Question

Arrange the following in the increasing order of deviation from normal tetrahedral angle
A.${P_4} < P{H_3} < {H_2}O$
B.$P{H_3} < {H_2}O < {P_4}_{}$
C.${P_4} < {H_2}O < P{H_3}$
D.${H_2}O < P{H_3} < {P_4}$

Answer 1

Hint: A molecule having tetrahedral geometry is made up of 4 equally spaced $s{p^3}$ hybrid orbitals forming bond angle of ${109.5^ \circ }$. The shape of each orbital as well as the molecule is tetrahedral. As we all know the electrons repel each other, in a molecule, lone pair-lone pair repulsions are strongest followed by lone pair-bond pair repulsions and then bond pair-bond pair repulsions. The bond angles deviate from ideal values depending on the type of repulsions present in a molecule.

Complete step by step answer:
Tetraphosphorus $({P_4})$ exists as a molecule made up of four phosphorus atoms in a tetrahedral structure. Each phosphorus has 5 valence electrons, out of which they use 3 valence electrons in bonding with each other and two electrons are present as a lone pair on each of them. The presence of lone pairs in a tetrahedral geometry results in ring strain and instability. To counter this strain the bond angle deviates from ideal bond angle of ${109.5^ \circ }$ to ${60^ \circ }$
Phosphine $(P{H_3})$ molecule has Trigonal pyramidal geometry. Phosphorus exist as central atom and uses 3 of its valence electrons in bonding with 3 hydrogen atoms out of total 5 valence electrons and the rest of two unbonded electrons exists as a lone pair, the $H - P - H$ bond angles are ${93.5^ \circ }$.
In water$({H_2}O)$ molecule oxygen is the central atom and is $s{p^3}$ hybridized with three $2p$ orbitals and one $2s$ orbital. Out of six valence electrons of oxygen 2 electrons are used in bonding with 2 hydrogen atoms and four electrons are present as two lone pairs of oxygen. Because of repulsions in lone pairs ${H_2}O$ has distorted tetrahedron geometry and the bond angle deviates from ${109.5^ \circ }$to ${104.5^ \circ }$
Comparing the bond angles of ${P_4}$,$P{H_3}$ and ${H_2}O$
${H_2}O({104.5^ \circ }) < P{H_3}({93.5^ \circ }) < {P_4}({60^ \circ })$

Therefore, the correct option is option D ${H_2}O < P{H_3} < {P_4}$.

Note:
Phosphine $(P{H_3})$ is a drago molecule. In other words it follows drago’s rule which states that if the central atom has at least one lone pair and has high electronegativity difference with surrounding atoms then the hybridization can’t take place. Thus, pure $p$ orbitals are formed.