How to compare P=O bond length of POF3, POCl3, POBr3 and PO(CH3)3 molecules?

All the given molecules have same hybridization (sp3) and same central atom (P) the we can apply Bent’s rule of hybridization. And according to Bent’s rule more electronegative atoms attach in those hybrid orbital’s which have less s-character or more electronegative atoms decreases s-character and increases p-character.

In case of POF₃  fluorine is more electronegative hence s-character  is increase more and more in P=O bond  as compare to other given molecules so P=O bond is shortest in POF₃  than others and also increase with decreasing electronegativity.  Hence P=O bond length order is x₁ < x₂ < x₃ < x₄

Related Questions:

What is Bent’s rule of hybridization?

Which of the following compound have longest (S=O)bond length , O=SF2, O=SCl2, O=SBr2.

Why Bond length of O-O is greater in H2O2 than O2F2?

How to arrange increasing (C-H) bond length in increasing order and H-C-F bond angle in the given compounds, CH4, CH3F, CH2F2 and CHF3 ?

Dipole moment of PCl3F2 molecule is zero while dipole moment of PCl2F3 molecule is non zero why?

Dipole moment of P(CH3)2(CF3)2 molecule is zero while dipole moment of P(CH3)2(CF)3 molecule is non zero why?

How to compare equatorial bond angle (F-S-F) and S-F bond length of SOF4 and SF4 molecules ?

How to compare equatorial bond angle (F-S-F) and S-F bond length of SOF4 and SF4 molecules ?

According to Bent's rule lone pair containing hybrid orbital have more s-character than double bond. 

In case of SF4 molecule has sp3d hybridization and one of the equatorial hybrid orbital containing lone pair it means it has more s-character and less p-character while  other remaining  equatorial hybrid orbitals have less s-character and more p-character hence its bond angle decreases and bond length increases.

Related Questions:

What is Bent’s rule of hybridization?

Which of the following compound have longest (S=O)bond length , O=SF2, O=SCl2, O=SBr2.

Why Bond length of O-O is greater in H2O2 than O2F2?

How to arrange increasing (C-H) bond length in increasing order and H-C-F bond angle in the given compounds, CH4, CH3F, CH2F2 and CHF3 ?

Dipole moment of PCl3F2 molecule is zero while dipole moment of PCl2F3 molecule is non zero why?

Dipole moment of P(CH3)2(CF3)2 molecule is zero while dipole moment of P(CH3)2(CF)3 molecule is non zero why?

 

Dipole moment of P(CH3)2(CF3)2 molecule is zero while dipole moment of P(CH3)2(CF)3 molecule is non zero why?

 According to bent’s rule more electronegative atom or group attached those hybrids orbital have minimum s-character.

There is sp3d (Sp2+pd) hybridization trigonal bipiramidal(TBP) geometry in P(CH₃)₂(CF₃)₂ molecule. we known that axial orbital (pd) have no s-character so –CF₃  groups are attached with axial positions while –CH₃ groups attach with equatorial (Sp2) position. Hence net dipole moment becomes zero. While in case of P(CH₃)₂(CF)₃ molecule one of the –CF₃ group is also present at equatorial position hence there is net dipole moment.

Related Questions:

What is Bent’s rule of hybridization?

Which of the following compound have longest (S=O)bond length , O=SF2, O=SCl2, O=SBr2.

Why Bond length of O-O is greater in H2O2 than O2F2?

How to arrange increasing (C-H) bond length in increasing order and H-C-F bond angle in the given compounds, CH4, CH3F, CH2F2 and CHF3 ?

Dipole moment of PCl3F2 molecule is zero while dipole moment of PCl2F3 molecule is non zero why?

According to bent’s rule more electronegative atom or group attached those hybrids orbital have minimum s-character.

There is sp3d (Sp2+pd) hybridization trigonal bipiramidal(TBP) geometry in PCl₃F₂ molecule. we known that axial orbital (pd) have no s-character so F  atoms are attached with axial positions while chlorine atoms attach with equatorial (Sp2) position. Hence net dipole moment becomes zero. While in case of PCl₂F₃ one of the fluorine atom is also present at equatorial position hence there is net dipole moment.

Related Questions:

What is Bent’s rule of hybridization?

Which of the following compound have longest (S=O)bond length , O=SF2, O=SCl2, O=SBr2.

Why Bond length of O-O is greater in H2O2 than O2F2?

How to arrange increasing (C-H) bond length in increasing order and H-C-F bond angle in the given compounds, CH4, CH3F, CH2F2 and CHF3 ?

Consider Al2(OH)6 compound and calculate the value of (X+Y)/Z Where X= total number of two centre-two electrons (2C-2e) bond, Y= total number of three centre-two electron(3C-2e)bonds, and Z=total number of three centre-four electron (3C-4e) bond;

 

Total number of two centre-two electrons (2C-2e) bonds(X)=6

Total number of three centre-two electron(3C-2e) bonds(Y)=4

Total number of three centre-four electron (3C-4e) bond(Z)=2

Hence (X+Y)/Z=(6+4)/2=5

Related Questions:

(1) What is the difference between the structure of AlCl3 and diborane?

(2) What is structural feature of monomeric and polymeric Beryllium halides ?

(3) What is bridge bond ? explaine 3C-4e bridge bond with suitable examples .

(4) What is structure of Beryllium chloride ?

(5) What is the structure of Aluminium tetrahydridoborate (Al(BH4)3) ?

(6) What is structure of Al(CH3)3 and also explain the bridge bonding in Al(CH3)3?

(7) The type of overlap in the bridge bond exist in Al2(CH3)6 is

(8) What is structure of diborane?

(9) What is structure of BeH2 in solid state ?

(10) Consider the structure of Al(CH3)3 given below and select the correct option?

(11) Consider Al2(OH)6 compound and calculate the value of (X+Y)/Z Where X= total number of two centre-two electrons (2C-2e) bond, Y= total number of three centre-two electron(3C-2e)bonds, and Z=total number of three centre-four electron (3C-4e) bond;