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PASSAGE - 2There are some deposits of nitrates and phosphates in earth’s crust. Nitrates are more soluble in water. Nitrates are difficult to reduce under the laboratory conditions but microbes do it easily.
Ammonia forms large number of complexes with transition metal ions. Hybridization easily explains the ease of sigma donation capability of NH3 and PH3. Phosphine is a flammable gas and is prepared from white phosphorous.Q. Among the following, the correct statement is- a)Phosphates have no biological significance in humans
- b)Between nitrates and phosphates, phosphates are less abundant in earth’s crust
- c)Between nitrates an d phosphates, n itr ates are less abundant in earth’s crust
- d)Oxidation of nitrates is possible in soil
Correct answer is option 'C'. Can you explain this answer?
PASSAGE - 2
There are some deposits of nitrates and phosphates in earth’s crust. Nitrates are more soluble in water. Nitrates are difficult to reduce under the laboratory conditions but microbes do it easily.
Ammonia forms large number of complexes with transition metal ions. Hybridization easily explains the ease of sigma donation capability of NH3 and PH3. Phosphine is a flammable gas and is prepared from white phosphorous.
Ammonia forms large number of complexes with transition metal ions. Hybridization easily explains the ease of sigma donation capability of NH3 and PH3. Phosphine is a flammable gas and is prepared from white phosphorous.
Q. Among the following, the correct statement is
a)
Phosphates have no biological significance in humans
b)
Between nitrates and phosphates, phosphates are less abundant in earth’s crust
c)
Between nitrates an d phosphates, n itr ates are less abundant in earth’s crust
d)
Oxidation of nitrates is possible in soil
|
Sahana Joshi answered |
We know that phosphates have a biological significance in human, therefore statement (a) is not correct.
Since nitrates are more soluble in water so they are less abundant in earth's crust where as phosphates are less soluble in water and so they are more abundant in earth's crust. Thus statement (b) is False and statement (c) is correct.
Since nitrates are more soluble in water so they are less abundant in earth's crust where as phosphates are less soluble in water and so they are more abundant in earth's crust. Thus statement (b) is False and statement (c) is correct.
NOTE : In nitrates (NO–3) nitrogen is in + 5 oxidation state which is the highest oxidation state exhibited by nitrogen. Because of this nitrates can not be oxidized (oxidation means increase in oxidation state). Hence statement (d) is not correct.
The correct answer is (c).
The correct answer is (c).
The structure of diborane (B2H6) contains- a)four 2c-2e bonds and four 3c-2e bonds
- b)two 2c-2e bonds and two 3c-3e bonds
- c)two 2c-2e bonds and four 3c-2e bonds
- d)four 2c-2e bonds and two 3c-2e bonds
Correct answer is option 'D'. Can you explain this answer?
The structure of diborane (B2H6) contains
a)
four 2c-2e bonds and four 3c-2e bonds
b)
two 2c-2e bonds and two 3c-3e bonds
c)
two 2c-2e bonds and four 3c-2e bonds
d)
four 2c-2e bonds and two 3c-2e bonds
|
Gauri Chauhan answered |
In diborane structure B2H6 there are two 2c-2e bonds and two 3c–2e bonds (see structure of diborane).
Structure of B2H6 :
Structure of B2H6 :
Ammonia, on reaction with hypochlorite anion, can form- a)NO
- b)NH4Cl
- c)N2H4
- d)HNO2
Correct answer is option 'C'. Can you explain this answer?
Ammonia, on reaction with hypochlorite anion, can form
a)
NO
b)
NH4Cl
c)
N2H4
d)
HNO2
|
Sagar Mukherjee answered |
2NH3 + OCl–→ NH2.NH2 + H2O + Cl–
Among the following oxoacids, the correct decreasing order of acid strength is:- a)HOCl > HClO2 > HClO3> HClO4
- b)HClO4 > HOCl > HClO2> HClO3
- c)HClO4 > HClO3 > HClO2> HOCl
- d)HClO2 > HClO4 > HClO3> HOCl
Correct answer is option 'C'. Can you explain this answer?
Among the following oxoacids, the correct decreasing order of acid strength is:
a)
HOCl > HClO2 > HClO3> HClO4
b)
HClO4 > HOCl > HClO2> HClO3
c)
HClO4 > HClO3 > HClO2> HOCl
d)
HClO2 > HClO4 > HClO3> HOCl
|
Sahana Ahuja answered |
Acidic strength increases as the oxidation number of central atom increases.
Hence acidic strength order is
Hence acidic strength order is
(+7) (+5) (+3) (+1)
HClO4 > HClO3 > HClO2 > HClO
HClO4 > HClO3 > HClO2 > HClO
Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen?- a)HNO3, NO, NH4Cl, N2
- b)HNO3, NO, N2, NH4 Cl
- c)HNO3, NH4Cl, NO, N2
- d)NO, HNO3, NH4Cl, N2
Correct answer is option 'B'. Can you explain this answer?
Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen?
a)
HNO3, NO, NH4Cl, N2
b)
HNO3, NO, N2, NH4 Cl
c)
HNO3, NH4Cl, NO, N2
d)
NO, HNO3, NH4Cl, N2
|
Priyanka Sen answered |
Explanation:
The oxidation state of an atom in a compound is a measure of the degree of oxidation (loss of electrons) of that atom. In order to determine the decreasing order of the oxidation state of nitrogen in the given compounds, we need to analyze the oxidation states of nitrogen in each compound.
Analysis of oxidation states:
1. HNO3 (Nitric acid):
In nitric acid, the oxygen atom is more electronegative than nitrogen, and hydrogen is less electronegative. Therefore, the oxidation state of hydrogen is +1 and the oxidation state of oxygen is -2. Since the overall charge of nitric acid is 0, the sum of the oxidation states must be zero. Let the oxidation state of nitrogen be x. Therefore,
(+1) + x + (-2) × 3 = 0
x - 6 = -1
x = +5
2. NO (Nitric oxide):
In nitric oxide, oxygen is more electronegative than nitrogen. Therefore, the oxidation state of oxygen is -2. Since the overall charge of nitric oxide is 0, the sum of the oxidation states must be zero. Let the oxidation state of nitrogen be x. Therefore,
x + (-2) = 0
x = +2
3. N2 (Nitrogen gas):
In nitrogen gas, the atoms are held together by a triple bond. Since the overall charge of nitrogen gas is 0, the sum of the oxidation states must be zero. Let the oxidation state of nitrogen be x. Therefore,
2x = 0
x = 0
4. NH4Cl (Ammonium chloride):
In ammonium chloride, hydrogen is less electronegative than nitrogen, and chlorine is more electronegative than nitrogen. Therefore, the oxidation state of hydrogen is +1 and the oxidation state of chlorine is -1. Since the overall charge of ammonium chloride is 0, the sum of the oxidation states must be zero. Let the oxidation state of nitrogen be x. Therefore,
(+1) × 4 + x + (-1) = 0
4 + x - 1 = 0
x = -3
Descending order of oxidation states:
Based on the analysis of oxidation states, the decreasing order of the compounds according to the oxidation state of nitrogen is:
HNO3 (+5) > NO (+2) > N2 (0) > NH4Cl (-3)
Therefore, the correct ordering of compounds according to the decreasing order of the oxidation state of nitrogen is option B: HNO3, NO, N2, NH4Cl.
The oxidation state of an atom in a compound is a measure of the degree of oxidation (loss of electrons) of that atom. In order to determine the decreasing order of the oxidation state of nitrogen in the given compounds, we need to analyze the oxidation states of nitrogen in each compound.
Analysis of oxidation states:
1. HNO3 (Nitric acid):
In nitric acid, the oxygen atom is more electronegative than nitrogen, and hydrogen is less electronegative. Therefore, the oxidation state of hydrogen is +1 and the oxidation state of oxygen is -2. Since the overall charge of nitric acid is 0, the sum of the oxidation states must be zero. Let the oxidation state of nitrogen be x. Therefore,
(+1) + x + (-2) × 3 = 0
x - 6 = -1
x = +5
2. NO (Nitric oxide):
In nitric oxide, oxygen is more electronegative than nitrogen. Therefore, the oxidation state of oxygen is -2. Since the overall charge of nitric oxide is 0, the sum of the oxidation states must be zero. Let the oxidation state of nitrogen be x. Therefore,
x + (-2) = 0
x = +2
3. N2 (Nitrogen gas):
In nitrogen gas, the atoms are held together by a triple bond. Since the overall charge of nitrogen gas is 0, the sum of the oxidation states must be zero. Let the oxidation state of nitrogen be x. Therefore,
2x = 0
x = 0
4. NH4Cl (Ammonium chloride):
In ammonium chloride, hydrogen is less electronegative than nitrogen, and chlorine is more electronegative than nitrogen. Therefore, the oxidation state of hydrogen is +1 and the oxidation state of chlorine is -1. Since the overall charge of ammonium chloride is 0, the sum of the oxidation states must be zero. Let the oxidation state of nitrogen be x. Therefore,
(+1) × 4 + x + (-1) = 0
4 + x - 1 = 0
x = -3
Descending order of oxidation states:
Based on the analysis of oxidation states, the decreasing order of the compounds according to the oxidation state of nitrogen is:
HNO3 (+5) > NO (+2) > N2 (0) > NH4Cl (-3)
Therefore, the correct ordering of compounds according to the decreasing order of the oxidation state of nitrogen is option B: HNO3, NO, N2, NH4Cl.
Under hydrolytic conditions, the compounds used for preparation of linear polymer and for chain termination, respectively, are- a)CH3SiCl3 and Si(CH3)4
- b)(CH3)2SiCl2 and (CH3)3SiCl
- c)(CH3)2SiCl2 and CH3SiCl3
- d)SiCl4 and (CH3)3SiCl
Correct answer is option 'B'. Can you explain this answer?
Under hydrolytic conditions, the compounds used for preparation of linear polymer and for chain termination, respectively, are
a)
CH3SiCl3 and Si(CH3)4
b)
(CH3)2SiCl2 and (CH3)3SiCl
c)
(CH3)2SiCl2 and CH3SiCl3
d)
SiCl4 and (CH3)3SiCl
|
Aryan Sharma answered |
Explanation:
Linear Polymer Preparation:
- The compound used for the preparation of a linear polymer is (CH3)2SiCl2.
- (CH3)2SiCl2 is a linear siloxane monomer that can undergo hydrolysis to form linear polymer chains.
Chain Termination:
- The compound used for chain termination is (CH3)3SiCl.
- (CH3)3SiCl can act as a chain terminator by reacting with the growing polymer chain and terminating its further growth.
Therefore, the compounds used for the preparation of a linear polymer and for chain termination under hydrolytic conditions are (CH3)2SiCl2 and (CH3)3SiCl, respectively.
Linear Polymer Preparation:
- The compound used for the preparation of a linear polymer is (CH3)2SiCl2.
- (CH3)2SiCl2 is a linear siloxane monomer that can undergo hydrolysis to form linear polymer chains.
Chain Termination:
- The compound used for chain termination is (CH3)3SiCl.
- (CH3)3SiCl can act as a chain terminator by reacting with the growing polymer chain and terminating its further growth.
Therefore, the compounds used for the preparation of a linear polymer and for chain termination under hydrolytic conditions are (CH3)2SiCl2 and (CH3)3SiCl, respectively.
In XeF2, XeF4, XeF6 the number of lone pairs on Xe are respectively- a)2, 3, 1
- b)1, 2, 3
- c)4, 1, 2
- d)3, 2, 1.
Correct answer is option 'D'. Can you explain this answer?
In XeF2, XeF4, XeF6 the number of lone pairs on Xe are respectively
a)
2, 3, 1
b)
1, 2, 3
c)
4, 1, 2
d)
3, 2, 1.
|
Shilpa Saha answered |
In the formation of XeF2, sp3d hybridisation occurs which gives the molecule a trigonal bipyramidal structure.
In the formation of XeF4, sp3d2 hybridization occurs which gives the molecule an octahedral structure.
In the formation of XeF6, sp3d3 hybridization occurs which gives the molecule a pentagonal bipyramidal structure.
Assertion: Nitrogen and oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen.
Reason: The reaction between nitrogen and oxygen requires high temperature.- a)The assertion is incorrect, but the reason is correct
- b)Both the assertion and reason are incorrect
- c)Both assertion and reason are correct, and the reason is the correct explanation for the assertion
- d)Both assertion and reason are correct, but the
Correct answer is option 'C'. Can you explain this answer?
Assertion: Nitrogen and oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen.
Reason: The reaction between nitrogen and oxygen requires high temperature.
Reason: The reaction between nitrogen and oxygen requires high temperature.
a)
The assertion is incorrect, but the reason is correct
b)
Both the assertion and reason are incorrect
c)
Both assertion and reason are correct, and the reason is the correct explanation for the assertion
d)
Both assertion and reason are correct, but the
|
|
Gauri Chauhan answered |
Nitrogen and oxgen in air do not react to form oxides of nitrogen in atmosphere because the reaction between nitrogen and oxygen requires high temperature.
Number of sigma bonds in P4O10 is- a)6
- b)7
- c)17
- d)16.
Correct answer is option 'D'. Can you explain this answer?
Number of sigma bonds in P4O10 is
a)
6
b)
7
c)
17
d)
16.
|
|
Kajal Mehta answered |
Number of sigma bonds in P4O10 is 16.
Explanation:
P4O10 is a compound composed of four phosphorus (P) atoms and ten oxygen (O) atoms. To determine the number of sigma bonds in P4O10, we need to consider the bonding between the atoms.
1. Phosphorus-Phosphorus (P-P) Bonds:
Each phosphorus atom is bonded to three other phosphorus atoms in the P4O10 molecule, forming a tetrahedral arrangement. This means that there are a total of four P-P bonds in the molecule.
2. Phosphorus-Oxygen (P-O) Bonds:
Each oxygen atom is bonded to one phosphorus atom in the molecule. Since there are ten oxygen atoms, there are ten P-O bonds.
Therefore, the total number of sigma bonds in P4O10 can be calculated by adding the number of P-P bonds and the number of P-O bonds:
Number of P-P bonds = 4
Number of P-O bonds = 10
Total number of sigma bonds = Number of P-P bonds + Number of P-O bonds = 4 + 10 = 14
However, we need to consider the two double bonds present in the molecule.
3. Double Bonds:
In P4O10, there are two double bonds between one phosphorus atom and two oxygen atoms. These double bonds are formed by sharing two pairs of electrons between the atoms. Each double bond consists of one sigma bond and one pi bond. Since we are only concerned with sigma bonds in this question, we count each double bond as one sigma bond.
Therefore, the two double bonds contribute an additional two sigma bonds to the molecule.
Total number of sigma bonds = 14 + 2 = 16
Hence, the correct answer is option D, which states that the number of sigma bonds in P4O10 is 16.
Explanation:
P4O10 is a compound composed of four phosphorus (P) atoms and ten oxygen (O) atoms. To determine the number of sigma bonds in P4O10, we need to consider the bonding between the atoms.
1. Phosphorus-Phosphorus (P-P) Bonds:
Each phosphorus atom is bonded to three other phosphorus atoms in the P4O10 molecule, forming a tetrahedral arrangement. This means that there are a total of four P-P bonds in the molecule.
2. Phosphorus-Oxygen (P-O) Bonds:
Each oxygen atom is bonded to one phosphorus atom in the molecule. Since there are ten oxygen atoms, there are ten P-O bonds.
Therefore, the total number of sigma bonds in P4O10 can be calculated by adding the number of P-P bonds and the number of P-O bonds:
Number of P-P bonds = 4
Number of P-O bonds = 10
Total number of sigma bonds = Number of P-P bonds + Number of P-O bonds = 4 + 10 = 14
However, we need to consider the two double bonds present in the molecule.
3. Double Bonds:
In P4O10, there are two double bonds between one phosphorus atom and two oxygen atoms. These double bonds are formed by sharing two pairs of electrons between the atoms. Each double bond consists of one sigma bond and one pi bond. Since we are only concerned with sigma bonds in this question, we count each double bond as one sigma bond.
Therefore, the two double bonds contribute an additional two sigma bonds to the molecule.
Total number of sigma bonds = 14 + 2 = 16
Hence, the correct answer is option D, which states that the number of sigma bonds in P4O10 is 16.
Aluminium chloride exists as dimer, Al2Cl6 in solid state as well as in solution of non-polar solvents such as benzene.
When dissolved in water, it gives- a)[Al(OH)6]3- + 3HCl
- b)[Al(H2O)6]3+ + 3Cl-
- c)Al3+ + 3Cl-
- d)Al2 O3 + 6HCl
Correct answer is option 'B'. Can you explain this answer?
Aluminium chloride exists as dimer, Al2Cl6 in solid state as well as in solution of non-polar solvents such as benzene.
When dissolved in water, it gives
When dissolved in water, it gives
a)
[Al(OH)6]3- + 3HCl
b)
[Al(H2O)6]3+ + 3Cl-
c)
Al3+ + 3Cl-
d)
Al2 O3 + 6HCl
|
Rishika Mishra answered |
Which one of the following is the correct statement?- a)Boric acid is a protonic acid
- b)Beryllium exhibits coordination number of six
- c)Chlorides of both beryllium and aluminium have bridged chloride structures in solid phase
- d)B2H6.2NH3 is known as ‘inorganic benzene’
Correct answer is option 'C'. Can you explain this answer?
Which one of the following is the correct statement?
a)
Boric acid is a protonic acid
b)
Beryllium exhibits coordination number of six
c)
Chlorides of both beryllium and aluminium have bridged chloride structures in solid phase
d)
B2H6.2NH3 is known as ‘inorganic benzene’
|
Sarthak Khanna answered |
The correct formula of inorganic benzene is B3N3H6 so (d) is incorrect statement
Boric acid 
is a lewis acid so (a) is
incorrect statement.
The coordination number exhibited by beryllium is 4 and not 6 so statement (b) is incorrect.
Both BeCl2 and AlCl3 exhibit bridged structures in solid state so (c) is correct statement.
is a lewis acid so (a) isincorrect statement.
The coordination number exhibited by beryllium is 4 and not 6 so statement (b) is incorrect.
Both BeCl2 and AlCl3 exhibit bridged structures in solid state so (c) is correct statement.
Which of the following is coloured- a)NO
- b)N2O
- c)SO3
- d)None
Correct answer is option 'D'. Can you explain this answer?
Which of the following is coloured
a)
NO
b)
N2O
c)
SO3
d)
None
|
Sanchita Reddy answered |
All are colourless gases.
PASSAGE - 1The noble gases have closed-shell electronic configuration and are monoatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions.The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers +2, +4 and +6. XeF4 reacts violently with water to given XeO3. The compounds of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell.Q. Argon is used in arc welding because of its- a)low reactivity with metal
- b)ability to lower the melting point of metal
- c)flammability
- d)high calorific value
Correct answer is option 'A'. Can you explain this answer?
PASSAGE - 1
The noble gases have closed-shell electronic configuration and are monoatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions.
The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers +2, +4 and +6. XeF4 reacts violently with water to given XeO3. The compounds of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell.
Q. Argon is used in arc welding because of its
a)
low reactivity with metal
b)
ability to lower the melting point of metal
c)
flammability
d)
high calorific value
|
Anisha Deshpande answered |
Argon, being a noble gas, will not react with the metals, thus, can be used in arc welding.
The temporary hardness of water due to calcium carbonate can be removed by adding –- a)CaCO3
- b)Ca(OH)2
- c)CaCl2
- d)HCl
Correct answer is option 'B'. Can you explain this answer?
The temporary hardness of water due to calcium carbonate can be removed by adding –
a)
CaCO3
b)
Ca(OH)2
c)
CaCl2
d)
HCl
|
Sanaya Menon answered |
Temporary hardness of water is due to presence of bicarbonates of Ca and Mg and it is removed by adding Ca(OH)2 to hard water and precipitating these soluble bicarbonates in the form of insoluble salts.
Nitrogen dioxide cannot be obtained by heating :- a)KNO3
- b)Pb(NO3)2
- c)Cu(NO3)2
- d)AgNO3
Correct answer is option 'A'. Can you explain this answer?
Nitrogen dioxide cannot be obtained by heating :
a)
KNO3
b)
Pb(NO3)2
c)
Cu(NO3)2
d)
AgNO3
|
Atharva Pillai answered |
Only nitrates of heavy metals and lithium decompose on heating to produce NO2.
H3BO3 is:- a)Monobasic and weak Lewis acid
- b)Monobasic and weak Bronsted acid
- c)Monobasic and strong Lewis acid
- d)Tribasic and weak Bronsted acid
Correct answer is option 'A'. Can you explain this answer?
H3BO3 is:
a)
Monobasic and weak Lewis acid
b)
Monobasic and weak Bronsted acid
c)
Monobasic and strong Lewis acid
d)
Tribasic and weak Bronsted acid
|
Pranav Pillai answered |
The central boron atom in boric acid, H3BO3 is electrondeficient.
NOTE : Boric acid is a Lewis acid with one p–orbital vacant. There is no d-orbital of suitable energy in boron atom. So, it can accommodate only one additional electron pair in its outermost shell.
NOTE : Boric acid is a Lewis acid with one p–orbital vacant. There is no d-orbital of suitable energy in boron atom. So, it can accommodate only one additional electron pair in its outermost shell.
HBr and HI reduce sulphuric acid, HCl can reduce KMnO4 and HF can reduce- a)H2SO4
- b)KMnO4
- c)K2Cr2O7
- d)none of the above
Correct answer is option 'D'. Can you explain this answer?
HBr and HI reduce sulphuric acid, HCl can reduce KMnO4 and HF can reduce
a)
H2SO4
b)
KMnO4
c)
K2Cr2O7
d)
none of the above
|
Tejas Chawla answered |
HI and HBr (in that order) are the strongest reducing hydracids and hence they reduce H2SO4. HCl is quite stable and hence is oxidised by strong oxidising agent like KMnO4. HF is not a reducing agent. In the smallest F– ion, the electron which is to be removed during oxidation is closest to the nucleus and therefore most difficult to be removed. Therefore, HF is a poor reducing agent.
In P4O10 each P atom is linked with .......... O atoms- a)2
- b)3
- c)4
- d)5
Correct answer is option 'C'. Can you explain this answer?
In P4O10 each P atom is linked with .......... O atoms
a)
2
b)
3
c)
4
d)
5
|
Rithika Mishra answered |
In P4O10, each P atom is linked to 4 oxygen atoms as can be confirmed by its structure. It is linked to three oxygen atoms by single bond and with one oxygen atom by double bond. [For structure refer to Q.11 of Section A]
PASSAGE - 1The noble gases have closed-shell electronic configuration and are monoatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions.The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers +2, +4 and +6. XeF4 reacts violently with water to given XeO3. The compounds of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell.Q. The structure of XeO3 is- a)linear
- b)planar
- c)pyramidal
- d)T-shaped
Correct answer is option 'C'. Can you explain this answer?
PASSAGE - 1
The noble gases have closed-shell electronic configuration and are monoatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions.
The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers +2, +4 and +6. XeF4 reacts violently with water to given XeO3. The compounds of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell.
Q. The structure of XeO3 is
a)
linear
b)
planar
c)
pyramidal
d)
T-shaped
|
Swara Ahuja answered |
In XeO3 there are total of 4 electron pairs around central atom. Out of which 3 are bonding electron pair and one is non-bonding electron pair. This combination provides sp3-hybridization and pyramidal shape.
Which one of the following species is not a pseudohalide?- a)CNO–
- b)RCOO–
- c)OCN–
- d)NNN–
Correct answer is option 'B'. Can you explain this answer?
Which one of the following species is not a pseudohalide?
a)
CNO–
b)
RCOO–
c)
OCN–
d)
NNN–
|
|
Pranav Pillai answered |
The species called as pseudohalide ions are these are monovalent and made by electronegative atoms. They possess properties similar to halide ion. The corresponding dimers of these pseudohalide ions are called pseudohalogens. RCOO– is not is pseudohalide.
Which of the following statements is true?- a)HClO4 is a weaker acid than HClO3
- b)HNO3 is a stronger acid than HNO2
- c)H3PO3 is a stronger acid than H2SO3
- d)In aqueous medium HF is a stronger acid than HCl
Correct answer is option 'B'. Can you explain this answer?
Which of the following statements is true?
a)
HClO4 is a weaker acid than HClO3
b)
HNO3 is a stronger acid than HNO2
c)
H3PO3 is a stronger acid than H2SO3
d)
In aqueous medium HF is a stronger acid than HCl
|
Prasenjit Malik answered |
The 
is stronger than 
The more the oxidation state of N, the more is the acid character.
is stronger than The more the oxidation state of N, the more is the acid character.In compounds of type ECl3, where E = B, P, As or Bi, the angles Cl – E– Cl for different E are in the order- a)B > P = As = Bi
- b)B > P > As > Bi
- c)B < P = As = Bi
- d)B < P < As < Bi
Correct answer is option 'B'. Can you explain this answer?
In compounds of type ECl3, where E = B, P, As or Bi, the angles Cl – E– Cl for different E are in the order
a)
B > P = As = Bi
b)
B > P > As > Bi
c)
B < P = As = Bi
d)
B < P < As < Bi
|
Advait Ghoshal answered |
In BCl3, H = 1/2 (3 + 3 + 0 - 0) = 3 ; sp2 hybridization (bond angle = 120°) similarly PCl3 AsCl3 and BiCl3 are found to have sp3 hybridized central atom with one lone pair of electrons on the central atom. The bond angle < 109°28', since the central atoms belong to the same group, the bond angle of the chlorides decreases as we go down the group. Thus the order of bond angle is, BCl3 > PCl3 > AsCl3 > BiCl3.
Which among the following factors is the most important in making fluorine the strongest oxidizing halogen ?
- a)Bond dissociation energy
- b)Ionization enthalpy
- c)Electron affinity
- d)Hydration enthalpy
Correct answer is option 'D'. Can you explain this answer?
Which among the following factors is the most important in making fluorine the strongest oxidizing halogen ?
a)
Bond dissociation energy
b)
Ionization enthalpy
c)
Electron affinity
d)
Hydration enthalpy
|
|
Gauri Chauhan answered |
The fluorine has low dissociation energy of F - F bond and reaction of atomic fluorine is exothermic in nature
PASSAGE - 1The noble gases have closed-shell electronic configuration and are monoatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions.The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers +2, +4 and +6. XeF4 reacts violently with water to given XeO3. The compounds of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell.Q. XeF4 and XeF6 are expected to be- a)oxidizing
- b)reducing
- c)unreactive
- d)strongly basic
Correct answer is option 'A'. Can you explain this answer?
PASSAGE - 1
The noble gases have closed-shell electronic configuration and are monoatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions.
The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers +2, +4 and +6. XeF4 reacts violently with water to given XeO3. The compounds of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell.
Q. XeF4 and XeF6 are expected to be
a)
oxidizing
b)
reducing
c)
unreactive
d)
strongly basic
|
Sameer Saha answered |
Answer:
The compounds XeF4 and XeF6 are expected to be oxidizing agents.
Explanation:
1. Noble Gases and their Electronic Configuration:
- Noble gases are a group of elements that have closed-shell electronic configurations, meaning that their outermost energy level is fully occupied with electrons.
- This configuration makes noble gases highly stable and unreactive under normal conditions.
2. Boiling Points of Noble Gases:
- The boiling points of noble gases increase with increasing atomic number.
- This is because the strength of the dispersion forces (also known as London forces or Van der Waals forces) increases with increasing molecular weight.
- Dispersion forces are the attractive forces between temporary dipoles that occur due to the constant motion of electrons.
- In the lighter noble gases, such as helium and neon, the dispersion forces are weak, resulting in low boiling points.
3. Xenon Compounds:
- Xenon (Xe) is a noble gas that can form compounds with other elements, particularly fluorine (F).
- The direct reaction of xenon with fluorine leads to the formation of compounds with oxidation numbers of 2, 4, and 6.
- XeF4 and XeF6 are examples of such compounds.
4. Oxidizing Agents:
- Oxidizing agents are substances that have a tendency to gain electrons or cause the oxidation of other substances.
- XeF4 and XeF6 are strong oxidizing agents because xenon has a high electronegativity and can readily accept electrons from other substances.
- In the case of XeF4, it reacts violently with water to give XeO3, indicating its ability to oxidize water.
Conclusion:
- Based on the above information, it can be concluded that XeF4 and XeF6 are expected to be oxidizing agents due to the high electronegativity of xenon and their ability to accept electrons from other substances.
The compounds XeF4 and XeF6 are expected to be oxidizing agents.
Explanation:
1. Noble Gases and their Electronic Configuration:
- Noble gases are a group of elements that have closed-shell electronic configurations, meaning that their outermost energy level is fully occupied with electrons.
- This configuration makes noble gases highly stable and unreactive under normal conditions.
2. Boiling Points of Noble Gases:
- The boiling points of noble gases increase with increasing atomic number.
- This is because the strength of the dispersion forces (also known as London forces or Van der Waals forces) increases with increasing molecular weight.
- Dispersion forces are the attractive forces between temporary dipoles that occur due to the constant motion of electrons.
- In the lighter noble gases, such as helium and neon, the dispersion forces are weak, resulting in low boiling points.
3. Xenon Compounds:
- Xenon (Xe) is a noble gas that can form compounds with other elements, particularly fluorine (F).
- The direct reaction of xenon with fluorine leads to the formation of compounds with oxidation numbers of 2, 4, and 6.
- XeF4 and XeF6 are examples of such compounds.
4. Oxidizing Agents:
- Oxidizing agents are substances that have a tendency to gain electrons or cause the oxidation of other substances.
- XeF4 and XeF6 are strong oxidizing agents because xenon has a high electronegativity and can readily accept electrons from other substances.
- In the case of XeF4, it reacts violently with water to give XeO3, indicating its ability to oxidize water.
Conclusion:
- Based on the above information, it can be concluded that XeF4 and XeF6 are expected to be oxidizing agents due to the high electronegativity of xenon and their ability to accept electrons from other substances.
Excess of KI reacts with CuSO4 solution and then Na2S2O3 solution is added to it. Which of the statements is incorrect for this reaction ?- a)Na2S2O3 is oxidised
- b)CuI2 is formed
- c)Cu2I2 is formed
- d)Evolved I2 is reduced
Correct answer is option 'B'. Can you explain this answer?
Excess of KI reacts with CuSO4 solution and then Na2S2O3 solution is added to it. Which of the statements is incorrect for this reaction ?
a)
Na2S2O3 is oxidised
b)
CuI2 is formed
c)
Cu2I2 is formed
d)
Evolved I2 is reduced
|
|
Ruchi Kumar answered |
Incorrect Statement: b) CuI2 is formed
Explanation:
- The given reaction involves the reaction between excess KI (potassium iodide) and CuSO4 (copper sulfate) solution followed by the addition of Na2S2O3 (sodium thiosulfate) solution.
- When excess KI is added to CuSO4 solution, a redox reaction takes place. Cu2+ ions in CuSO4 are reduced to Cu+ ions, while I- ions in KI are oxidized to I2 (elemental iodine).
- The overall reaction can be represented as follows:
CuSO4 + 2KI -> CuI2 + K2SO4
Analysis of the given options:
a) Na2S2O3 is oxidized:
- Na2S2O3 is a reducing agent and can undergo oxidation. However, in this reaction, Na2S2O3 is not oxidized. It acts as a reducing agent to reduce the CuI2 formed in the previous step.
- Na2S2O3 reduces CuI2 to form Cu2+ ions and I- ions.
- The reaction can be represented as follows:
CuI2 + 2Na2S2O3 -> 2Cu2+ + S4O6^2- + 2NaI
b) CuI2 is formed:
- This statement is incorrect. CuI2 is not formed in this reaction. Instead, CuI (copper(I) iodide) is formed.
- CuI is a yellow precipitate that forms when the Cu+ ions from CuSO4 react with the I- ions from excess KI.
- The reaction can be represented as follows:
2Cu+ + 4I- -> 2CuI
c) Cu2I2 is formed:
- This statement is correct. Cu2I2 (copper(II) iodide) is formed as a result of the reaction between CuI and I2.
- CuI reacts with I2 to form Cu2I2, which is a white precipitate.
- The reaction can be represented as follows:
2CuI + I2 -> Cu2I2
d) Evolved I2 is reduced:
- This statement is correct. The I2 that is evolved during the reaction is reduced by Na2S2O3.
- Na2S2O3 acts as a reducing agent and reduces I2 to I- ions.
- The reaction can be represented as follows:
I2 + 2Na2S2O3 -> 2NaI + Na2S4O6
In conclusion, the incorrect statement in this reaction is b) CuI2 is formed. The correct product formed is CuI (copper(I) iodide).
Explanation:
- The given reaction involves the reaction between excess KI (potassium iodide) and CuSO4 (copper sulfate) solution followed by the addition of Na2S2O3 (sodium thiosulfate) solution.
- When excess KI is added to CuSO4 solution, a redox reaction takes place. Cu2+ ions in CuSO4 are reduced to Cu+ ions, while I- ions in KI are oxidized to I2 (elemental iodine).
- The overall reaction can be represented as follows:
CuSO4 + 2KI -> CuI2 + K2SO4
Analysis of the given options:
a) Na2S2O3 is oxidized:
- Na2S2O3 is a reducing agent and can undergo oxidation. However, in this reaction, Na2S2O3 is not oxidized. It acts as a reducing agent to reduce the CuI2 formed in the previous step.
- Na2S2O3 reduces CuI2 to form Cu2+ ions and I- ions.
- The reaction can be represented as follows:
CuI2 + 2Na2S2O3 -> 2Cu2+ + S4O6^2- + 2NaI
b) CuI2 is formed:
- This statement is incorrect. CuI2 is not formed in this reaction. Instead, CuI (copper(I) iodide) is formed.
- CuI is a yellow precipitate that forms when the Cu+ ions from CuSO4 react with the I- ions from excess KI.
- The reaction can be represented as follows:
2Cu+ + 4I- -> 2CuI
c) Cu2I2 is formed:
- This statement is correct. Cu2I2 (copper(II) iodide) is formed as a result of the reaction between CuI and I2.
- CuI reacts with I2 to form Cu2I2, which is a white precipitate.
- The reaction can be represented as follows:
2CuI + I2 -> Cu2I2
d) Evolved I2 is reduced:
- This statement is correct. The I2 that is evolved during the reaction is reduced by Na2S2O3.
- Na2S2O3 acts as a reducing agent and reduces I2 to I- ions.
- The reaction can be represented as follows:
I2 + 2Na2S2O3 -> 2NaI + Na2S4O6
In conclusion, the incorrect statement in this reaction is b) CuI2 is formed. The correct product formed is CuI (copper(I) iodide).
The major role of fluorspar (CaF2), which is added in small quantities in the electrolytic reduction of alumina dissolved in fused cryolite (Na3AlF6), is- a)as a catalyst
- b)to make the fused mixture very conducting
- c)to lower the temperature of the melt
- d)to decrease the rate of oxidation of carbon at the anode.
Correct answer is option 'B,C'. Can you explain this answer?
The major role of fluorspar (CaF2), which is added in small quantities in the electrolytic reduction of alumina dissolved in fused cryolite (Na3AlF6), is
a)
as a catalyst
b)
to make the fused mixture very conducting
c)
to lower the temperature of the melt
d)
to decrease the rate of oxidation of carbon at the anode.
|
Nandita Basak answered |
To make the fused mixture very conducting and to reduce the temperature of the melt.
Aqueous solution of Na2S2O3 on reaction with Cl2 gives –- a)Na2S4O6
- b)NaHSO4
- c)NaCl
- d)NaOH
Correct answer is option 'B'. Can you explain this answer?
Aqueous solution of Na2S2O3 on reaction with Cl2 gives –
a)
Na2S4O6
b)
NaHSO4
c)
NaCl
d)
NaOH
|
Athira Datta answered |
The following reaction occurs
Na2S2O3 + 4Cl2 + 5H2O → 2 NaHSO4 + 8HCl.
A solution of colourless salt H on boiling with excess NaOH produces a non-flammable gas. The gas evolution ceases after sometime. Upon addition of Zn dust to the same solution, the gas evolution restarts. The colourless salt (s) H is (are)- a)NH4NO3
- b)NH4NO2
- c)NH4Cl
- d)(NH4)2SO4
Correct answer is option 'A,B'. Can you explain this answer?
A solution of colourless salt H on boiling with excess NaOH produces a non-flammable gas. The gas evolution ceases after sometime. Upon addition of Zn dust to the same solution, the gas evolution restarts. The colourless salt (s) H is (are)
a)
NH4NO3
b)
NH4NO2
c)
NH4Cl
d)
(NH4)2SO4
|
Sharmila Chavan answered |
When ammonium salt NH4NO3 or NH4NO2 (ammonium salts are colourless) is boiled with excess of NaOH, ammonia (NH3) gas is evolved as follows:
NH4NO2 + NaOH → NaNO2 + NH3 + H2O
NH4NO3 + NaOH → NaNO3 + NH3 + H2O
NH4NO3 + NaOH → NaNO3 + NH3 + H2O
The NH3 gas evolved is non-flammable gas.
When the gas evolution ceases we are left with NaNO2 or NaNO3 in solution.
These salts get reduced when Zn is added to this solution containing salt (NaNO2 or NaNO3) and excess NaOH and NH3 gas is evolve.
When the gas evolution ceases we are left with NaNO2 or NaNO3 in solution.
These salts get reduced when Zn is added to this solution containing salt (NaNO2 or NaNO3) and excess NaOH and NH3 gas is evolve.
Thus the colourless salt [H] is either NH4NO2 or NH4NO3.
Thus (a) and (b) are correct answers. [NOTE : NaCl formed has no reaction with NaOH]
Thus (a) and (b) are correct answers. [NOTE : NaCl formed has no reaction with NaOH]
White phosphorus (P4) has- a)six P-P single bonds
- b)four P-P single bonds
- c)four lone pairs of electrons
- d)PPP angle of 60°
Correct answer is option 'A,C,D'. Can you explain this answer?
White phosphorus (P4) has
a)
six P-P single bonds
b)
four P-P single bonds
c)
four lone pairs of electrons
d)
PPP angle of 60°
|
Poulomi Datta answered |
The four atoms in a P4 molecule are situated at the corners of a tetrahedron. There are six P - P single bonds with PPP bond angle equal to 60o. Each phosphorus has a lone pair of electrons.
For H3PO3 and H3PO4 the correct choice is:- a)H3PO3 is dibasic and reducing
- b)H3PO3 is dibasic and non-reducing
- c)H3PO4 is tribasic and reducing
- d)H3PO3 is tribasic and non-reducing
Correct answer is option 'A'. Can you explain this answer?
For H3PO3 and H3PO4 the correct choice is:
a)
H3PO3 is dibasic and reducing
b)
H3PO3 is dibasic and non-reducing
c)
H3PO4 is tribasic and reducing
d)
H3PO3 is tribasic and non-reducing
|
|
Swara Joshi answered |
H3PO3 and H3PO4
H3PO3 and H3PO4 are both phosphoric acids, but they have different chemical properties.
H3PO3
- Dibasic: H3PO3 is a dibasic acid, meaning it can donate two protons (H+) per molecule. This is because it has two acidic hydrogen atoms.
- Reducing: H3PO3 is a reducing agent. It can donate electrons to other substances and undergo oxidation itself.
- Structure: H3PO3 has a structure with a central phosphorus atom bonded to three hydrogen atoms and one oxygen atom. It also has a P-OH group, which is responsible for its acidic properties.
- Reaction: When H3PO3 reacts with an oxidizing agent, it gets oxidized to H3PO4. This reaction involves the loss of hydrogen atoms and the gain of oxygen atoms.
H3PO4
- Tribasic: H3PO4 is a tribasic acid, meaning it can donate three protons (H+) per molecule. This is because it has three acidic hydrogen atoms.
- Non-reducing: H3PO4 is not a reducing agent. It does not readily donate electrons to other substances.
- Structure: H3PO4 has a structure with a central phosphorus atom bonded to four oxygen atoms and one hydrogen atom. It does not have any P-OH groups like H3PO3.
- Reaction: H3PO4 does not undergo oxidation-reduction reactions like H3PO3. It can, however, react with bases to form salts, such as sodium phosphate (Na3PO4).
Summary
In summary, H3PO3 is dibasic and reducing, while H3PO4 is tribasic and non-reducing. The key differences between them lie in their acidic properties and reactivity towards oxidation-reduction reactions. H3PO3 can donate two protons and act as a reducing agent, while H3PO4 can donate three protons but does not possess reducing properties.
H3PO3 and H3PO4 are both phosphoric acids, but they have different chemical properties.
H3PO3
- Dibasic: H3PO3 is a dibasic acid, meaning it can donate two protons (H+) per molecule. This is because it has two acidic hydrogen atoms.
- Reducing: H3PO3 is a reducing agent. It can donate electrons to other substances and undergo oxidation itself.
- Structure: H3PO3 has a structure with a central phosphorus atom bonded to three hydrogen atoms and one oxygen atom. It also has a P-OH group, which is responsible for its acidic properties.
- Reaction: When H3PO3 reacts with an oxidizing agent, it gets oxidized to H3PO4. This reaction involves the loss of hydrogen atoms and the gain of oxygen atoms.
H3PO4
- Tribasic: H3PO4 is a tribasic acid, meaning it can donate three protons (H+) per molecule. This is because it has three acidic hydrogen atoms.
- Non-reducing: H3PO4 is not a reducing agent. It does not readily donate electrons to other substances.
- Structure: H3PO4 has a structure with a central phosphorus atom bonded to four oxygen atoms and one hydrogen atom. It does not have any P-OH groups like H3PO3.
- Reaction: H3PO4 does not undergo oxidation-reduction reactions like H3PO3. It can, however, react with bases to form salts, such as sodium phosphate (Na3PO4).
Summary
In summary, H3PO3 is dibasic and reducing, while H3PO4 is tribasic and non-reducing. The key differences between them lie in their acidic properties and reactivity towards oxidation-reduction reactions. H3PO3 can donate two protons and act as a reducing agent, while H3PO4 can donate three protons but does not possess reducing properties.
The increasing order of atomic radii of the following Group 13 elements is- a)Aℓ < Ga < In < Tℓ
- b)Ga < Aℓ < In < Tℓ
- c)Aℓ < In < Ga < Tℓ
- d)Aℓ < Ga < Tℓ < In
Correct answer is option 'B'. Can you explain this answer?
The increasing order of atomic radii of the following Group 13 elements is
a)
Aℓ < Ga < In < Tℓ
b)
Ga < Aℓ < In < Tℓ
c)
Aℓ < In < Ga < Tℓ
d)
Aℓ < Ga < Tℓ < In
|
Saikat Dasgupta answered |
Atomic r adii increases on moving down a gr oup.
However due to poor shielding effect of d orbit, atomic radius of Ga is smaller than Al (anomaly). Thus the correct order is Ga < Al < In < Tl
However due to poor shielding effect of d orbit, atomic radius of Ga is smaller than Al (anomaly). Thus the correct order is Ga < Al < In < Tl
The shape of XeO2F2 molecule is- a)trigonal bipyramidal
- b)square planar
- c)tetrahedral
- d)see-saw
Correct answer is option 'D'. Can you explain this answer?
The shape of XeO2F2 molecule is
a)
trigonal bipyramidal
b)
square planar
c)
tetrahedral
d)
see-saw
|
Akshay Sharma answered |
XeO2F2 has tr igonal bipyramidal geometry, due to presence of lone pair of electrons on equitorial position, its shape is see-saw.
The correct statement(s) about O3 is(are)- a)O—O bond lengths are equal
- b)Thermal decomposition of O3 is endothermic
- c)O3 is diamagnetic in nature
- d)O3 has a bent structure
Correct answer is option 'A,C,D'. Can you explain this answer?
The correct statement(s) about O3 is(are)
a)
O—O bond lengths are equal
b)
Thermal decomposition of O3 is endothermic
c)
O3 is diamagnetic in nature
d)
O3 has a bent structure
|
Nandini Nair answered |
Ozone is diamagnetic in nature (due to presence of paired electron) and both the O – O bond length are equal. It has a bent structure.
Which one of the following reactions of xenon compounds is not feasible?- a)3Xe F4 + 6H2O → 2 Xe + XeO3 + 12HF + 1.5O2
- b)2Xe F2 + 2H2O → 2Xe + 4HF + O2
- c)Xe F6 + RbF → Rb[Xe F7]
- d)XeO3 + 6HF → Xe F6 + 3H2O
Correct answer is option 'D'. Can you explain this answer?
Which one of the following reactions of xenon compounds is not feasible?
a)
3Xe F4 + 6H2O → 2 Xe + XeO3 + 12HF + 1.5O2
b)
2Xe F2 + 2H2O → 2Xe + 4HF + O2
c)
Xe F6 + RbF → Rb[Xe F7]
d)
XeO3 + 6HF → Xe F6 + 3H2O
|
|
Shilpa Saha answered |
The products of the concerned reaction react each other forming back the reactants.
XeF6 + 3H2O → XeO3+ 6HF .
Chlorine acts as a bleaching agent only in presence of- a)dry air
- b)moisture
- c)sunlight
- d)pure oxygen
Correct answer is option 'B'. Can you explain this answer?
Chlorine acts as a bleaching agent only in presence of
a)
dry air
b)
moisture
c)
sunlight
d)
pure oxygen
|
Rohit Chakraborty answered |
Explanation:
Chlorine (Cl2) is a highly reactive gas and it can act as a bleaching agent due to its strong oxidizing properties. It oxidizes the colored substances and converts them into colorless compounds, thus bleaching them. However, this oxidation process requires the presence of moisture in the surroundings.
Effect of Moisture:
When chlorine gas comes in contact with moisture, it forms hypochlorous acid (HClO) and hydrochloric acid (HCl) as shown below:
Cl2 + H2O → HClO + HCl
Hypochlorous acid is a highly reactive oxidizing agent and it reacts with the colored substances to bleach them. For example, when hypochlorous acid reacts with the colored pigment in a cloth, it oxidizes the pigment and breaks it down into colorless compounds, thus bleaching the cloth.
Effect of Dry Air:
In the absence of moisture, chlorine gas does not react with the colored substances and hence, it cannot bleach them. Therefore, chlorine does not act as a bleaching agent in dry air.
Effect of Sunlight:
Sunlight can enhance the bleaching effect of chlorine as it provides energy for the oxidation process. However, sunlight is not essential for the bleaching action of chlorine.
Effect of Pure Oxygen:
Pure oxygen can also act as an oxidizing agent and bleach the colored substances. However, chlorine is a stronger oxidizing agent than pure oxygen and hence, it is more effective as a bleaching agent.
Therefore, the correct answer is option B - Chlorine acts as a bleaching agent only in the presence of moisture.
Chlorine (Cl2) is a highly reactive gas and it can act as a bleaching agent due to its strong oxidizing properties. It oxidizes the colored substances and converts them into colorless compounds, thus bleaching them. However, this oxidation process requires the presence of moisture in the surroundings.
Effect of Moisture:
When chlorine gas comes in contact with moisture, it forms hypochlorous acid (HClO) and hydrochloric acid (HCl) as shown below:
Cl2 + H2O → HClO + HCl
Hypochlorous acid is a highly reactive oxidizing agent and it reacts with the colored substances to bleach them. For example, when hypochlorous acid reacts with the colored pigment in a cloth, it oxidizes the pigment and breaks it down into colorless compounds, thus bleaching the cloth.
Effect of Dry Air:
In the absence of moisture, chlorine gas does not react with the colored substances and hence, it cannot bleach them. Therefore, chlorine does not act as a bleaching agent in dry air.
Effect of Sunlight:
Sunlight can enhance the bleaching effect of chlorine as it provides energy for the oxidation process. However, sunlight is not essential for the bleaching action of chlorine.
Effect of Pure Oxygen:
Pure oxygen can also act as an oxidizing agent and bleach the colored substances. However, chlorine is a stronger oxidizing agent than pure oxygen and hence, it is more effective as a bleaching agent.
Therefore, the correct answer is option B - Chlorine acts as a bleaching agent only in the presence of moisture.
Graphite is a soft solid lubricant extremely difficult to melt. The reason for this anomalous behaviour is that graphite- a)is an allotropic form of diamond
- b)h as molecules of var iable molecular masses like polymers
- c)has carbon atoms arranged in large plates of rings of strongly bound carbon atoms with weak interplate bonds
- d)is a non-crystalline substance
Correct answer is option 'C'. Can you explain this answer?
Graphite is a soft solid lubricant extremely difficult to melt. The reason for this anomalous behaviour is that graphite
a)
is an allotropic form of diamond
b)
h as molecules of var iable molecular masses like polymers
c)
has carbon atoms arranged in large plates of rings of strongly bound carbon atoms with weak interplate bonds
d)
is a non-crystalline substance
|
|
Sahana Ahuja answered |
In graphite, carbon is sp2 hybridized. Each carbon is thus linked to three other carbon atoms forming hexagonal rings. Since only three electrons of each carbon are used in making hexagonal ring, fourth electron of each carbon is free to move. This makes graphite a good conductors of heat and electricity.
Further graphite has a two dimensional sheet like structure. These various sheets are held together by weak van der Waal’s force of attraction. due to these weak forces of attraction, one layer can slip over the other. This makes graphite soft and a good lubricating agent.
Further graphite has a two dimensional sheet like structure. These various sheets are held together by weak van der Waal’s force of attraction. due to these weak forces of attraction, one layer can slip over the other. This makes graphite soft and a good lubricating agent.
Hydrolysis of one mole of peroxodisulphuric acid produces- a)two moles of sulphuric acid
- b)two moles of peroxomonosulphuric acid
- c)one mole of sulphuric acid and one mole of peroxomonosulphuric acid
- d)one mole of sulphuric acid, one mole of peroxomonosulphuric acid and one mole of hydrogen peroxide.
Correct answer is option 'D'. Can you explain this answer?
Hydrolysis of one mole of peroxodisulphuric acid produces
a)
two moles of sulphuric acid
b)
two moles of peroxomonosulphuric acid
c)
one mole of sulphuric acid and one mole of peroxomonosulphuric acid
d)
one mole of sulphuric acid, one mole of peroxomonosulphuric acid and one mole of hydrogen peroxide.
|
|
Athira Datta answered |
Concentrated hydrochlor ic acid when kept in open air sometimes produces a cloud of white fumes. The explanation for it is that- a)oxygen in air reacts with the emitted HCl gas to form a cloud of chlorine gas
- b)strong affinity of HCl gas for moisture in air results in forming of droplets of liquid solution which appears like a cloudy smoke.
- c)due to strong affinity for water, concentrated hydrochloric acid pulls moisture of air towards itself.This moisture forms droplets of water and hence the cloud.
- d)concentrated hydrochloric acid emits strongly smelling HCl gas all the time.
Correct answer is option 'A'. Can you explain this answer?
Concentrated hydrochlor ic acid when kept in open air sometimes produces a cloud of white fumes. The explanation for it is that
a)
oxygen in air reacts with the emitted HCl gas to form a cloud of chlorine gas
b)
strong affinity of HCl gas for moisture in air results in forming of droplets of liquid solution which appears like a cloudy smoke.
c)
due to strong affinity for water, concentrated hydrochloric acid pulls moisture of air towards itself.This moisture forms droplets of water and hence the cloud.
d)
concentrated hydrochloric acid emits strongly smelling HCl gas all the time.
|
Sneha Sengupta answered |
Introduction
When concentrated hydrochloric acid (HCl) is exposed to open air, it may produce a cloud of white fumes. This phenomenon is primarily due to the interaction of HCl gas with moisture in the air.
Understanding the Reaction
- HCl is a strong acid that readily vaporizes, releasing HCl gas.
- When this gas comes into contact with water vapor in the atmosphere, it reacts to form hydrochloric acid mist.
Why Option A is Incorrect
- Option A suggests that oxygen in the air reacts with emitted HCl gas to form chlorine gas. This is misleading as the primary reaction occurring is not between HCl gas and oxygen, but rather between HCl gas and moisture.
- The formation of chlorine gas from HCl would require different conditions and is not relevant in this context.
Correct Explanation for the White Fumes
- The white fumes observed are actually tiny droplets of hydrochloric acid solution formed when HCl gas interacts with water vapor.
Key Points
- Strong Affinity for Water: HCl gas has a high affinity for water, leading to its absorption from the air.
- Formation of Mist: This absorption forms droplets of hydrochloric acid, which appear as a white cloud.
- Visual Appearance: The cloud is a visible mist, primarily composed of tiny liquid HCl droplets suspended in the air.
Conclusion
In summary, the cloud of white fumes from concentrated hydrochloric acid is due to its reaction with moisture in the air, forming hydrochloric acid mist. This highlights its strong affinity for water rather than any reaction with oxygen.
When concentrated hydrochloric acid (HCl) is exposed to open air, it may produce a cloud of white fumes. This phenomenon is primarily due to the interaction of HCl gas with moisture in the air.
Understanding the Reaction
- HCl is a strong acid that readily vaporizes, releasing HCl gas.
- When this gas comes into contact with water vapor in the atmosphere, it reacts to form hydrochloric acid mist.
Why Option A is Incorrect
- Option A suggests that oxygen in the air reacts with emitted HCl gas to form chlorine gas. This is misleading as the primary reaction occurring is not between HCl gas and oxygen, but rather between HCl gas and moisture.
- The formation of chlorine gas from HCl would require different conditions and is not relevant in this context.
Correct Explanation for the White Fumes
- The white fumes observed are actually tiny droplets of hydrochloric acid solution formed when HCl gas interacts with water vapor.
Key Points
- Strong Affinity for Water: HCl gas has a high affinity for water, leading to its absorption from the air.
- Formation of Mist: This absorption forms droplets of hydrochloric acid, which appear as a white cloud.
- Visual Appearance: The cloud is a visible mist, primarily composed of tiny liquid HCl droplets suspended in the air.
Conclusion
In summary, the cloud of white fumes from concentrated hydrochloric acid is due to its reaction with moisture in the air, forming hydrochloric acid mist. This highlights its strong affinity for water rather than any reaction with oxygen.
Which of the following statements about anhydrous aluminium chloride is correct?- a)it exists as AlCl3 molecules
- b)it is not easily hydrolysed
- c)it sublimes at 100ºC under vacuum
- d)it is a strong Lewis base
Correct answer is option 'C'. Can you explain this answer?
Which of the following statements about anhydrous aluminium chloride is correct?
a)
it exists as AlCl3 molecules
b)
it is not easily hydrolysed
c)
it sublimes at 100ºC under vacuum
d)
it is a strong Lewis base
|
|
Advait Ghoshal answered |
AlCl3 exists as a dimer (Al2Cl6). It is a strong Lewis acid as it has an incomplete octet and has a tendency to gain electrons. AlCl3 undergoes hydrolysis easily and forms an acidic solution.
AlCl3 + 3H2O → Al(OH)3 + 3HCl
AlCl3 + 3H2O → Al(OH)3 + 3HCl
Option (c) is true that AlCl3 sublimes at 100ºC under vacuum.
AlCl3 is a Lewis acid.
AlCl3 is a Lewis acid.
The crystalline form of borax has- a)tetranuclear [B4O5(OH)4]2– unit
- b)all boron atoms in the same plane
- c)equal number of sp2 and sp3 hybridized boron atoms
- d)one terminal hydroxide per boron atom
Correct answer is option 'A,C,D'. Can you explain this answer?
The crystalline form of borax has
a)
tetranuclear [B4O5(OH)4]2– unit
b)
all boron atoms in the same plane
c)
equal number of sp2 and sp3 hybridized boron atoms
d)
one terminal hydroxide per boron atom
|
Juhi Das answered |
The crystalline form of borax is actually represented by the chemical formula Na2B4O7·10H2O. It consists of boron, oxygen, sodium, and water molecules. The structure of borax can be described as a network of interconnected borate ions (B4O5(OH)4) linked together by sodium ions (Na+). It also contains 10 water molecules (H2O) that are associated with the sodium ions and the borate ions.
Identify the incorrect statement among the following.- a)Br2 reacts with hot and strong NaOH solution to give NaBr and H2O.
- b)Ozone reacts with SO2 to give SO3.
- c)Silicon reacts with NaOH(aq) in the presence of air to give Na2SiO3 and H2O.
- d)Cl2 reacts with excess of NH3 to give N2 and HCl.
Correct answer is option 'D'. Can you explain this answer?
Identify the incorrect statement among the following.
a)
Br2 reacts with hot and strong NaOH solution to give NaBr and H2O.
b)
Ozone reacts with SO2 to give SO3.
c)
Silicon reacts with NaOH(aq) in the presence of air to give Na2SiO3 and H2O.
d)
Cl2 reacts with excess of NH3 to give N2 and HCl.
|
|
Shilpa Saha answered |
Chlorine reacts with excess of ammonia to produce ammonium chloride and nitrogen.
This question contains STATEMENT-1 (Assertion/ Statement ) and STATEMENT-2 (Reason/Explanation) and has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.Q. Statement-1 : Although PF5 , PCl5 and PBr5 are known,the pentahalides of nitrogen have not been observed Statement-2 : Phosphorus has lower electronegativity than nitrogen.- a)Statement-1 is True, Statement-2 isTrue; Statement-2 is a correct explanation for Statement-1
- b)Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
- c)Statement-1 is True, Statement-2 is False
- d)Statement-1 is False, Statement-2 is True.
Correct answer is option 'B'. Can you explain this answer?
This question contains STATEMENT-1 (Assertion/ Statement ) and STATEMENT-2 (Reason/Explanation) and has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.
Q.
Statement-1 : Although PF5 , PCl5 and PBr5 are known,the pentahalides of nitrogen have not been observed Statement-2 : Phosphorus has lower electronegativity than nitrogen.
a)
Statement-1 is True, Statement-2 isTrue; Statement-2 is a correct explanation for Statement-1
b)
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
c)
Statement-1 is True, Statement-2 is False
d)
Statement-1 is False, Statement-2 is True.
|
|
Sahana Joshi answered |
Nitrogen cannot form pentahalides because it cannot expand its octet due to non-availability of d-orbitals.
So E is not correct explanation of S.
So E is not correct explanation of S.
The correct statement(s) regarding,(i) HClO, (ii) HClO2, (iii) HClO3 and (iv) HClO4, is(are)- a)The number of Cl=O bonds in (ii) and (iii) together is two
- b)The number of lone pairs of electrons on Cl in (ii) and (iii) together is three
- c)The hybridization of Cl in (iv) is sp3
- d)Amongst (i) to (iv), the strongest acid is (i)
Correct answer is option 'B,C'. Can you explain this answer?
The correct statement(s) regarding,
(i) HClO, (ii) HClO2, (iii) HClO3 and (iv) HClO4, is(are)
a)
The number of Cl=O bonds in (ii) and (iii) together is two
b)
The number of lone pairs of electrons on Cl in (ii) and (iii) together is three
c)
The hybridization of Cl in (iv) is sp3
d)
Amongst (i) to (iv), the strongest acid is (i)
|
Ishita Reddy answered |
Number of Cl = O bonds in (ii) and (iii) together is 3 Number of lone pairs on Cl in (ii) and (iii) together is 3 Hybridisation of Cl in all the four is sp3
Strongest acid is HClO4 (iv)
Strongest acid is HClO4 (iv)
Which one of the following statement regarding helium is incorrect ?- a)It is used to produce and sustain powerful superconducting magnets
- b)It is used as a cryogenic agent for carrying out experiments at low temperatures
- c)It is used to fill gas balloons instead of hydrogen because it is lighter and non-inflammable
- d)It is used in gas-cooled nuclear reactors
Correct answer is option 'C'. Can you explain this answer?
Which one of the following statement regarding helium is incorrect ?
a)
It is used to produce and sustain powerful superconducting magnets
b)
It is used as a cryogenic agent for carrying out experiments at low temperatures
c)
It is used to fill gas balloons instead of hydrogen because it is lighter and non-inflammable
d)
It is used in gas-cooled nuclear reactors
|
|
Sanchita Patel answered |
Helium is heavier than hydrogen although it is non inflammable
On heating ammonium dichromate, the gas evolved is- a)oxygen
- b)ammonia
- c)nitrous oxide
- d)nitrogen
Correct answer is option 'D'. Can you explain this answer?
On heating ammonium dichromate, the gas evolved is
a)
oxygen
b)
ammonia
c)
nitrous oxide
d)
nitrogen
|
Amrutha Chaudhary answered |
(NH4)2Cr2O7 → N2 + Cr2O3 + 4H2O
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