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30051.

The higher oxidation states of transition elements are found to be in the combination with A and B which are :

Answer»

F, O
O, N
O, S
F, Cl

Answer :A
Discussion
30052.

The higher the critical temperature of the gas

Answer»

GREATER is its EXTENT of adsorption
Lower its adsorption
Lesser is the CASE if liquification
Lesser is its VOLATILE nature

Answer :A
Discussion
30053.

The higher B.P of Glycerol than that of Propanol is because of

Answer»

hybridization
resonance
H-bond
All of these

Solution :`UNDERSET("ETHANOL")(CH_(3)CH_(2)OH)"" underset("PROPANE 1,2,3 tri-ol")(underset(OH)underset(|)CH_(2)-underset(OH)underset(|)CH-underset(OH)underset(|)CH_(2))`
Having more -OH group.
Discussion
30054.

The high temperature polymers of Si are called :

Answer»

Silicates
Silicones
Silanes
Silicon halides.

Answer :B
Discussion
30055.

the high viscosity and high boiling point of HF is due to :

Answer»

low DISSOCIATION energy of `F_(2)` molecules
associated nature due to HYDROGEN bonding
ionic CHARACTER of HF
high ELECTRONEGATIVITY of fluorine.

Answer :B
Discussion
30056.

The high reactivity of fluorine is mainly due to

Answer»

HIGH HEAT of hydration
small size
low bond DISSOCIATION energy of F-F bond
high ionisationpotential

Answer :C
Discussion
30057.

The high reactivity of alkyl halides can be explained in terms of nature of C-X bond which is a highly polarised covalent bond. This polarity is responsible for the nucleophilic substitution reaction of alkyl halides which jmostly occur by S_(N^(1)) and S_(N^(2)) mechanisms. S_(N^(1)) reaction is a two step process and in the first step R-X ionises to give carbocation (slow process). IN the second step, the nucleophile attacks the carbocation from either side to form the product (fast process). IN S_(N^(1)) reaction, there can be racemization and inversion. S_(N^(1)) reaction is favoured by heavy (bulky) group on the carbon atom attached to halogens. IN S_(N^(2)) reaction, the strong nucleophile OH^(-) attacks fromt he opposite side Of the halogen atom to give an intermediate (transition state), which breaks to yields to product (alcohol) and leaving group (X^(-)). The alcohol has a configuratio opposite to that of the halide and is said to proceed with inversion of configuration. S_(N^(2)) reaction is favoured by small groups on the carbon atom attached to halogen. Q. Which of the following is an example of S_(N^(2)) reaction?

Answer»

`CH_(3)Br+OH^(-)toCH_(3)OH+Br^(-)`
`CH_(3)CH_(2)OHtoCH_(2)=CH_(2)`
`(CH_(3))_(2)CHBr+OH^(-)to(CH_(3))_(2)CHOH+Br^(-)`
`(CH_(3))_(3)CBr+OH^(-)to(CH_(3))_(3)COH+Br^(-)`.

SOLUTION :It is an example of `S_(N^(2)) ` REACTION.
Discussion
30058.

The high reactivity of alkyl halides can be explained in terms of nature of C-X bond which is a highly polarised covalent bond. This polarity is responsible for the nucleophilic substitution reaction of alkyl halides which jmostly occur by S_(N^(1)) and S_(N^(2)) mechanisms. S_(N^(1)) reaction is a two step process and in the first step R-X ionises to give carbocation (slow process). IN the second step, the nucleophile attacks the carbocation from either side to form the product (fast process). IN S_(N^(1)) reaction, there can be racemization and inversion. S_(N^(1)) reaction is favoured by heavy (bulky) group on the carbon atom attached to halogens. IN S_(N^(2)) reaction, the strong nucleophile OH^(-) attacks fromt he opposite side Of the halogen atom to give an intermediate (transition state), which breaks to yields to product (alcohol) and leaving group (X^(-)). The alcohol has a configuratio opposite to that of the halide and is said to proceed with inversion of configuration. S_(N^(2)) reaction is favoured by small groups on the carbon atom attached to halogen. Q. The main product formed in the following reaction is : CH_(3)-underset(CH_(3))underset(|)(C)H-CH_(2)Br+CH_(3)CH_(2)O^(-)overset(S_(N^(1))to

Answer»

`CH_(3)-UNDERSET(CH_(3))underset(|)(C)H-CH_(2)OCH_(2)CH_(3)`
`CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-OCH_(2)CH_(3)`
`CH_(3)-underset(CH_(3))underset(|)(C)H-CH_(3)OH`
`CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-OH`

Solution :It is expected to be the major product because the nucleophile `(C_(2)H_(5)O^(-))` will prefer to COMBINE with `CH_(3)underset(CH_(3))underset(|)(C)Hoversete(o+)(CH_(2))` carbocation (primary) due to minimum steric hindrance involved.
Discussion
30059.

The high reactivity of alkyl halides can be explained in terms of nature of C-X bond which is a highly polarised covalent bond. This polarity is responsible for the nucleophilic substitution reaction of alkyl halides which jmostly occur by S_(N^(1)) and S_(N^(2)) mechanisms. S_(N^(1)) reaction is a two step process and in the first step R-X ionises to give carbocation (slow process). IN the second step, the nucleophile attacks the carbocation from either side to form the product (fast process). IN S_(N^(1)) reaction, there can be racemization and inversion. S_(N^(1)) reaction is favoured by heavy (bulky) group on the carbon atom attached to halogens. IN S_(N^(2)) reaction, the strong nucleophile OH^(-) attacks fromt he opposite side Of the halogen atom to give an intermediate (transition state), which breaks to yields to product (alcohol) and leaving group (X^(-)). The alcohol has a configuratio opposite to that of the halide and is said to proceed with inversion of configuration. S_(N^(2)) reaction is favoured by small groups on the carbon atom attached to halogen. Q. Which one of the following compounds is most readily hydrolysed by S_(N^(1)) mechanism.?

Answer»

`CH_(3)CH=CHCl`
`(C_(6)H_(5))_(3)C Cl`
`ClCH_(2)CH=CH_(2)`.
.

Solution :This is linked with the relative stabilities of the CARBOCATION intermediates formed in the slow or rate determining step. The increasing order of STABILITY is:
(i). `CH_(2)=CH=overset(o+)(C)H_(@2)`.
(B).
Discussion
30060.

The high reactivity offluorine is due to _______

Answer»

high ionisation energy
LOW bond dissociation energy
low electron affinity
high electronegativity

Answer :A::B::C::D
Discussion
30061.

The high reactivity of alkyl halides can be explained in terms of nature of C-X bond which is a highly polarised covalent bond. This polarity is responsible for the nucleophilic substitution reaction of alkyl halides which jmostly occur by S_(N^(1)) and S_(N^(2)) mechanisms. S_(N^(1)) reaction is a two step process and in the first step R-X ionises to give carbocation (slow process). IN the second step, the nucleophile attacks the carbocation from either side to form the product (fast process). IN S_(N^(1)) reaction, there can be racemization and inversion. S_(N^(1)) reaction is favoured by heavy (bulky) group on the carbon atom attached to halogens. IN S_(N^(2)) reaction, the strong nucleophile OH^(-) attacks fromt he opposite side Of the halogen atom to give an intermediate (transition state), which breaks to yields to product (alcohol) and leaving group (X^(-)). The alcohol has a configuratio opposite to that of the halide and is said to proceed with inversion of configuration. S_(N^(2)) reaction is favoured by small groups on the carbon atom attached to halogen. Q. Which among the following will not give S_(N^(1)) reaction?

Answer»

`CH_(3)-underset(C_(6)H_(5))underset(|)(C)H-Br`
`(CH_(3))_(3)C-Br`
`CH_(3)CH_(2)-I`
`(C_(6)H_(5))_(2)underset(CH_(3))underset(|)(C)-Cl`.

SOLUTION :Primary alkyl halides PREFERABLE REACT by `S_(N^(2))` mechanism.
Discussion
30062.

The high melting point and insolubility of sulphanilic acid in organic solvents are due to its ………… structure .

Answer»


ANSWER :ZWITTER ION
Discussion
30063.

The high oxidizing power of F_(2) is due to :

Answer»

HIGH heat of HYDROGEN
Low heat of dissociation
Both
High heat of dissociation and low hieat of hydration

SOLUTION :Due to low heat of dissociation of `F_(2)` MOLECULE, and high hydration ENERGY of `F^(-)` ion.
Discussion
30064.

The high oxidising power of F_2 is due to:

Answer»

HIGH ELECTRON affinity
High heat of DISSOCIATION and low heat of HYDRATION
High heat of hydration and low heat of dissociation
High electronegativity

Answer :C
Discussion
30065.

The high ionisation enthalpies of noble gases is due to:

Answer»

their large sizes
their stable VALENCE SHELL configuration
high NUCLEAR charge
their inertness TOWARDS reactions.

Answer :B
Discussion
30066.

The high density of water compared to ice is due to :

Answer»

dipole - dipole interactions
HYDROGEN bonding interactions
dipole induced dipole interactions
none of the above

Answer :B
Discussion
30067.

The high boiling points of carboxylic acids is due to

Answer»

WEAK VANDERWAAL's forces.
intermolecular HYDROGEN bonding.
intramolecular hydrogen bondling.
delocalisation of `PI` electrons.

Answer :B
Discussion
30068.

DeltaH_(t)^(@) for CO_(2)(g) and H_(2)O(g) are -393.5, -110.5 and -241.8 kJ"mol"^(-1) respectively. The standard enthalpy change (in kJ) for the reaction. CO_(2)(g) + H_(2)(g) to CO(g) + H_(2)O(g) is:

Answer»

524.1
41.2
-262.5
-41.2

Answer :B
Discussion
30069.

The high boiling point and viscosity of H_(2)SO_4 is due

Answer»

HYDROGEN BONDING
covalent bonding
IONIC bonding
vander WAAL's forces

Answer :A
Discussion
30070.

The high amount of E.coli in water is the indicator of

Answer»

HARDNESS of WATER
INDUSTRIAL pollution
Sewage pollution
Presence of CHLORINE in water

Answer :C
Discussion
30071.

The hexaaquomanganese (II) ion contains five unpaired electrons while the hexacyano ion contains only one unpaired electron. Explain using Crystal Field Theory.

Answer»

Solution :Mn in +2 state has the CONFIGURATION `3d^(5)`. In presence of `H_(2)O` as ligand, the distribution of these five electrons is `t_(2g)^(3)` `e_(g)^(2)`, i.e., all the electrons remain unpaired. In presence of `CN^(-)` as ligand, the distribution is `t_(2g)^(5) e_(g)^(0)`, i.e., TWO `t_(2g)` orbitals contain PAIRED electrons while the third `t_(2g)` ORBITAL contains one unpaired electron.
Discussion
30072.

The hexaaquo manganese (II) ion contains five unpaired electrons while the hexacyanoion contains only one unpaired electron. Explain using Crystal Field Theory.

Answer»

SOLUTION :Mn in +2 state has the configuration `3d^(5)`. In PRESENCE of `H_(2)O)` as weak ligand PAIRING does not take place. Distribution of these five electrons is `t_(2g)^(3)e_(g)^(2)`, i.e., all the electrons REMAIN unpafred. In presence of `CN^(-)` as strong ligand, the distribution is `t_(2g)^(5)e_(g)^(0)`, because of pairing and hence two `t_(2g)` orbitals contain paired electrons while the third `t_(2g)` orbital contains one unpaired electron.
Discussion
30073.

The hexaaqua complex[Ni(H_(2)O)_(6)]^(2+) is green, whereas the corresponding ammonia complex [Ni(NH_(3))_(6)]^(2+) is violet. Explain.

Answer»

SOLUTION :`NH_(3)` is a stronger ligand than `H_(2)O`, SO `[Ni(NH_(3))]^(2+)` will absorb higher energy LIGHT than `[Ni(H_(2)O)_(6)]^(2+)`. Being green, `[Ni(H_(2)O)_(6)]^(2+)` is PROBABLY absorbing higher energy YELLOW green light
Discussion
30074.

The heterolytic cleavage of a carbon-chlorine bond produces :

Answer»

Two free RADICALS
Two carbanions
Two CARBONIUM ions
ONE cation and one ANION.

ANSWER :D
Discussion
30075.

The Henry's law constant for the solubility of N_(2) gas in water at 298 K is 1.0xx10^(5) atm. The mole fraction of N_(2) in air is 0.8. The number of moles of N_(2) from air dissolved in 10 moles of water at 298 K and 5 atm pressure is

Answer»

`4.0xx10^(-4)`
`5.0xx10^(-5)`
`5.0xx10^(-4)`
`4.0xx10^(-6)`

Solution :`P_(N_(2))=K_(H)xx X_(N_(2))`
`X_(N_(2))=(1)/(10^(5))xx0.8xx5=4XX10^(-5)` per mole
In 10 mole, solubility is `4xx10^(-4)`.
Discussion
30076.

The Henry's law constant for the solubility of nitrogen gas in water at 298 K is 1.0 xx 10^5 a t m. The mole fraction of N_2 in air is 0.8. The number of moles of N_2 from air dissolved in 10 moles fo water at 298 K and 5 atm. Pressure is :

Answer»

`4.0 xx 10^(-4)`
`4.0 xx 10^(-5)`
`5.0 xx 10^(-4)`
`4.0 xx 10^(-6)`

Answer :A
Discussion
30077.

The Henry's law constant for the solubility of N_(2) gas in water at 298 K is 1.0 xx 10^(5) atm. The mole fraction of N_(2) in air is 0.8. The number of moles of N_(2) from air dissolved in 10 moles of water at 298 K and 5 atm. Pressure is:

Answer»

`4.0xx10^(-4)`
`4.0xx10^(-5)`
`5.0xx10^(-4)`
`4.0xx10^(-6)`

Solution :`P_(N_(2))=K_(H)x_(N_(2)) ("in air")`
But `P_(N_(2))=X_(N_(2))xxPT=0.8xx5` ATM (in WATER)
`(4 atm)= K_(H)x_(N_(2))`
`x_(N_(2))=((4atm))/K_(H)=((4atm))/((1.0xx10^(5)atm))`
`=4xx10^(-5)`
`"Now" n_(N_(2))/(n_(H_(2))+n_(H_(2)O))=x_(N_(2))=4xx10^(-5)`
`n_(H_(2))/n_(H_(2)O)=4xx10^(-5)`
`n_(H_(2))=4xx10^(-5)xxn_(H_(2)O)`
` 4xx10^(-5)XX(10 "mol")`
`=4xx10^(-4)"mol"`
Discussion
30078.

The Henry's law constant for oxygen dissolved in water is 4.34 xx10^(4) atm at 25^(@)C. If the the partial pressure of oxygen in air is 0.2 atm under atmospheric conditions, calculate the concentration (in moles per litre) of dissolved oxygen in water in equilibrium with air at 25^(@)C.

Answer»


Solution :`P_(O_(2))=K_(H)xxx_(O_(2))"or"x_(O_(2))=(P_(O_(2)))/(K_(H))=("0.2 atm")/(4.34xx10^(4)" atm")=4.6xx10^(-6)`
`"To convert it into MOLARITY."x_(O_(2))=(n_(O_(2)))/(n_(O_(2))+n_(H_(2)O))`
`"For 1 LITRE of water, "m_(H_(2)O)=1000//18="55.5 moles "therefore (m_(O_(2)))/(55.5)=4.6xx10^(-6)"or"n_(O_(2))=2.55xx10^(-4)" mole"`
`"Hence,molarity "=2.55xx10^(-4)"mol L"^(-1)`
Discussion
30079.

The Henry's constant for solubility of N_(2) gas in water at 298K is 1.0xx10^(5) atm. The mole fraction of N_(2) in air is 0.8 . The number of moles of N_(2) from air dissolved in 10 moles of water at 298K and 5 atm pressure is :

Answer»

`4XX10^(-4)`
`4.0xx10^(-5)`
`5.0xx10^(-4)`
`4.0xx10^(-6)`

ANSWER :A
Discussion
30080.

The hemiaceltel form of glucose is indicatred by

Answer»

reaction with `(CH_(3)CO)_(2)O`/pyridine
OXIDATION withTollens' REAGENT
reduction with HI/P
glycoside formation

Solution :
Glycoside is funtionallyacetal which is FORMED from hemiacetal .
Discussion
30081.

The helium nucleus contains

Answer»

4 PROTONS
4 neutrons
2 neutrons and 2 protons
3 protons and 2 electrons

Answer :C
Discussion
30082.

The helical structure or a secondary structure of proteins is stabilized by:

Answer»

PEPTIDE bonds
Dipeptide bonds
H-bond
None

Answer :B
Discussion
30083.

The helical structure of proteins is stabilized by

Answer»

H-bonding 
van der Waals' FORCES 
IONIC BOND 
PEPTIDE bond 

Answer :A
Discussion
30084.

Thehelical structure of proteinis stabilzed by …………. .

Answer»

oxygenbonds
PEPTIDE BONDS
dipeptidebonds
HYDROGEN bonds

ANSWER :D
Discussion
30085.

The helical structure of protein is stabilized by

Answer»

DIPEPTIDE bonds
HYDROGEN bonds
ether bonds
peptide bonds.

ANSWER :B
Discussion
30086.

Helical structure of protein is stabilised by :

Answer»


ANSWER :HYDROGEN BONDS
Discussion
30087.

The helical structure of protein is stabilised by …………...........

Answer»

Ionic BOND
H - bond
VAN DAR WAAL's forces
Covalent bond

SOLUTION :H - bond
Discussion
30088.

The helical structure of protein is stabilised by _____.

Answer»

PEPTIDE BONDS
disulphide bonds
HYDROGEN bonds
Vanderwaals INTER action

Solution :hydrogen bonds
Discussion
30089.

The helical form of the periodic table was given by De Chancourtouis and was called telluric helix after the ele- ment ‘tellurium’ because

Answer»

Tellurium was the last element in his GIVEN table
Tellurium FELL NEAR the center of his table
Tellurium was the most ABUNDANT of all elements featured in his table
Tellurium was the only semi-metal in his table

Answer :B
Discussion
30090.

The Heisenberg's uncertainty principle can be applied to:

Answer»

A CRICKET ball
A football
A JET aeroplane
An electron

Answer :D
Discussion
30091.

The height of a boy who is 5 feet and 9 inches may be written with three significant figures as (1 inch = 2.54 cm)

Answer»

175.3 CM
175.26 cm
`0.1753 XX 10^3 cm`
`1.75 xx 10^2 cm `

ANSWER :A
Discussion
30092.

The heaviest atom amongst the following is

Answer»

U
Ra
Pb
Hg

Solution :`UgtRagtPbgtHg`
Discussion
30093.

The heavisest atom oamongst the following is

Answer»

URANIUM
raidum
LEAD
mercury

Answer :A
Discussion
30094.

The heavier elements of group 13, 14 and 15 besides their group oxidation state exhibit another oxidation state which is two units lower than the group oxidation state. The stability of lower oxidation state increases down the group. The display of lower oxidation state is due to inert pair effect.QThe strongest reducing agent among the following is:

Answer»

`GeCl_(2)`
`SnCl_(2)`
`PbCl_(2)`
 TICI

Answer :A
Discussion
30095.

The heavier elements of group 13, 14 and 15 besides their group oxidation state exhibit another oxidation state which is two units lower than the group oxidation state. The stability of lower oxidation state increases down the group. The display of lower oxidation state is due to inert pair effect.QWhich of the following statements is incorrect?

Answer»

 Boron shows only +3 OXIDATION state
n `GA^(+3)` oxidation state is more STABLE than + 1 oxidation state
In `Sn^(+3)` oxidation state is more stable than +4 oxidation state
in `Ti^(+1)` oxidation state is more stable than +3 oxidation state

Answer :C
Discussion
30096.

The heavier elements of group 13, 14 and 15 besides their group oxidation state exhibit another oxidation state which is two units lower than the group oxidation state. The stability of lower oxidation state increases down the group. The display of lower oxidation state is due to inert pair effect.QWhich of the halide does not exist?

Answer»

 TiCI
`PbBr_(4)`
 `PbCl_(2)`
`SnCI_(2)`

ANSWER :B
Discussion
30097.

The heats of solution of one mole of Na and that of Na_2Oin water under standardconditions are -183.79 kJ/mol and -237.94 kJ/mol respectively, water being taken in large excess in both the cases. Calculate the standard heat of formation of sodium oxide if the standard heat of formation of water is –285.84 kJ/mol.

Answer»

SOLUTION :`NA + H_2O = NAOH+ 1/2 H_2`
`Na_2 O + H_2O = 2NAOH`
-415.48kJ/mol
Discussion
30098.

The heats of oxidation of Mg and Fe are given (a) Mg + 1/2O_(2) to MgO , DeltaH = -145700 cals (b) 2Fe + 3/2O_(2) to Fe_(2)O_(3), DeltaH=-193500 cals The het produced in the reaction 3Mg + Fe_(2)O_(3) to 3MgO + 2Fe, is

Answer»

`-243600` CALS
`-257800` cals
`-271400` cals
`-222800` cals

ANSWER :A
Discussion
30099.

The heats of vaporization of noble gases vary in the order

Answer»

He > NE > Ar > Kr > XE > Rn
He < Ne < Ar < Kr < Xe < Rn
Xe < K < Ne < He < Rn < AI
He < Ne = Ar > KI < Xe < Rn

Answer :B
Discussion
30100.

The heats of neutralisation of four acids A, B, C, D are -13.7, -9.4, -11.2 and -12.4 kcal respectively when they are neutralised by a common base. The acidic character obeys the order:

Answer»

AgtBgtCgtD
AgtDgtCgtB
DgtCgtBgtA
DgtBgtCgtA

Answer :B
Discussion