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The most stable canonical structure among all of above is:
- a)Only I
- b)Only II
- c)Only III
- d)All are equally stable
Correct answer is option 'C'. Can you explain this answer?
The most stable canonical structure among all of above is:
a)
Only I
b)
Only II
c)
Only III
d)
All are equally stable
|
Vikram Kapoor answered |
The structure which has a maximum double bond is more stable.
Equivalent weight of H3PO4 in the above reaction is:
Correct answer is '49'. Can you explain this answer?
Equivalent weight of H3PO4 in the above reaction is:
|
Bhavana Pillai answered |
H3PO4 is an acid and the equivalent mass of an acid is calculated by using the formula:
From the given reaction:
Ca(OH)2 + H3PO4 → CaHPO4 + 2H2O
As, the replaceable hydrogen ions in acid is 2, by putting the values in above equation, we get:
Hence, The equivalent mass of H3PO4 for the given reaction is 49 g eq.
Strength of following bases decrease in the order?
(I) Br- (II) F- (III) NH2 (IV) CH3-
- a)IV > III > I>II
- b)III > IV > I > II
- c)II > I > III > IV
- d)IV > I > II > III
Correct answer is option 'A'. Can you explain this answer?
Strength of following bases decrease in the order?
(I) Br- (II) F- (III) NH2 (IV) CH3-
a)
IV > III > I>II
b)
III > IV > I > II
c)
II > I > III > IV
d)
IV > I > II > III
|
|
Vedika Singh answered |
-
Br⁻ (bromide ion): This is a halide ion. Halide ions become stronger bases as the size of the ion increases (i.e., as we move down the periodic table). Bromide is quite large, but it is not as large as iodide, making it a relatively weaker base compared to other larger halides.
-
F⁻ (fluoride ion): Fluoride is a small ion with high electronegativity, which means it holds onto its electron pair tightly and is less inclined to share it to form a bond with a proton. Thus, in an aqueous solution, fluoride is a weaker base than larger halide ions.
-
NH₂⁻ (amide ion): This ion has a nitrogen atom, which is less electronegative than fluorine, holding the negative charge. Nitrogen can share its electron pair more easily than fluorine. Also, the negative charge on the amide ion is not as effectively stabilized because nitrogen is less electronegative than fluorine.
-
CH₃⁻ (methide ion): This ion has a carbon atom bearing the negative charge. Carbon is less electronegative than nitrogen and fluorine, so it is more ready to share its electron pair compared to NH₂⁻ and F⁻. However, the negative charge is not delocalized and is concentrated on a single carbon atom, making it a strong base.
Considering these factors, the order of increasing base strength from weakest to strongest should be:
F⁻ < Br⁻ < NH₂⁻ < CH₃⁻
This order takes into account that fluoride is very electronegative and holds onto its electron pair tightly, bromide is a larger halogen and can distribute its negative charge over a larger volume, the amide ion is less stable due to the higher energy of nitrogen-based anions, and the methide ion is a strong base due to carbon's low electronegativity and the localized negative charge.
The correct order of stability among the following canonical structure is: - a)I > II > III
- b)II > I > III
- c)III > I > II
- d)I > III > II
Correct answer is option 'D'. Can you explain this answer?
The correct order of stability among the following canonical structure is:
a)
I > II > III
b)
II > I > III
c)
III > I > II
d)
I > III > II
|
Sinjini Nair answered |
Negative charge are stable at more Electronegative atom so the order of Electronegative is
O>N>C
stability order
1>3>2
O>N>C
stability order
1>3>2
Which of the following is the most stable conformer of the following molecule:
- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Which of the following is the most stable conformer of the following molecule:
a)
b)
c)
d)
|
Asf Institute answered |
Correct Answer :- B
Explanation : When the methyl group in the structure above occupies an axial position it suffers steric crowding by the two axial hydrogens located on the same side of the ring. The conformation in which the methyl group is equatorial is more stable, and thus the equilibrium lies in this direction.
Therefore,the equatorial bonds are the most stable in chair form.
How many of the following inter halogen species have 2 lone pairs of electrons on the central atom? ClF3, ClF2-, CIF5 ICl2- and XeF2
Correct answer is '2'. Can you explain this answer?
How many of the following inter halogen species have 2 lone pairs of electrons on the central atom? ClF3, ClF2-, CIF5 ICl2- and XeF2
|
Dronacharya Institute answered |
Chlorine trifluoride has 5 regions of electron density around the central chlorine atom (3 bonds and 2 lone pairs). These are arranged in a trigonal bipyramidal shape with a 175° F(axial)-Cl-F(axial) bond angle.
In a ClF5 molecule, chlorine is the central atom featuring five single bonds with fluorine atoms. Since chlorine has 7 electrons in the valence shell, the remaining two electrons form a lone pair.
In a ClF5 molecule, chlorine is the central atom featuring five single bonds with fluorine atoms. Since chlorine has 7 electrons in the valence shell, the remaining two electrons form a lone pair.
If a particle has linear momentum 
at position 
, then its angular momentum is:
- a)
- b)
- c)-15i -2j +39k
- d)
Correct answer is option 'C'. Can you explain this answer?
If a particle has linear momentum at position , then its angular momentum is:
at position , then its angular momentum is:a)
b)
c)
-15i -2j +39k
d)
|
Veda Institute answered |
Angular momentum is equal to the cross product of the position vector and the linear mamentum.
J=r×p
(4i + 9j + 2k) × (3i - 3j + k)
= -15i-2j+39k
XeO2F2 has shape: a) See-Sawb)TBPc)Octahedrald) Square pyramidalCorrect answer is option 'A'. Can you explain this answer?
|
|
Sinjini Nair answered |
XeO2F2 molecular geometry is originally said to be trigonal bipyramidal but due to the presence of lone pair on equatorial position, the actual shape will be see-saw. The repulsion between bond pair and lone pair of electrons will be more.
Find the corresponding subshell utilizing the information from graph. 
- a)3s
- b)3p
- c)3d
- d)2p
Correct answer is option 'A'. Can you explain this answer?
Find the corresponding subshell utilizing the information from graph.
a)
3s
b)
3p
c)
3d
d)
2p
|
Chirag Verma answered |
According to radial nodes formula:
Radial Nodes = n - l – 1
= 3 – 0 – 1
=2 nodes.
Therefore 3s.
Hence, A is the correct answer.
Therefore 3s.
Hence, A is the correct answer.
The correct order of stability for the given canonical structures is:- a)I > III > II
- b)III > I > II
- c)II > III > I
- d)II > I > III
Correct answer is option 'B'. Can you explain this answer?
The correct order of stability for the given canonical structures is:
a)
I > III > II
b)
III > I > II
c)
II > III > I
d)
II > I > III
|
|
Veda Institute answered |
There are a certain set of rules that must be followed in order to check the stability of resonating structures.
These rules have a preference so we have to follow the order:
These rules have a preference so we have to follow the order:
- Uncharged species is more stable than charged species.
- Charged species having more number of covalent bond are more stable.
- The species in which the octet rule is followed by every atom is more stable than the species in which the octet rule is violated.
- Species in which opposite charges are closer are more stable than species having same charges closer.
- Species with negative charge on more electronegative atom while positive charge on more electropositive atom is more stable.
According to above discussion, B is the correct answer.
Consider two molecules A & B.
∠α = ∠HCH; ∠β = ∠FCFWhich of following is true?- a)α = β
- b)α > β
- c)β > α
- d)Insufficient information
Correct answer is option 'A'. Can you explain this answer?
Consider two molecules A & B.
∠α = ∠HCH; ∠β = ∠FCF
Which of following is true?
a)
α = β
b)
α > β
c)
β > α
d)
Insufficient information
|
|
Vedika Singh answered |
Though the substituting atoms are different in both cases but the hybridization of carbon is the same that is sp3, having the bond angle of 109′28″.
Examine the following resonating structures of formic acid for their individual stability and then answer the question given below:
Which of the following arrangements gives the correct order of decreasing stability of the above- mentioned resonance contributors?- a)II > I > III > IV
- b)I > II > III > IV
- c)IV > III > I > II
- d)IV > III > II > I
Correct answer is option 'B'. Can you explain this answer?
Examine the following resonating structures of formic acid for their individual stability and then answer the question given below:
Which of the following arrangements gives the correct order of decreasing stability of the above- mentioned resonance contributors?
a)
II > I > III > IV
b)
I > II > III > IV
c)
IV > III > I > II
d)
IV > III > II > I
|
|
Dronacharya Institute answered |
There are certain sets of rules that must be followed in order to check the stability of resonating structures.
These rules have a preference so we have to follow the order:
These rules have a preference so we have to follow the order:
- Uncharged species are more stable than charged species.
- Charged species having more number of covalent bond are more stable.
- The species in which octet rule is followed by every atom is more stable than the species in which octet rule is violated.
- Species in which opposite charges are closer are more stable than species having same charges closer.
- Species with negative charge on more electronegative atoms while positive charge on more electropositive atoms are more stable.
According to the above discussion, B is the correct answer.
Select incorrect statement(s):- a)At very low pressure, real gases show minimum deviation from ideal behaviour.
- b)The compressibility factor for an ideal gas is zero.
- c)At Boyle temperature, real gas behave as ideal gas in high pressure region
- d)Real gas show minimum deviation at high pressure and low temperature
Correct answer is option 'B,C'. Can you explain this answer?
Select incorrect statement(s):
a)
At very low pressure, real gases show minimum deviation from ideal behaviour.
b)
The compressibility factor for an ideal gas is zero.
c)
At Boyle temperature, real gas behave as ideal gas in high pressure region
d)
Real gas show minimum deviation at high pressure and low temperature
|
Edurev.iitjam answered |
(b) Compressibility factor of an ideal gas is 1.
(c) Real gas behaves like ideal gas at high temperature and low pressure.
(c) Real gas behaves like ideal gas at high temperature and low pressure.
Lattice energy (numerical value) of chloride of alkali metals is in order:- a)LiCl > NaCl > KCl > RbCl > CsCl
- b)LiCl < NaCl < KCl < RbCl < CsCl
- c)NaCl < KCl < LiCl < RbCl < CsCl
- d)NaCl < KCl < RbCl < CsCl < LiCl
Correct answer is option 'A'. Can you explain this answer?
Lattice energy (numerical value) of chloride of alkali metals is in order:
a)
LiCl > NaCl > KCl > RbCl > CsCl
b)
LiCl < NaCl < KCl < RbCl < CsCl
c)
NaCl < KCl < LiCl < RbCl < CsCl
d)
NaCl < KCl < RbCl < CsCl < LiCl
|
Milan Majumdar answered |
Lattice energy is the attraction force acting between the cation and anion.
Greater the size of cation, lower is the attractive force.
On moving down the group, size of cation increases then,
- Attractive force decreases.
- Lattice energy decreases.
What is the value of PV/nRT for ideal gas?
Correct answer is '1'. Can you explain this answer?
What is the value of PV/nRT for ideal gas?
|
Neha Choudhury answered |
In SI units, P is measured in pascals, V is measured in cubic metres, n is measured in moles, and T in kelvins (the Kelvin scale is a shifted Celsius scale, where 0.00 K = −273.15 degree C, the lowest possible temperature). R has the value 8.314 J/(K. mol) ≈ 2 cal/(K. mol), or 0.08206 L.
The table indicates the value of van-der Waal’s constant ‘a’. The gas which can be liquefied most easily is:- a)02
- b)N2
- c)NH3
- d)CH4
Correct answer is option 'C'. Can you explain this answer?
The table indicates the value of van-der Waal’s constant ‘a’. The gas which can be liquefied most easily is:
a)
02
b)
N2
c)
NH3
d)
CH4
|
Aryan Gupta answered |
The ease of liquification of a gas depends on their intermolecular force of attraction which in turn is measured in terms of van der Waals’ constant ‘a’. Hence, higher the value of ‘a’, greater the intermolecular force of attraction, easier the liquification. In the present case, NH3 has highest ‘a’, can most easily be liquefied.
Total number of delocalized lone pairs (combined) in [Adenine & Guanine]:
Correct answer is '4'. Can you explain this answer?
Total number of delocalized lone pairs (combined) in [Adenine & Guanine]:
|
Pawan Kajal answered |
The line pair of an atom will undergo conjugation if it is attached by single bond to that atom which has further multiple bond in conjugation.
Consider AB2 molecule 
Calculate % s-character in A—B bond:
Correct answer is '46-47'. Can you explain this answer?
Consider AB2 molecule Calculate % s-character in A—B bond:
Calculate % s-character in A—B bond:|
|
Saikat Ghoshal answered |
From Bent’s Rule equation, cosx = s/s-1
where x is an bond angle and s is the % s character in the bond, it is clear that if we increase bond angle, the %s Character will be maximum.
Here angle is 150,
⇒ Cosx = s/(s-1)
⇒ Cos 150° = s/(s-1)
⇒ -0.86 = s/(s-1)
⇒ -0.86s +0.86 = s
⇒ 0.86 = 1.86s
⇒ s = 0.86/1.86
⇒ s = 0.462
⇒ %s = 0.462 X 10
⇒ %s = 46.2%
Total number of nucleophiles among following molecules is/are___________.
CH3O-, carbene, Nitrene, H2O , ROH, ZnCl2, NHR2, SnCl4, -NH2 , Cl-.
Correct answer is '8'. Can you explain this answer?
Total number of nucleophiles among following molecules is/are___________.
CH3O-, carbene, Nitrene, H2O , ROH, ZnCl2, NHR2, SnCl4, -NH2 , Cl-.
CH3O-, carbene, Nitrene, H2O , ROH, ZnCl2, NHR2, SnCl4, -NH2 , Cl-.
|
Bijoy Patel answered |
-
Nucleophiles are molecules or ions that are attracted to electron-deficient sites and have a tendency to donate electrons. Let's identify the nucleophiles among the given molecules:
- CH3O- (methoxide ion) - Nucleophile
- Carbene - Nucleophile
- Nitrene - Nucleophile
- H2O - Nucleophile
- ROH (alcohol) - Nucleophile
- ZnCl2 - Not a nucleophile (does not have lone pairs to donate)
- NHR2 - Nucleophile
- SnCl4 - Not a nucleophile (does not have lone pairs to donate)
- -NH2 (amide ion) - Nucleophile
- Cl- - Nucleophile
-
So, the total number of nucleophiles among the given molecules is 8.
Some or more species give belowhave pH less than 7.- a)10-8 M HCl
- b)10-8 M NaOH
- c)10 mL of 10-6 M HCl diluted to 100 mL
- d)10 mL of 0.01 M CH3COOH at the end point after addition of 10 mL 0.01 M NaOH
Correct answer is option 'A,C'. Can you explain this answer?
Some or more species give belowhave pH less than 7.
a)
10-8 M HCl
b)
10-8 M NaOH
c)
10 mL of 10-6 M HCl diluted to 100 mL
d)
10 mL of 0.01 M CH3COOH at the end point after addition of 10 mL 0.01 M NaOH
|
|
Edurev.iitjam answered |
The pH of 10-8M HCl solution is 6.98
The pH of NaoH solution will be 8
The pH of 10 mL of 10-6 M HCl diluted to 100 mL is 6.99
The pH of NaoH solution will be 8
The pH of 10 mL of 10-6 M HCl diluted to 100 mL is 6.99
Of the following acids, the one that is strongest is:- a)HBrO4
- b)HOC1
- c)HNO2
- d)H3PO3
Correct answer is option 'A'. Can you explain this answer?
Of the following acids, the one that is strongest is:
a)
HBrO4
b)
HOC1
c)
HNO2
d)
H3PO3
|
Swara Dasgupta answered |
So HCl is strongest of all. There are some super acids, like fluroantimonic acid, that are 100 times more stronger than sulfuric acid.
Which of the following is not resonating structure of each other?- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Which of the following is not resonating structure of each other?
a)
b)
c)
d)
|
|
Rishabh Mehta answered |
Molecules in option (a) are linkage isomers of each other.
According to rules of Resonance, no change in position of atom should take place.
What is the strength (in g/litre) of a solution of H2SO4, 12 mL of which neutralized 15 mL of N/10 NaOH solution:
Correct answer is between '6.12,6.13'. Can you explain this answer?
What is the strength (in g/litre) of a solution of H2SO4, 12 mL of which neutralized 15 mL of N/10 NaOH solution:
|
Kiran Punj answered |
N1 V1 = N2 V2
(H2SO4) (NAOH)
N1*12 = 1/10*15
N2 = 0.125
WE KNOW THAT
N = XM WHERE X FOR H2SO4 IS 2
0.125 = 2*W/MW*V
0.125*98/2 = W/V
W/V= 6.125 g/litre
1 mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27°C. The work done is 3 kJ. The final temperature (K) of the gas is equal to (CV = 20 J mol–1K–1):
Correct answer is '150'. Can you explain this answer?
1 mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27°C. The work done is 3 kJ. The final temperature (K) of the gas is equal to (CV = 20 J mol–1K–1):
|
Vivek Nain answered |
answer is correct
like in carnot cycle when we go for expansion adiabatically nd reversibly then temp always dec. i.e. T(H) to T(L)
so in this also temp goes on decrease coz on expansion work done should be negative then
dT = 150 is literally as
T(L)-T(H) = -150
so T(L) should be 150 to make work done negative
Distribution of molecules with velocity is represented by curve as shown:
Velocity at point A is:- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Distribution of molecules with velocity is represented by curve as shown:
Velocity at point A is:
a)
b)
c)
d)
|
Dharshini Raghavan answered |
We know that RMS>Average speed>Most probable. Point B has highest velocity and has to be RMS. Next,
the lowest speed is also the one possessed by the highest fraction of molecules and this is the Most probable speed, which is point C.
Now average speed has values in between RMS and MPS. So, point A has got to be average speed.
Strongest base among the following species is:- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Strongest base among the following species is:
a)
b)
c)
d)
|
|
Anshul Mehra answered |
Nitrogen atom is more basic than other two oxygen and carbon . Also it has an available lone pair.
The set representing the correct order of ionic radius is:- a)Na+ > Mg2+ > Al3+ > Li+ > Be2+
- b)Na+ > Li+ > Mg2+ > Al3+ > Be2+
- c)Na+ > Mg2+ > Li+ > Al3+ > Be2+
- d)Na+ > Mg2+ > Li+ > Be2+ > Al3+
Correct answer is option 'B'. Can you explain this answer?
The set representing the correct order of ionic radius is:
a)
Na+ > Mg2+ > Al3+ > Li+ > Be2+
b)
Na+ > Li+ > Mg2+ > Al3+ > Be2+
c)
Na+ > Mg2+ > Li+ > Al3+ > Be2+
d)
Na+ > Mg2+ > Li+ > Be2+ > Al3+
|
Rising Star answered |
When positive ion increase,radious is decreases due to effective nuclear charge.
Solubility of BaSO4 in aqueous solution is 1 × 10–5 M. Hence, solubility in 0.1 M BaCl2 is: - a)1 × 10–10 M
- b)1 × 10–9 M
- c)1 × 10–4 M
- d)1 × 10–5 M
Correct answer is option 'C'. Can you explain this answer?
Solubility of BaSO4 in aqueous solution is 1 × 10–5 M. Hence, solubility in 0.1 M BaCl2 is:
a)
1 × 10–10 M
b)
1 × 10–9 M
c)
1 × 10–4 M
d)
1 × 10–5 M
|
|
Vedika Singh answered |
BaSO₄ ---------> Ba²⁺ + SO₄²⁻
According to ICE table
BaSO₄ -----------------> Ba²⁺ + SO₄²⁻
I 0.1 M 0
C +s +s
E s + 0.1 s
therefore we can write
ksp = ( s + 0.01) . s
as the value of ksp is very small as compared to silver ions initial concentration
we can use approximation
s + 0.1 ≈ 0.1
Ksp = 0.01 . s
s = 1 X 10⁻⁵ / 0.1
s = 0.0001 mol/ L
s = 1 X 10⁻4 mol/ L
According to ICE table
BaSO₄ -----------------> Ba²⁺ + SO₄²⁻
I 0.1 M 0
C +s +s
E s + 0.1 s
therefore we can write
ksp = ( s + 0.01) . s
as the value of ksp is very small as compared to silver ions initial concentration
we can use approximation
s + 0.1 ≈ 0.1
Ksp = 0.01 . s
s = 1 X 10⁻⁵ / 0.1
s = 0.0001 mol/ L
s = 1 X 10⁻4 mol/ L
The ground state electronic configuration of valence shell electrons in a molecule A2 is written as
• Hence bond order is_____________________.- a)0
- b)1
- c)2
- d)3
Correct answer is option 'C'. Can you explain this answer?
The ground state electronic configuration of valence shell electrons in a molecule A2 is written as• Hence bond order is_____________________.
• Hence bond order is_____________________.a)
0
b)
1
c)
2
d)
3
|
|
Nilotpal Basu answered |
Bond Order = 
Number of electrons in bonding MO = 8
Number of electrons in anti-bonding MO = 4
Bond order = (8 - 4)/2
Bond order = 4/2
Bond order = 2
Select the correct statement:- a)
is more basic than
- b)
is more basic than 
- c)
is more basic than 
- d) All of them
Correct answer is option 'C'. Can you explain this answer?
Select the correct statement:
a)
is more basic thanb)
is more basic than c)
is more basic than d)
All of them
|
Bhavana Dasgupta answered |
Benzoic acid is a stronger acid than phenol because the benzoate ion is stabilised by two equivalent resonance structures in which the negative charge is present at the more electronegative oxygen atom.
The conjugate base of phenol, a phenoxide ion, has non-equivalent resonance structures in which the negative charge is at the less electronegative carbon atom. Thus, the benzoate ion is more stable than phenoxide ion. Hence, benzoic acid is a stronger acid than phenol.
Find ΔE for states n = 3,4 in Bohr’s model of hydrogenic atoms (in eV):
Correct answer is '0.64-0.66'. Can you explain this answer?
Find ΔE for states n = 3,4 in Bohr’s model of hydrogenic atoms (in eV):
|
|
Vandana Chopra answered |
ΔE = 13.6 [ 1/9 - 1/16]
ΔE = 13.6 [7/144]
ΔE = 13.6 x 0.048
ΔE = 0.66
The ratio of van der Waal’s constant a and b, 
has the dimensions of:- a)Atm b-1
- b)L atm mol-1
- c)L mol-1
- d)atm L mol-2
Correct answer is option 'B'. Can you explain this answer?
The ratio of van der Waal’s constant a and b, has the dimensions of:
has the dimensions of:a)
Atm b-1
b)
L atm mol-1
c)
L mol-1
d)
atm L mol-2
|
|
Partho Gupta answered |
The ratio of van der Waals’ constants a and b has the dimensions of atm L mole-1
Predict the shape of 
molecule using VSEPR theory:- a)TBP
- b)See-Saw
- c)Octahedral
- d)Sq. Planar
Correct answer is option 'B'. Can you explain this answer?
Predict the shape of molecule using VSEPR theory:
molecule using VSEPR theory:a)
TBP
b)
See-Saw
c)
Octahedral
d)
Sq. Planar
|
|
Mihir Singh answered |
The electron geometry of IF4+ is trigonal bipyramidal. This is because it has five total electron groups. The central atom is bonded to four other atoms and has one lone pair of electrons. The lone pair of electrons will affect the molecular geometry of the compound as well as the bond angles.
Which of the following is an electrophile?- a) PCl5
- b)ZnCl2
- c)
- d)Carbene
Correct answer is option 'A,B,C,D'. Can you explain this answer?
Which of the following is an electrophile?
a)
PCl5
b)
ZnCl2
c)
d)
Carbene
|
|
Anisha Pillai answered |
Electrophile is an electron pair acceptor. Electrophiles are positively charged or neutral species having vacant orbitals that are attracted to an electron rich centre. It participates in a chemical reaction by accepting an electron pair in order to bond to a nucleophile.
Correct order of stability of the following carbocation is:
- a)I > II > III > IV
- b)II > I > IV > III
- c)I > II > IV > III
- d)I > IV > II > III
Correct answer is option 'B'. Can you explain this answer?
Correct order of stability of the following carbocation is:
a)
I > II > III > IV
b)
II > I > IV > III
c)
I > II > IV > III
d)
I > IV > II > III
|
Lavanya Menon answered |
At metaposition, hyperconjugation do not take place.
Inductive effect order −H<−D i.e R−CH3<R−CD3 (Stability)
Hyperconjugation order −H<−D i.e R−CH3>R−CD3 (Stability)
and Hyperconjugated compound is always more stable than inductive effect.
Therefore, Order is I>II>IV>III
The correct stability order of the given resonating structures is:- a)I > II > III
- b)III > I > II
- c)I > III > II
- d)II > III > I
Correct answer is option 'B'. Can you explain this answer?
The correct stability order of the given resonating structures is:
a)
I > II > III
b)
III > I > II
c)
I > III > II
d)
II > III > I
|
|
Shivani Mehta answered |
Here 2 rule of resonance stabilising rule works
1 More the number of bond more stable in resonance
2. + ve charge in electronegative decreasing resonance stability .Now for 1 2 3 , 1 and 3 has greater number of bond so 1 and 3 is greater stable than 2 . Compare 1 and 3 for 1 positive charge on oxygen which is more electronegative than N so 1 has less stable than 3 . So order is 3 > 1 >2
Hydrogen bonding plays a central role in the following phenomena:- a)Ice floats in water
- b)Higher Lewis basicity of primary amines than tertiary amines in aqueous solutions
- c)Formic acid is more acidic than acetic acid
- d)Dimerisation of acetic acid in benzene
Correct answer is option 'A,B,D'. Can you explain this answer?
Hydrogen bonding plays a central role in the following phenomena:
a)
Ice floats in water
b)
Higher Lewis basicity of primary amines than tertiary amines in aqueous solutions
c)
Formic acid is more acidic than acetic acid
d)
Dimerisation of acetic acid in benzene
|
|
Dronacharya Institute answered |
Which of the following is most basic?- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Which of the following is most basic?
a)
b)
c)
d)
|
|
Tejas Goyal answered |
Basicity is the tendency of an atom to donate their lone pair.
In case of (a), (b),(d), lone pair of Nitrogen atom is in conjugation which is not available to be donated.
Hence (c) is more basic.
Which of these structures is practically not a valid resonance structure for formaldehyde?- a)I
- b)II
- c)III
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
Which of these structures is practically not a valid resonance structure for formaldehyde?
a)
I
b)
II
c)
III
d)
None of these
|
|
Ameya Reddy answered |
The lowest energy form is I, because every atom has a complete octet. It is the major contributor.
II is lower in energy than III, because the negative charge is in the more electronegative O atom.
Which of the following drawings is not a resonance structure of 1 -nitrocyclohexane:
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Which of the following drawings is not a resonance structure of 1 -nitrocyclohexane:
a)
b)
c)
d)
|
Study Sesh answered |
In structure C, oxygen - the atom more electronegative than carbon - has a positive charge. This would be a highly unstable form and hence does not contribute to the resonating structure.
Another way to look at it would be that -NO2, a -I and -R group appears to show +R effect in structure C, which is incorrect.
Boric acid is dissolved in D2O. Hence, which of the following is/are true - a)Acidity of it is due to H+
- b)Acidity of it is due to D+
- c)Reaction in D2O is reversible
- d)Boric acid is not an Arrhenius acid.
Correct answer is option 'B,C'. Can you explain this answer?
Boric acid is dissolved in D2O. Hence, which of the following is/are true
a)
Acidity of it is due to H+
b)
Acidity of it is due to D+
c)
Reaction in D2O is reversible
d)
Boric acid is not an Arrhenius acid.
|
|
Aditi Basak answered |
Acidity of boric acid in D2O
Explanation:
Acidity of it is due to H:
- This statement is incorrect because the acidity of boric acid in D2O is not due to hydrogen (H).
- Boric acid (H3BO3) contains three hydroxyl groups (OH groups) attached to the boron atom.
- In water, these hydroxyl groups can undergo ionization to form borate ions (B(OH)4-) and hydronium ions (H3O+).
- However, in the presence of deuterium oxide (D2O), the hydroxyl groups can exchange their hydrogen atoms with deuterium atoms, resulting in the formation of deuterium oxide (D3O+).
- Therefore, the acidity of boric acid in D2O is not due to hydrogen (H).
Acidity of it is due to D:
- This statement is correct because the acidity of boric acid in D2O is due to deuterium (D).
- As mentioned earlier, the hydroxyl groups in boric acid can exchange their hydrogen atoms with deuterium atoms in the presence of D2O.
- The resulting deuterium oxide (D3O+) is responsible for the acidity of boric acid in D2O.
- Therefore, the presence of deuterium (D) in the solution makes the boric acid acidic.
Reaction in D2O is reversible:
- This statement is correct because the reaction between boric acid and D2O is reversible.
- When boric acid is dissolved in D2O, it undergoes ionization to form borate ions (B(OH)4-) and deuterium ions (D3O+).
- However, these ions can also react with each other to reform boric acid and D2O.
- This reversible reaction allows for the establishment of an equilibrium between the boric acid and its ionized forms.
- Therefore, the reaction in D2O is reversible.
Boric acid is not an Arrhenius acid:
- This statement is incorrect because boric acid is considered an Arrhenius acid.
- Arrhenius acids are substances that ionize in water to produce hydrogen ions (H+).
- Although the acidity of boric acid in D2O is due to deuterium ions (D3O+) instead of hydrogen ions (H+), it still exhibits acidic properties in aqueous solutions.
- Therefore, boric acid can be classified as an Arrhenius acid.
Summary:
- The acidity of boric acid in D2O is due to deuterium (D) and not hydrogen (H).
- The reaction between boric acid and D2O is reversible.
- Boric acid is considered an Arrhenius acid.
Explanation:
Acidity of it is due to H:
- This statement is incorrect because the acidity of boric acid in D2O is not due to hydrogen (H).
- Boric acid (H3BO3) contains three hydroxyl groups (OH groups) attached to the boron atom.
- In water, these hydroxyl groups can undergo ionization to form borate ions (B(OH)4-) and hydronium ions (H3O+).
- However, in the presence of deuterium oxide (D2O), the hydroxyl groups can exchange their hydrogen atoms with deuterium atoms, resulting in the formation of deuterium oxide (D3O+).
- Therefore, the acidity of boric acid in D2O is not due to hydrogen (H).
Acidity of it is due to D:
- This statement is correct because the acidity of boric acid in D2O is due to deuterium (D).
- As mentioned earlier, the hydroxyl groups in boric acid can exchange their hydrogen atoms with deuterium atoms in the presence of D2O.
- The resulting deuterium oxide (D3O+) is responsible for the acidity of boric acid in D2O.
- Therefore, the presence of deuterium (D) in the solution makes the boric acid acidic.
Reaction in D2O is reversible:
- This statement is correct because the reaction between boric acid and D2O is reversible.
- When boric acid is dissolved in D2O, it undergoes ionization to form borate ions (B(OH)4-) and deuterium ions (D3O+).
- However, these ions can also react with each other to reform boric acid and D2O.
- This reversible reaction allows for the establishment of an equilibrium between the boric acid and its ionized forms.
- Therefore, the reaction in D2O is reversible.
Boric acid is not an Arrhenius acid:
- This statement is incorrect because boric acid is considered an Arrhenius acid.
- Arrhenius acids are substances that ionize in water to produce hydrogen ions (H+).
- Although the acidity of boric acid in D2O is due to deuterium ions (D3O+) instead of hydrogen ions (H+), it still exhibits acidic properties in aqueous solutions.
- Therefore, boric acid can be classified as an Arrhenius acid.
Summary:
- The acidity of boric acid in D2O is due to deuterium (D) and not hydrogen (H).
- The reaction between boric acid and D2O is reversible.
- Boric acid is considered an Arrhenius acid.
The standard enthalpy of formation of NH3 is -4 6 .0 kJ mol -1. If the enthalpy of formation of H2 from its atoms is —436 kJ mol -1 and that of N2 is -712 kJ mol the average bond enthalpy of N—H bond in NH3 is: (in kJ/mol)
Correct answer is '352'. Can you explain this answer?
The standard enthalpy of formation of NH3 is -4 6 .0 kJ mol -1. If the enthalpy of formation of H2 from its atoms is —436 kJ mol -1 and that of N2 is -712 kJ mol the average bond enthalpy of N—H bond in NH3 is: (in kJ/mol)
|
Kaavya Sengupta answered |
The balanced chemical equation for formation of NH3 would be as follows:
1/2N2(g) + 3/2H2(g)-->NH3(g)
The energy required to break required bonds for the formation of product is calculated as:
1/2 mole of N2 will require 712 kJmol^-1/2 = 356 kJ/mol
3/2 mole of H2 will require 3x 436 kJmol^-1/2 = 654 kJ/mol
Total energy require to break the bonds = (356 + 654) kJ/mol = 1010 kJ/mol
total energy of reaction = (total energy of bonds broken) - (total energy of bonds formed)
-46.3 kJ/mol =1010) kJ/mol - 3(N-H)
3(N-H) = 1010 kJ/mol + 46.3kJ/mol = 1056.3 kJ/mol
(N-H) = 1056.3 kJ/mol /3 = 352.1kJ/mol
Which solution will show maximum vapor pressure at 300K- a)1M Sucrose
- b)1M Acetic acid
- c)1M Calcium Chloride
- d)1M Sodium Chloride
Correct answer is option 'A'. Can you explain this answer?
Which solution will show maximum vapor pressure at 300K
a)
1M Sucrose
b)
1M Acetic acid
c)
1M Calcium Chloride
d)
1M Sodium Chloride
|
|
Kiran Punj answered |
Vapor pressure =k/vantt of factor
where k is constant
Which of the following set(s) of quantum numbers is(are) not allowed:- a)n = 3, l = 2, ml = -1
- b)n = 4, l = 0, ml = -1
- c)n = 3, / = 3, ml = -3
- d)n = 5, / = 3, ml = +2
Correct answer is option 'B,C'. Can you explain this answer?
Which of the following set(s) of quantum numbers is(are) not allowed:
a)
n = 3, l = 2, ml = -1
b)
n = 4, l = 0, ml = -1
c)
n = 3, / = 3, ml = -3
d)
n = 5, / = 3, ml = +2
|
Sinjini Singh answered |
Forbidden Quantum Numbers in Atoms
Forbidden quantum numbers are those that do not conform to the rules of quantum mechanics. The rules of quantum mechanics dictate that the values of n, l, ml, and ms must be within specific ranges. If any of these quantum numbers are outside the allowed range, they are considered forbidden.
Forbidden Quantum Numbers
a) n = 3, l = 2, ml = -1
According to the rules of quantum mechanics, the values of n, l, and ml must be such that l ≤ n-1 and -l ≤ ml ≤ l. In this case, n = 3 and l = 2. Therefore, the allowed values of ml are -2, -1, 0, 1, and 2. However, ml is given as -1, which is an allowed value. Therefore, this set of quantum numbers is allowed.
b) n = 4, l = 0, ml = -1
In this case, l = 0, which means that ml must also be 0. Therefore, ml cannot be -1. This set of quantum numbers is not allowed.
c) n = 3, l = 3, ml = -3
For a given value of n, the maximum value of l is n-1. Therefore, in this case, l = 3 is not allowed because n = 3. Therefore, this set of quantum numbers is not allowed.
d) n = 5, l = 3, ml = 2
For a given value of l, ml must be such that -l ≤ ml ≤ l. In this case, l = 3, which means that the allowed values of ml are -3, -2, -1, 0, 1, 2, and 3. Therefore, ml = 2 is an allowed value. This set of quantum numbers is allowed.
Conclusion
The sets of quantum numbers that are not allowed are b) n = 4, l = 0, ml = -1 and c) n = 3, l = 3, ml = -3. These sets of quantum numbers violate the rules of quantum mechanics.
Forbidden quantum numbers are those that do not conform to the rules of quantum mechanics. The rules of quantum mechanics dictate that the values of n, l, ml, and ms must be within specific ranges. If any of these quantum numbers are outside the allowed range, they are considered forbidden.
Forbidden Quantum Numbers
a) n = 3, l = 2, ml = -1
According to the rules of quantum mechanics, the values of n, l, and ml must be such that l ≤ n-1 and -l ≤ ml ≤ l. In this case, n = 3 and l = 2. Therefore, the allowed values of ml are -2, -1, 0, 1, and 2. However, ml is given as -1, which is an allowed value. Therefore, this set of quantum numbers is allowed.
b) n = 4, l = 0, ml = -1
In this case, l = 0, which means that ml must also be 0. Therefore, ml cannot be -1. This set of quantum numbers is not allowed.
c) n = 3, l = 3, ml = -3
For a given value of n, the maximum value of l is n-1. Therefore, in this case, l = 3 is not allowed because n = 3. Therefore, this set of quantum numbers is not allowed.
d) n = 5, l = 3, ml = 2
For a given value of l, ml must be such that -l ≤ ml ≤ l. In this case, l = 3, which means that the allowed values of ml are -3, -2, -1, 0, 1, 2, and 3. Therefore, ml = 2 is an allowed value. This set of quantum numbers is allowed.
Conclusion
The sets of quantum numbers that are not allowed are b) n = 4, l = 0, ml = -1 and c) n = 3, l = 3, ml = -3. These sets of quantum numbers violate the rules of quantum mechanics.
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