PH3}$$ is about $$92-93^\circ$$, and by the percentage $$s$$ character formula i.e $$\cos(\theta) = s/(s-1)$$ you can calculate the $$s$$ character and that comes out to be 6%. And that where drago rule holds true , that is the orbitals involved in bonding are pure $$p_x$$ ,$$p_y$$, $$p_z$$ and as a character is high the lone pair is present in pure s orbital and highly penetrated towards phosphorus. Hence acts as a poor Lewis base.