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Showing posts with label BENT'S AND DRAGO'S RULES. Show all posts
Showing posts with label BENT'S AND DRAGO'S RULES. Show all posts

Tuesday, January 19, 2021

Which of the molecule have strongest P=O bond present the given molecules are POF3, POCl3, POBr3 and PO(CH3)3 ?

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  and Bond strength order is reverse of bond length.

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


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 ?


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?

 


Monday, January 18, 2021

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)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.

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.

Sunday, January 17, 2021

What is the structure of (O2F2) oxygen fluoride

The compound id produced as result of glow dischage through amixture of O2 and F2 at -180 to -190 degree (c). As should be expected. O2F2 is an extremely reactive fluorinating agent. Under the conditions that produce OF2, small and O4F2 are produced, but these unstable compounds decomposed at liquid nitrogen temperatures.

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

Electronegativity of F is much more than hydrogen and also hybridization of oxygen atoms in of H2O2 and O2F2 both have (sp3) same and hence we can apply bent’s rule.

According bent’s rule more eletronegative atoms reduce % s-character (or increases % p-character vice versa) of those hybride orbital in which they attach. So in case of O2F2 % s-character of those hybride orbital decrease which have  flourine while % p-character increases in same way % s-character of  remaing hybride orbital inceases and p-character decreases hence its bond length also (bond length is directly proportional to % p-character) decreases.  hence O-O bond length in O2F2 is shorter thane H2O2.

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.

Wednesday, January 6, 2021

Why N-N bond length of H2N-NH2 is greater than N-N bond length of F2N-NF2 molecules?

Electronegativity of F is much more than hydrogen and also hybridization of nitrogen atoms in of H2N-NH2 and F2N-NF2 both have (sp3) same . According bent’s rule those hybrid orbitals containing flourine atom have less % s-characterand more p-character while other orbitals have more % s-character and less % s-character hence N-N bond length in F2N-NF2 is shorter than H2N-NH2

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.

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

The central atoms of all the given conpund (O=SF2, O=SCl2, O=SBr2) are same and also have same hybidization (Sp3) hence we can applied  Bent’s rule of Hybidization . According to bent rule more electronegative elements increases % P-character  and decreases % S-character  of  that orbital in its  attach. While  in other orbital % S-character increase that why bond length decrease. Hence S=O bond length depends upon electronegativitiy of other attach atoms (F>Cl>I).

The overall S=O bond length order is   O=SBr2 > O=SCl2 > O=SF2O

Related Questions: 

Why aqueous solution of AlCl3 is acidic in nature ?

What happen when aq AlCl3 react with Acid or Base?

Although anhydrous aluminium chloride is covalent but its aqueous solution is ionic in nature. Why?

Why BF3 do not exist as dimer?. Explain.

Why B-F bond length in BF3 is shorter (130 pm) than B-F bond Iength in BF4- (143 pm)?. Explain.

B-F bond length in BF3 is shorter than B-F bond length in (BF4)- why?

When B2H6 is allowed to react with following Lewis bases, then how may given Lewis base form adduct through symmetrical Cleavage of B2H6.

What is product of reaction between diborane (B2H6) and ammmonia (NH3)?

Why methylation of Diborane (B2H6) replace four hydrogen only ?

What is Use of Boric Acid?

What is use of Orthoboric acids?

What is basicity of "Boric acid" ?

Why Boric acid exist in solid state ?

What is structure of solid Ortho Boric acid ?

What is effect of heat on Borax?

What is the structure of trimetaboric acid and trimetaborate ion?

What is the Sodium per borate ,give the structure and its uses?

Why aqueous solution of borax reacts with two moles of acids ?

What is the molecular formula of Borax ?

Why Boric acid become strong acid in the presence of cis 1,2-diol or 1,3-diol ?

Why Borazine is more reactive than benzene towards Electrophic Aromatic substitution reactions ?

Why Borazine (B3N3H6) is also known as inorganic benzene ?.

Four-center two-electron bond (4C-2e Bond): Structure of AlCl3:

What is the difference between the structure of AlCl3 and diborane?


Monday, December 28, 2020

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

Electronegativity of F is much more than hydrogen and also hybridization of oxygen atoms in of H2O2 and O2F2 both have (sp3) same . According bent’s rule those hybrid orbitals containing flourine atom have less % s-characterand more p-character while other orbitals have more % s-character and less % s-character hence O-O bond length in O2F2 is shorter thane H2O2.

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.





What is Bent’s rule of hybridization?

According to the bent's rule the more electronegative atom not only prefer to stay in that Hybride orbital which having less  % S character (more p-character) but it also decreases % S-character and increases % P-character in its attached orbital from the central atom.


On increasing % s character in hybrid orbital , the bond length  decreases while bond angle increases.

For example:C-H bond of CH4 is longer than C-H bond of floromethan (CH3F) because In CH4 all the Sp3 hybrid orbitals are equal in term of  s-% character(25%) and % p-charater(75%)  so that all C-H bond in CH4  are equal while in case of CH3F the hybrid orbital has more p-character (more than 75%) and less s-character (less then 25%) containg fluorine atom and other hybrid orbitals have less p-character (less than75%) and more s-character (more than 25%) hence C-H bond length of CH3F decreases.

EXCEPTIONS OF BENT’S RULE:

 (1): Bent’s rule is applicable in those molecules where central atoms are same and they are also in same Hybridization.

For example N-N bond length cannot be compared in N2H4 and N2O4 using Bent rule.

(2): Bent’s rule violets in those molecules where steric factor’s plays dominating rule.

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.

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 ?

According to the bent's rule the more electronegative atom not only prefer to stay in that Hybride orbital which having less  % S character (more p-character) but it also decreases % S-character and increases % P-character in its attached orbital from the central atom depending on circumstance.



Sunday, May 24, 2020

Why the C-C bond length in graphite is shorter than C-C bond length of diamond?

Graphite has carbon in sp2 hybrid state (33.3 % s character) but diamond has carbon in sp3 hybrid state (25 % s characrer). More is the percentage of s characters, (by Bent rule) more is the bond multiplicity and hence, shorter the bond is. Thus carbon-carban bond in graphite has double bond character but has a single bond character in diamond. Hence, C-C bond length in graphite is shorter than than in diamond.

Related Questions:

Is all the C-C bond length in fullerene is equal ?

Explain C-H bond length of CH4 is longer than C-H bond length of Difloromethan (CH2F2) ?

Explain C-H bond length of CH4 is longer than C-H bond length of Difloromethan (CH2F2)  because according to bent rule more electronegative atoms decreases S-character in hybrid orbitals and the same time P- Character increases as results bond length increases for those hybrid orbital have more electronegative atoms and bond length of other orbitals decrease having less electronegative atom present hence C-H bond length of CH4 is longer than C-H bond length of Difloromethan (CH2F2

Dipole moment of PCl3F2 and P(CH3)3(CF)2 molecules are zero while dipole moment of PCl2F3 and P(CH3)2(CF3)3 are non zero why?

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


There is in trigonal bipiramidal (TBP) Geometry we known that axial orbital hare no s-character so F and -CFgroup are attached with equatorial positions.  Hence dipole moment of P(CH3)2(CF3)2  is zero. While in case of PCl2F3 and P(CH3)2(CF)3 molecules one of the F and –CF3 group are also present at equatorial position hence there is net dipole moment.



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

According to bent rule more electronegative atom or group attached those orbital have minimum S- character. There is in Trigonal bipiramidal (TBP) Geometry we known that axial orbital hare no S- character so F atom attached with axial positions only. Hence PCl3F2 has zero dipole moment.

Sunday, February 9, 2020

What are the structural difference between oxides ( P4O6 and P4O10) of phosphorous?

According to “Bent Rule” loan pair prefer to stay to in those atomic orbital have more (S) character. And bond length is directly proportional to p-character. 

(1) In case of P4O6 molecules atomic orbital containing loan pair have more (s)-character and less (P) character hence shorter bond length while remaining orbitals have less (S) character and more (P) character hence longer bond length (X1) than (X2) in P4O10 .

(2) We know that On increasing % s character in hybrid orbital, the bond length decreases while bond angle increases.

(3) In Case of P4O6   (O-P-O) bond is Smaller due to higher P character in atomic orbital than orbital containing (O-P-O) in P4O10(have more s-character )

For reading more Details about  click on  Bent Rule andDrago's rule

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.


Sunday, December 16, 2018

Bent's rule of hybridization:


Conceptual facts: The more electronegative atom prefer to stay in that hybrid orbital having less S-character or more p- character and more electropositive atom prefer to that hybrid orbital which have more S-character or less p-character.

Explaination: The more electropositive atom or group will withdraw the bond pair more from central atom. it is easy when hybrid orbital is having less S-character and more P-character .
 S-orbital is closer to the nucleus so it electronegativity is more than p-orbital.

ILLUSTRATIVE EXAMPLE (1): PCl3F2 (sp3d Hybridization) and TBP (Trigonal bi pyramidal)
Where angle < F-P-F =180 and <Cl-P-Cl =120 two P-Cl bond is axial while three P-Cl bond is equatorial                                  

(1) APICOPHIOLICITY IN TBP GEOMETRY:

(1) Trigonal bipiramidal geometry more electronegative atom prefers to stay in low electronegative PZdz2 orbital of sp3d hybridization. You can also say that more electronegative atom prefer that hybrid orbital having low s-character or no s-character.
(2) If electronegativity difference between central atom and surrounding atom is large then due to polarity some ionic character is developed and covalent character decreased.
(3)The poor covalence is not only due to electronegativity difference between bonding atom.  it is also generated due to poor overlapping, steric hindrance or mismatch of the overlapping orbitals.
(4) Position of lone pair: Bent rule is very important in predicting position of lone pair. lone pair is attracted by only one nucleus while bond pair is attracted by two nucleuses. Central atom hold lone pair cloud tightly if central atom is having more S-Character.

ILLUSTRATIVE EXAMPLE (2): SF4 (Sp3d) TBP and SF2Cl2 (Sp3d)
ILLUSTRATIVE EXAMPLE (3): XeF2 and XeO3F2
(2) ORBITAL ANALYSIS (CALCULATION OF (% S ) AND ( %P) CHARACTER:  
S% character is equatorial orbital at1 200c :
S% character for orbital at 900 C:



(3) Effect of the straingth of covalency: (Alternate Statement of bent):

RULE: The more electronegative atom not only prefer to stay in that orbital which having less  % S character (more p-character) but it also decreases % S-character and increases % P-character in its attached orbital from the central atom depending on circumstance.
On increasing % s character in hybrid orbital , the bond length  decreases while bond angle increases.

ILLUSTRATIVE EXAMPLE (3): : Explain C-H bond length of CH4 is longer than C-H bond length of Difloromethan (CH2F2) ?

EXCEPTIONS OF BENT’S RULE:

(1): Bent’s rule is applicable in those molecules where central atoms are same and they are also in same Hybridization. For example N-N bond length cannot be compared in N2H4 and N2O4 using Bent rule.

ILLUSTRATIVE EXAMPLE (4): Arrange PF3 ASH3, PH3, NH3, H2Se in decreasing order by Bent’s rule here we used DRAGO’S RUEE”
SOLUTION: NH3> NF3>PF3> PH3>AsH3>H2Se
(2): Bent’s rule violets in those molecules where steric factor’s plays dominating rule.

ILLUSTRATIVE EXAMPLE (5): Compared Bond angle among H2O OF2, OMe2, OCl2  
SOLUTION: 
ILLUSTRATIVE EXAMPLE (6):   IN CH2SF4 


Which of the following option is correct regarding ?

(1) 1800> 120                                                       (2) 1800> >120
(3) 1200> >900                                                         (4) 900 > >00

SOLUTION: S -character present in equatorial so more decrease in Bond angle in equatorial orbital than in axial. Because no S-character in present in axial orbital.
So 1200 > > 90 and 1800 > > 120