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Why The phenoxide ion is more stable than phenol ?

The phenoxide ion is more stable than phenol, due to greater dispersal of negative charge, therefore. Equilibrium get shifted to forward direction. whereas alkoxide ion is less stable than alcohols due to intensification of negative charge, so equilibrium get shifted towards backward direction. Hence phenols are more acidic than alcohols. 

Related Questions:

Phenoxide ion is more stable than an alkaoxide. why?

Which is more stable? CH3-CH2+ (Ethyl carbocation) or CH2=CH+ (Vinyl carbocation)? And why?

No Doubt, Ethyl carbocation (CH3-CH2+) is more stable than vinyl carbocation (CH2=CH+). The greater stability of CH3-CH2+ carbocation over vinyl carbocation because the positive charge on the CH2 is stabilised by the inductive effect as well as hyperconjugation  from the neighbouring carbon atom.
But in case of CH2=CH+. The two carbon atoms involved are sp2 hybridised. An sp2 hybridised carbon atom is more electronegative which increases positive charge and increases as the s-character of a hybridised carbon atom increases. As the the result inductive effect decreases by the neighbouring carbon atom. 

Why Guanidine behave as the strongest base among neutral compounds?. explaine the remarkable basicity of guanidine.

Basicity of Guanidine:
Guanidine is the strongest base among neutral compounds:


The remarkable basicity of guanidine is attributed to the fact that the positive charge on the guanidinium ion is delocalized equally over the three nitrogen atoms, as shown by
these three equivalent resonating  structures:

Basicity of nitrogen can be increased by attachment topi-donors (NH2) group. These two pi-donating NH2groups donate electron density to the (pi-accepting) C=NH. Hence, the guanidinium ion is a highly stable cation.

What is Guanidine and its structure , name amino acid which contains guanidine group ?

Guanidine is strongest organic nitrogenous compound with the formula HNC(NH2)2. Guanidine is analogue of carbonic acid. That is, the C=O group in carbonic acid is replaced by a C=NH group, and each OH is replaced by a NH2 group.

A guanidine group also appears in larger organic molecules, including on the side chain of arginine (a basic amino acid).


It is a colourless solid that dissolves in polar solvents. It is a strong base that is used in the production of plastics and explosives. It is found in urine as a normal product of protein metabolism.

What is correct acidic strengths order of the haloforms acids ? Give correct explanation.

We know that acidic strength of the acid also depends upon stability of conjugate bases, so for relative strength of acid, we need to check the relative stabilities of their conjugate bases.

                                                     CF3-, CCl3-, CBr3- CI3-

We are expecting the acidic strength haloform acids asCHF3, CHCl3, CHBr3, CHI3 in decreasing order. Because Fluorine is most electronegative atom so it would be stabilize CF3- more, as electronegativity decreases from F to I the stability of conjugate -ve ion would be but that is not correct the actual order isCHCl3 > CHF3 > CHBr3 > CHI3.

This is because there is effective back bonding in CCl3-and hence the negative charge partially gets stabilised by back donation to the vacant 3d orbitals of Cl. Thus, CHCl3 is a stronger acid than CHF3 and also among them due to 2pπ-3dπ back bonding.


The acidic strengths of the other three haloforms can be compared the inductive effects of their anions. F is very electronegative and hence stabilises the negative charge on the C atom. So, CHF3 is a better acid than CHBr3, and the least acidic is CHI3.

 The overall acidic strength order is:
     CHCl3 > CHF3 > CHBr3 > CHI3.

Why are bridge head carbocations unstable?

According to Bredt’s rule a bridgehead carbon atom of bicyclo compound cannot be sp2 hybridised or in other word a bridgehead carbon atom cannot be form double bond. unless the ring that contains at least eight atoms.

Related Questions:

Why is cyclopropyl methyl carbocation exceptionally stable?

The exceptional stability of cyclopropane methyl cation can be explained by the concept of dancing resonance concept. The stability of additional cyclopropyl group , is result of more conjugation between the bent orbital of cyclopropyl ring and cationic carbon.
The most stable carbocation known till date in organic chemistry is explain by Dancing resonance.
Related Questions:

Azabicyclo[2,2,1]heptane is more basic thantriethylamine why?.

1-Azabicyclo[2,2,1]heptane is more basic than triethylamine . Because in case of triethylamine the lone pair of electrons is less available in the latter due to rapid nitrogen inversion. Nitrogen inversion is not possible in the bicyclic amine.

Which carbocation is more stable : Benzyl or Tertiary?

Actually answers of this question is always confusing, most of the authors believe benzyl carbocation is more stable than tertiary because benzyl carbocation involves in resonance.

But some of the authors believe that Tertiary carbocation is more stable as it involves maximum +I effect and maximum hyperconjuation +H (9-alpha hydrogens). Maximum +I and +H is more dominant than +M effect. Thus tertiary carbocation is more stable than benzyl carbocation.

Important note:

Stability of Benzyl , allylic and tertiary alkyl carbocation is practically almost same .so that stabilities infact cannot be compared.

Similar Questions:

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

Aqueous solution of borax reacts with two moles of acids this is because of formation of 2mol each of Na[B(OH)4]¯and B(OH)3of which only Na[B(OH)4]¯ reacts with acid. The chemical reaction given as below:
 Na2B4O7 + 7H2O ⟶ 2B(OH)3 + 2Na[B(OH)4]
B(OH)3 or H3BO3 is an acid and does not react with acid. Hence Na[B(OH)4] reacts with acid.
 

Related Questions:

Why Ga has small size than Al exceptionally

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

What is structure of solid Ortho Boric acid ?

What is the structure of trimetaboric acid and trimetaborate ion?  

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

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

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

Why B-F do not exist as dimer?. Explain.

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

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

 


How can derivatives relation between relative lovering of vapour pressure and molality?

A saturared solution of XCl3 has a vapour pressure 17.20 mm Hg at 20°C, while pure water vapour pressure is 17.25 mm Hg. Solubility product (Ksp) of XCl3 at 20°C is

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 ?

Silianol (SiH3OH) is more acidic than methanol (CH3OH) why?

We know that acidic strength of an acid also depend upon stability of its conjugate baseSo silianol is more acidic than methanol (H3C-OH) because conjugate base silianol (H3Si-OH) stablised by dispersal of negative charge in H3Si-O- ion by 2pπ—3dπ back bonding


Hence methanol is less acidic than silianol.
For more detail on Back bonding click here

Why Bond angle of NF3(102 degree) is lesser than in NH3 (107) ? explaine.

Bond angle of NF3(102 degree) is lesser than in NH3 (107) as per VSEPR theory which suggests that in case of less electronegative terminal atoms like H, Bond pairs would be closer to more electronegative central atom, N and hence bonds open up due to repulsion between  bond pairs electron density in vicinity. But bond angle of PF3(100) is greater than PH3, its due to possibility of back bonding in PF3 between lone pair of fluorine and vacant d-orbital of phosphorous (2pπ-3dπ) hence forth P-F bond acquires partial double character and we know well that multiple bonds causes more repulsion so bond angle is greater.

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

Which is more acidic Me3C-OH Or Me3Si-OH and why ?

Me3Si-OH is more acidic than Me3C-OH because after removal of H+ it's negative ions (Me3Si-O-)stabilized by 2pπ—3dπ back bonding. While Me3C-O- is unstable than Me3C-OH.

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

B-F bond length increases when BF3(130 pm) reacts with F- to form (BF4)- [143 pm]. Its due to absence of Back-bonding in (BF4)- hence B-F bond has completely single bond character

Dipole moment of P(CH3)2 (CF)3 is non zero while dipole moment of P(CH3)2(CF3)2 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 -CFgroup is  attached with axial positions.  Hence dipole moment of P(CH3)2(CF3)2  is zero.
(2)

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.

What is Banana bond (3C-2e bridge bond) ? Explaine with suitable examples.

3C-2e BOND OR BANANA BOND:
EXAMPLE FORMATION OF B2H6:
(1) Formation of 3C-2e bond in B2H6 is best explain by MOT and total number of bond in B2H6 is 6 (3C-2e=2 and 3C-4e=4)
(2) Bridge bonds are longer than terminal bond because at bridge bonds electrons are delocalized at three centres
(3)  Bond energy (441kj/mole) of B-H-B bond is greater than bond energy (381 K j/mole) of   B-H bond.
(4) Hybridization of B atom is sp3, so non planer, and non polar (U=0)
(5)  B2H6 Methylated up to B2H2 Me4
(6) B2H6 is hypovalent molecule hence act as Lewis acid and undergoes two type of cleavage when react with Lewis base

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

 3C-4e BOND or 3C-4e BRIDGE BOND:
Al2Cl6 Dimmerised by 3C-4e bond bridge bond:

Al2Cl6 is neither hypovalent nor hypovalent rather its octet is complete. We will used  MOT here  it cannot act as Lewis acid  due to crowding in spite   having vacant d orbital’s however Alcl3  act as Lewis acid .
Al2Cl6 contains six bond having two bridge bond(3c-4e) and four bond is (2C-2e)
Boron do not formed bridge bond because boron experience steric crowding.

If silver iodide crystallizes in a zinc blende structure with I- ions forming the lattice then calculate fraction of the tetrahedral voids occupied by Ag+ ions.

In AgI, if there are nI- ions, there will be nAg+ ions. As I- ions form the lattice, number of tetrahedral voids = 2n. As there are nAg+ ions to occupy these voids, therefore fraction of tetrahedral voids occupied by Ag+ ions = n/2n = ½ = 50%.

Arrange the silicon halides into decreasing order of Lewis acids Character? SiF3, SiCl3, SiBr3, SiI3

In case of silicone halides inductive effect dominate over back bonding hence lewis acid character decided by inductive effect.
Hence order of lewis acid character   SiF>SiCl3 > SiBr> SiI3

What is the d-Orbital resonance ?

D-ORBITAL RESONANCE:
It is a phenomenon in which electrons of ms and np get delocalized to vacant nd orbital because this availability of vacant d orbital to expect back bond get reduced .
In those molecules species where d orbital’s resonance exist of back Bonding is decreased.

N(CH3)3 is pyramidal while(SiH3)3N is trigonal planer why?


N(CH3)3 has sp3 hybridization & pyramidal shape at N, but in (SiH3)3N again there is 2pπ—3dπ back bonding between lone pair orbital of nitrogen into vacant orbital of silicon. Hence trisilyl amines is sp2, planer & is less basic than trimethyl amine.

Calculate the percentage degree of dissociation of an electrolyte XY2 ( normal molar mass=165) in water if the observed molar mass by measuring elevation in boiling point is 65.5.

Which of the following pair of solution (aq.) contain isotonic solution at same temperature? (Assume 100 % ionisation of electrolytes) (1) 0.1 M NaCI and 0.2 M CaCI2(2) 0.1 M NaCI and 0.3 M AICI3 (3) 0.3 M NaCI and 0.1 M AICI3 (4) 0.3 M NaCI and 0.2 CaCI2

Freezing point of pure liquid A is T K. If some amount of non-electrolyte non-volatile impurity is added in A, an ideal solution is formed. On cooling at 200 K, only 30% (by mass) liquid A is present and at 201 K, 60 % (by mass) liquid A is present. The value of T is .

Liquid A and B form ideal solution at temperature T. Mole fraction of A in liquid and vapour phase are 0.4 and 4/13 respectively, when total pressure is 130 torr. The vapour pressure (in torr) of A and B in pure state at temperature T are respectively.

We know that partial pressure of components of a binary solution containing A and B component.
(1) In liquid state:
P_A= P°_A X_A and  P_B= P°_B X_B
(2) In Vapour phase: 
P_A= P_T Y_A  and P_B= P_T Y_B

Where X_A and X_B is mole fraction in liquid state while Y_A and Y_B is mole fraction in vapour phase

What is the relation between Raoult' law and Dalton's ?

According to Raoult's Law: The partial pressure of any volatile component of a solution at any temperature is equal to the vapour pressure of the pure component multiplied by the mole fraction of that component in the solution.
      Where XA­ and XB is the mole fraction of the component A and B in liquid phase respectively

According to Dalton's Law:
The vapour behaves like an ideal gas, then according to Dalton's law of partial pressures, the total pressure PTis given by:
 Partial pressure of the gas = Total pressure x Mole fraction
                                        PA = PT YA and PB =PT YB
Where YA­ and YB is the mole fraction of the component A and B in gas phase respectively
Combination of Raoult's and Dalton's Law:
(3) Thus, in case of ideal solution the vapour phase is phase is richer with more volatile component i.e., the one having relatively greater vapour pressure

Graph Between 1/YA Vs 1/XA:
According to Dalton's law of partial pressures, the total pressure PT is given by:
 Partial pressure of the gas = Total pressure x Mole fraction
Where YA­ and YB is the mole fraction of the component A and B in gas phase respectively
According to Raoult's law:
On rearrangement of this equation we get a straight line equation:
Illustrative Examples:

Sodium metal crystallizes In body centered cubic lattice with the cell edge, a= 4.29 A.What is the radius of sodium atom? (III-.JEE 1994)

A metal crystallizes into two cubic phases, face centered cubic (fcc) and body centered cubic (bcc), whose unit cell lengths are 3.5 A° and 3.0 A°, rcspcctively. Calculate the ratio of densities of fcc and bcc. (IIT-JEE 1999)

The ratio of the densities of fcc to bcc is

A metallic element crystallizes into a lattice containing a sequence of layers ABABAB....Any packing of spheres leaves out voids in the lattice. What percentage by volume of this lattice is empty space? ( IIT-.IEF 2006)

The sequence ABABAB... indicates hcp unit cell. Now, there are 6 atoms per unit cell and volume of the umt cell is 24√2 r3.Thus, the packing fraction is:

The coordination number of Al in the crystalline state of AlCl3 is .... (IIT-JEE 2009)

AlCl3 exist in CCP unit cell with 6 co-ordinate layer structure and coordination number of Al is (6)

The number of hexagonal faces that are present in truncated octahedron is .... (IIT-JEE 2011)

The number of hexagonal faces that are present in truncated octahedron is  (6) this can seen if given fig.

In Diamond Carbon atom occupie FCC lattice point as well as alternate tetrahedral voids. If edge length of unit cell is 3.56 pm the find the radius of carbon atom is.

What are structural information of Dimomd ?

Only those atoms which form four covalent bond produce a repeated 3D structure using only covalent bonds i.g. diamond structure the latter is based  on of a face centred cubic lattice where four out of eight tetrahedral voids are occupied by canon atoms. Every atoms in this structure is surrounded tetrahedrally by four other. No discrete molecule can be dicemed in diamond .the entire crystal is giant molecule a unit cell of which is shown as below.

Lattice of Diamond is ZnS type structure.
(1) C- form FCC (4-atom)
(2) C- atoms present at the 
Alternative tetrahedral voids (4-atoms)
(3) Total Number of one lattice unit is eight(8) hence molecular formula of diamond is (C8)
(4) Number of C-C bond in lattice cell is = 4×4= 16
(5) Number of C-C bond per carbon atom is 16/8=2
(6) the distance between two Corbin atom is d_C-C = a√3/4
(7) PE = π√3/10= 0.34 or 34%
(8) Voids = 66 % 

Germanium, silicon and grey tin also crystallize in the same way as diamond.

2 millimolar solution of sodium ferrocyanide is 60% dissociated at 27°C. osmotic pressure of the slotution is .

Given that 2 millimolar solution of sodium ferrocyanide (Na4[Fe(CAN)6] ) is 60% dissociated at 27°C.
First find van't Hoff facter's (i) than Calculate osmotic pressure (π)= icRT as follow:

Decimolar solution of K4[Fe(CN)6] dissociate by 60% at 27°C. Determine osmotlc pressure in Nature/M2.

We know that osmotlc pressure is equal to the (π) = iCRT then  first Calculate the value of(i) then osmotlc pressure.

A complex is represented as CoCl3.xNH3.Freezing point of its 0.1 molar solution is - 0.558° C (Kf = 18.6) Assuming 100 % dissociation and coordination number of Co(iii) is six , the what is the complex formula ?

We can find the formula of Complex by :

A complex is wrltten as M(en)y.xBr. lts 0.05 molar solution shows 2.46 atm osmotlc pressure at 27°C. Assuming 100 % ionisaton and coordination number of metal (iii) is six, complex may be:

The Complex is M(en)y.xBr. and given c= 0.05 M, osmotlc pressure = 2.46 atm and The= 300K.

We know that osmotlc pressure (π)= icRT
Where To is universal gas constant = 0.0821j/mole per K
Then I = π/ cRT = 2.46/0.05×0.0821×300
                           =2 
  The vant Hoff facter's is (2) means their is two ions furnish after dissociation of Complex since  coordination number is (6) and van't hoff facter is (2) then formula of Complex is :
[M(en)2Br2]Br --> [M(en)2Br2]+  +Br- 

A non Stoichiometric oxide of iron represented as Fe_x O_1.0 , contains one Fe+3 for every three Fe+2 ions , then find the value of x is...

Given that one Fe+3 for every Fe+2 and total ions of Fe+3 and Fe+2 is equal to "x"  then number of Fe+3 is x/4 and Fe+2 is equal to 3x/4.
Finally we can calculate the value of x by balancing Oxidation state of Compound, given as :

Find the freezing point of a solution when 5 moI of Kl is mixed with 1 moI of HgCI2 in 3500 g of H20. Kf of H20 1.86 K/m. (KI + HgCI2 --> K2Hgl4 + KCI)

First find the composition of solution after reaction.

What is Gold number of protective colloidal solution?


Lyophilic sols are more stable than lyophobic sols.this is due to the fact that lyophilic colloids are extensively solvated, i.e. colloidal particles are covered by a sheath of the liquid in which they are dispersed.

Lyophilic colloids have a unique property of protecting lyophobic colloids. When a lyophilic sol is added to the lyophobic sol, the lyophilic particles form a layer around lyophobic particles and, thus, protect the latter from electrolytes.

Lyophilic colloids used for this purpose are called protective colloids.

The lyophilic colloids differ in their protective power. The protective power is measured in terms of Gold Number and is defined as the number of milligrams of a lyophilic colloid that will just prevent the precipitation
of 10 ml of a gold sol on the addition of 1 ml of 10% sodium chloride solution.

Lyophilic sols are more stable than lyophobic sols. Why ?

Lyophilic sols are more stable than lyophobic sols.this is due to the fact that lyophilic colloids are extensively solvated, i.e. colloidal particles are covered by a sheath of the liquid in which they are dispersed.

Lyophilic colloids have a unique property of protecting lyophobic colloids. When a lyophilic sol is added to the lyophobic sol, the lyophilic particles form a layer around lyophobic particles and, thus, protect the latter from electrolytes.
Lyophilic colloids used for this purpose are called protective colloids.

Give some examples of common negatively charged colloidal solution?

The common negatively charged colloids are given below:

As2S3 sol,
Sb2S3 sol, 
CdS sol, 
Au sol, 
Cu sol, 
Ag sol 
And acid dyes like
congo red.

Give some examples of common positively charged colloidal solution?

The common positively charged colloids are given below:
Fe(OH)3 sol, 
Cr(OH)3 sol,
Al(OH)3 sol, 
Ca(OH)2 sol, 
And also dyes like methylene blue and haemoglobin.

A mixture of weighing 228 g contain CaCl2 and NaCl . If this mixture is dissolved in 10 kg of water and form ideal solution that boil at 100.364 ℃ The mol % of NaCl in mixture is [ kb of water = 0.52 K per mol Kg]

Given that 228 g mixture of NaCl and CaCl2 consider that x g NaCl and (228-x)gm  CaCl2 present in this mixture:

NaCl and CaCl2 both are strong electrolytes hence bant Hoff facter are 2 and 3 respectively.

Moles of NaCl = x/ 58.5
Moles of CaCl2 = 228-x
And hence molality of NaCl and CaCl2 are 
m NaCl = ×/58.5×10kg
m CaCl2= 228-x/111×10kg

We know that 
∆Tb = i1m1+i2m2
100.364= 2(×/58.5×10kg)+3(228-x/111×10kg)

On solving we got x= 117 g 

Moles of NaCl= 117/58.5 =2
Moles of CaCl2= 111/111=1

Mol% 0f NaCl = 66.67%

What is fractional presence of Fe+2 in Fe0.94 O1.0 ?

Fe0.94 O 1.0 is non stachiometric compound hence Nickel contains both Fe+2 and Fe+3 .

We can calculate ℅  presence of Fe+2  and Fe+3 by oxidation method.

(F+2+Fe+3) O1.O
2x+3(0.94-x)-2=0
Fe+2=x=0.82 or 82% and Fe+3= 8%

What are coordination number (nearest neighbour) in a FCC at distance of (a/√2) and next nearest neighbour at distance (a) and next to next nearest neighbour at distance (a√3/√2) ?.

Coordination Numbers:

What are coordination number (nearest neighbour) in a BCC at distance of (a√3/2) and next nearest neighbour at distance (a) and next to next nearest neighbour at distance (a√2) ?.

Coordination Numbers:
(1) The nearest neighbour distance is half of body diagonal (a) root 3/2 (along body diagonal) thereforecoordination number for a given atom in BCC unit cell is 8.
(2) The next nearest neighbour are 6 at distance (a)Lattice parameter.
(3) 3rd neighbour (Next to Next nearest neighbour) are(12) at distance (a root 2) ( All corner along face diagonal in x,y and Z plane).



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