Welcome to Inam Jazbi’s ultimate guide to XI Chemistry, featuring chapter-wise textbook MCQs and the most updated 2026 past papers! 🚀 Whether you’re preparing for your board exams or competitive tests like MDCAT/ECAT, this resource has everything you need to excel in chemistry.
🔎💥 XI MCQs Chapter 1 ………… Stoichiometry
1.
If the volume occupied by oxygen gas (O2) at STP is 44.8dm3,
the number molecules of O2 in the vessels are:
(a) 3.01 x 1023
(b) 6.02 x 1023
(c) 12.04 x 1023
(d) 24.08 x 1023
Explanation;
(Answer, c)
22.4 dm3 (1
mole) of any gas at STP contains 6.02 x 1023 molecules. Hence
44.8 dm3 (2 mole) of any gas at STP would contain twice of 6.02
x 1023 molecules i.e. 12.04 x 1023 molecules.
No. of molecules = Vg/Vm x NA = 44.8/22.4 x 6.02 x 1023 = 12.04 x 1023 or 1.204 x 1024 molecules.
2. The
number of carbon atoms in 1 mole of sugar (C12H22O11)
are approximately:
(a) 6 x 1023
(b) 24 x 1023
(c) 60 x 1023
(d) 72 x 1023
Explanation;
(Answer, d)
1 mole of sugar (C12H22O11)
= 12 moles of carbon atoms
12 moles of carbon atoms = n x NA = 12 x 6.02 x 1023 = 72 x 1023 or 7.2 x 1024 carbon atoms.
3. In
the reaction 2Na + 2H2O → 2NaOH + H2, if 23g of Na reacts
with excess of water, the volume of hydrogen gas (H2) liberated at
STP should be
(a) 11.2dm3
(b) 22.4dm3
(c) 33.6dm3
(d) 44.8dm3
Explanation;
(Answer, a)
From stoichiometric ratio
in the given equation:
46g (2 mole) Na liberates
22.4 dm3 of hydrogen gas at STP
23g (1 mole) Na liberates 22.4/46 x 23 dm3 of hydrogen gas at STP = 11.2 dm3 of hydrogen gas
4.
Which of the following sample of substances contains the same number of atoms
as that of 20g calcium:
(a) 16g S
(b) 20g C
(c) 19 g K
(d) 24g Mg
Explanation;
(Answer, a)
Species containing same
number of moles will have same number of particles (atoms, molecules or formula
units). 20 g Ca and 16 g S constitute same number of moles (0.5 moles), hence
they have same number of atoms.
No. of moles in 20 g Ca =
mass/molar mass = 20/40 = 0.5 mol
No. of moles in 16 g S = mass/molar mass = 16/32 = 0.5 mol
5. The
minimum number of moles are present in:
(a) 1 dm3 methane gas at STP
(b) 5 dm3 of helium gas at STP
(c) 10 dm3 of hydrogen gas at STP
(d) 22.4 dm3 of chlorine gas at STP
Explanation;
(Answer, a)
For gases at STP, minimum number of moles are present in that gas which has least volume. Here 1 dm3 methane gas at STP has least number of moles as it has least volume.
6. Number
of atoms in 60g carbon are:
(a) 3.01x 1023
(b) 3.01 x 1024
(c) 6.02 x 1023
(d) 6.02 x 1024
Explanation;
(Answer, b)
No. of atoms = (Mass/molar mass) x NA = 60/12 x 6.02 x 1023 = 3.01 x 1024
7. Maximum
number of molecules present in the following sample of gas:
(a) 100g O2
(b) 100g CH4
(c) 100g CO2
(d) 100g Cl2
Explanation;
(Answer, b)
The species containing
greatest moles will have greatest number of molecules.
For species having same mass, the specie having least molar mass would have greatest number of moles. Here CH4 has least molar mass of 16, so its 100 g would contain maximum number of moles and also maximum number of molecules.
8. Which of the following statement is incorrect?
(a) The mass of 1 mole Cl2 gas is 35.
(b) One mole of H2 gas contains 6.02 x 1023 molecules of H2
(c) Number of atoms in
23g Na and 24 g Mg are
equal
(d) One moles of O2 at S.T.P occupy 22.4 dm3 volume
Explanation;
(Answer, a)
The statement a is incorrect as the mass of 1 mole of Cl2 is 71 g (not 35).
9. For
Avogadro's number, this statement is incorrect?
(a) It is the no. of
particles in one mole of any substance
(b) Its numerical value
is 6.02x 1023
(c) Its value change if
temperature increases
(d) Its value change if
number of moles increases.
Explanation;
(Answer, c)
Avogadro’s number is independent of temperature and pressure. Its value only depends upon the number of moles and volume of gas at STP.
10. Generally
actual yield is:
(a) Greater than
theoretical yield
(b) Less than theoretical
yield
(c) Equal to the
theoretical yield
(d) Some times greater
& some times less than theoretical yield
Explanation;
(Answer, c)
Generally actual yield is Less than theoretical yield mainly due to side reactions and mechanical loss of product.
🔎💥 XI MCQs Chapter 2 …… Atomic Structure
1. Bohr's
theory is not applicable to which of the following
(a) H
(b) H+
(c) He1+
(d) Li2+
Explanation;
(Answer, b)
One of the limitations of
Bohr's atomic model is that it does not explain the spectra of multi-electron
atoms. Bohr's theory is applicable to hydrogen like atoms or hydrogenic ions
(single electron system).
All the given species like H, He+ and Li2+ are isoelectronic and have only one electron. Their electronic configurations are same and so their spectra is explained by Bohr's atomic model. But H+ has no electron at all and hence cannot form spectrum.
2. Nitrogen
has the electronic configuration 1s2 2s2 2px1 2py1 2pz1 and
1s2 2s2 2px2 2py1.This
is determined by
(a) Aufbau
principle
(b) Pauli’s
rule
(c) Hund's rule
(d) n+l rule
Explanation;
(Answer, c)
The filling of electrons in degenerate orbitals like p, d and f is governed by Hund’s rule of maximum electron multiplicity accordingly electrons in degenerate orbitals tend to remain singly with same spin until all of the degenerate orbitals become half filled then pairing of electrons are allowed in them.
3. Quantum
number values for 3s orbital are
(a) n=0, l=1
(b) n=1,
l=0
(c) n=3,
l=1
(d) n=3, l=0
Explanation;
(Answer, d)
In the notation 3s (nl),
3 stands for n while s represent l.
For 3s orbital, n =3 and l=0.
4. The
radius of first orbit of hydrogen atom is .....
(a) 529Å
(b) 52.9 Å
(c) 5.29 Å
(d) 0.529 Å
Explanation;
(Answer, d)
Where ao is known as Bohr’s constant or Bohr radius and it the radius of first orbital of hydrogen atom. and its value is 0.529 x 10−10 m or 0.529 x 10−8 cm or 0.529Å or 0.0529 nm or 52.9 pm. This is the radius of the first orbit of H. This equation is used for the determination of nth orbit of hydrogen atom and hydrogen like ions like He+, Li2+ etc.
5. Line
spectrum is used as a tool for the identification of ....
(a) Colors
(c) Elements
(b) Electrons
(d) Molecules
Explanation;
(Answer, c)
Samples of same element always produces same characteristic line spectrum. Each element emit of light of specific wavelength therefore the number of lines and the distance between them depends upon the nature of element, so line spectra is used as “Finger Print” for the identification of elements. For example, line spectrum of sodium contains two yellow coloured lines separated by a definite distance.
6. The
shape of orbital for which l = 0 is
(a) Spherical
(b) Dumbbell
(c) Double dumbbell
(d) Complicated
Explanation;
(Answer, a)
Its values show shape of
orbitals. It has values l = 0 to (n – 1). e.g.
l = 0, used for s-orbital, spherical in
shape
l = 1, used for p-orbital, dumb-bell in
shape
l = 2, used for d-orbital, double dumb-bell in shape
l = 3, used for f-orbital, complicated shape
7. When
4d orbital is filled, the next electron enter into
(a) 5s
(b) 5p
(c) 5d
(d) 6s
Explanation;
(Answer, b)
According to n+ l rule,
electrons are filled in various orbitals in the increasing order of (n+l)
value. Those orbitals which have lower value of (n+l) value filled first. In
case, if two orbitals which have same (n+l) value, then orbital having lower
‘n’ value will be filled first.
The n+l value of 4d orbital is 4+2 =6. The next orbital to be filled must have n+l value either equal to 6 or greater than 6. The n+l value of 5d orbital is 5+2 =7, for 5s it is 5+0=5, for 5p it is 5+1=6 and for 6s it is 6+0=6. Since 5p and 6s orbitals have same n + l value, hence the electron first go to that orbital which has least n value i.e. 5p.
8.
Which of the following is not an iso electronic pair?
(a) Na+ and
Ne
(b) Na+ and
F−
(c) Na and Ca
(d) Na+ and
Mg2+
Explanation;
(Answer, b)
The species having same number of total electrons and hence have same electronic configuration are called isoelectronic pair. Na+ and F− are isoelectronic pair as both of them contains 10 electrons.
9. Balmer
series appears in the hydrogen spectrum if electron jumps from any appropriate
higher energy orbit to
(a) Second orbit
(b) Third orbit
(c) Fourth orbit
(d) Fifth orbit
Explanation;
(Answer, a)
Balmer noted first that the Hydrogen spectrum consisted of some well defined discrete lines in the visible region (i.e. having l between 4000 to 7000Å) of spectrum. This series is obtained by the transition of electrons from any higher orbit (n2 = 3, 4, 5, 6, 7 …. µ) to 2nd orbit (n1 = 2).
10. In
1935 A.D. James Chadwick was awarded Nobel Prize because ...
(a) He discovered proton
(b) He discovered neutron
(c) He determined the
radius of hydrogen atom
(d) He gave the rules for
electronic configuration
Explanation;
(Answer, b)
James Chadwick won the 1935 Nobel Prize in Physics for the discovery of the neutron.
🔎💥XI MCQs Chapter 3 … Chemical Bonding
1.
If the bond angle is AB2 type molecule is 104.5o, it geometry
should be;
(a)
Linear
(b)
Pyramidal
(c)
Bent
(d) Planar
Trigonal
Explanation;
(Answer, c)
The bond angle of 104.5o is
the characteristic of AB2 type of molecules with bent or
V-shaped or angular geometry.
2. The
highest bond energy in the following is:
(a)
Cl–Cl
(b)
H–F
(c)
H–O
(d) H–N
Explanation;
(Answer, b)
Bond energy is directly
proportional to the bond polarity or partial ionic character which in turn is
proportional to the difference in electronegativity of the bonded atoms.
Greater the ΔEN, greater is the bond energy. H – F bond has the greatest ΔEN
(1.9) and hence it has the highest bond energy (in fact highest among all
diatomic molecules containing single bond; 570 kJmol−1).
The second highest bond
energy is for H – O bond with second highest ΔEN (1.4). The next highest bond
energy is for H – N bond with third highest ΔEN (0.9). Cl–Cl being non-polar
has the least bond energy.
3.
The molecule which has zero dipole moment is:
(a) NH3
(b)
HCl
(c) H2O
(d) CCl4
Explanation;
(Answer, d)
Symmetrical molecules
with linear, trigonal or tetrahedral geometry show zero dipole moment. CCl4 has
symmetrical tetrahedral structure in which all four bond moments for C – Cl
bonds being directed in opposite directions are cancel out giving net zero
dipole moment.
4. The
molecule which has maximum bond angle
(a) CS2
(b) H2O
(c) NH3
(d) BF3
Explanation;
(Answer, a)
Linear molecules like CS2, CO2, BeCl2, C2H2, HgCl2, ZnCl2 etc. with AB2 formula has the largest bond angle of 180o.
5. The
shape and hybridization of BCl3 molecule is:
(a) Tetrahedral and sp3
(b) Linear and
sp
(c) Planar trigonal and
sp2
(d) Angular and sp3
Explanation;
(Answer, c)
Being AB3 molecule
with no non-bonding orbital, BCl3 has sp2 hybridization
with planar trigonal shape.
6.
Amongst the following molecules which one has trigonal pyramidal shape?
(a) SO2
(b) CO2
(c) NH3
(d) C2H4
Explanation;
(Answer, c)
The ÄB3 molecules like NH3, PH3, PCl3, NCl3 etc. with one lone pair and 3 bond pairs has trigonal pyramidal shape
7. A
simple covalent molecule possesses two bond pairs and two lone pairs around the
central atom, its shape should be:
(a)
Linear
(b) Planar
trigonal
(c)
Angular
(d) Tetrahedral
Explanation;
(Answer, c)
The presence of two bond
pairs and two lone pairs around the central atom gives rise to angular shape to
the molecules. Such molecules have :ÄB2 type general formula.
8. The
correct relation between debye and coulomb meter is:
(a) 1D = 3.33 x 10−30Cm
(b) 1D = 1.6 x 10−19Cm
(c) 1D = 1.88 x 10−12Cm
(d) 1D = 1.23 x 10−8Cm
Explanation;
(Answer, a)
The SI unit of dipole
moment C-M and cgs unit debye is inter related as
1D = 3.33x10−30Cm
9. The
bond order of N2 molecule is:
(a)
0
(b)
1
(c)
2
(d) 3
Explanation;
(Answer, d)
The bond order of N2 molecule
is 3 showing that it has triple bond between two nitrogen atoms.
10. The number of sigma and pi bonds in C2H4 molecules are respectively:
(a) 3 and
1
(b) 2 and
2
(c) 5 and
1
(d) 4 and 2
Explanation;
(Answer, c)
There are total 5 sigma and one pi bond in in C2H4 molecule.
🔎💥XI MCQs Chapter 4 State of Matter, Gas
1.
According to Graham’s Law of diffusion, the ratio of diffusion of H2 and
O2 are respectively:
(a)
1:2
(b)
2:1
(c)
1:4
(d) 4:1
Explanation;
(Answer, d)
r1/r2=
√M2/√M2 ⇒ r1/r2= √32/√2 ⇒ r1/r2=
√16 ⇒ r1/r2= 4:1
2. Collection
of gas over water is an example of:
(a) Graham’s
law
(b) Dalton’s law
(c) Avogadro’s
law
(d) Gay-Lussac
law
Explanation;
(Answer, b)
Collection of gas over
water is an example of Dalton’s law. The pressure of dry gas is calculated by
using Dalton’s law.
Pdry gas =
Pmoist gas – Pwater vapours
3. The
molar volume of oxygen gas is maximum at:
(a) 0°C and 1
atm
(b) 0°C and 2
atm
(c ) 25°C and 1
atm
(d) 25°C and 2
atm
Explanation;
(Answer, c)
The molar volume of a gas
varies directly with temperature and varies inversely with pressure. The molar
volume of a gas is maximum at highest temperature and least pressure. In option
c and d, temperature is highest hence deciding factor will be pressure which is
least in option c.
4. The
volume of gas would be theoretically zero at:
(a)
0°C
(b) 0
K
(c) 273
K
(d) 273°C
Explanation;
(Answer, b)
The volume of gas would
be theoretically zero at 0 K (-273°C).
5.
If the Kelvin temperature of ideal gas is increase to double and pressure is
reduced to one half, the volume of gas will:
(a) Remains
same
(b) Double
(c) Reduced to
half
(d) Four times
Explanation;
(Answer, d)
PV=nRT ⇒ PV = T (R and n
= constant) ⇒ V = T/P ⇒ V = 2/ ½ = 2 x 2 = 4
Thus, doubling the Kelvin
temperature and halving the pressure quadruples the volume.
6. The
molar volume of oxygen gas is 22.4 dm3 at:
(a) 0°C and 1
atm
(b) 25°C and 0.5
atm
(c) 0 K and 1
atm
(d) 25 K and 0.5
atm
Explanation;
(Answer, a)
The molar volume of a gas
is 22.4 dm3 at STP i.e. at 0C and 1 atm.
7. Under
similar condition CH4 gas diffuses........ times faster than SO2 gas:
(a) 1.5
times
(b) 2
times
(c) 4
times
(d) 16 times
Explanation;
(Answer, b)
r1/r2=
√M2/√M2 ⇒ r1/r2= √64/√16 ⇒ r1/r2=
√4 ⇒ r1/r2= 2
8. Which
one of the following statement is incorrect about the gas molecules?
(a) They have large
spaces
(b) They possess kinetic
energy
(c) Their collision is
elastic
(d) Their molar mass
depends upon temperature
Explanation;
(Answer, d)
The molar mass of gas is
independent of temperature.
9. The
diffusion rate of C3H8 and CO2 are
same because:
(a) Both are poly atomic gases
(b) Both are denser than
air
(c) Both have same molar mass
(d) Both contains carbon
atoms
Explanation;
(Answer, c)
The rate of diffusion
depends upon molar masses of gases. The gases with same molar masses would have
same rate of diffusion. The given gases ethane (C3H8) and
CO2 have same molar mass of 44 gmol−1.
10. Real
gas reaches the ideal behavior at:
(a) Low temperature and
low pressure
(b) High temperature and
high pressure
(c) Low temperature and
high pressure
(d) High temperature and
low pressure
Explanation;
(Answer, d)
High temperature and low pressure make the ideal.
🔎💥XI MCQs Chapter 5 - Liquid State
1.
Hydrogen bond is not found in:
(a) H2O
(b) CH4
(c) NH3
(d)
HF
1.
Explanation (Answer; b)
Hydrogen bonding is found
in those polar compounds which have polarized hydrogen. Since CH4 is
non-polar, it cannot form hydrogen bond.
2. The
boiling points of different liquids may be different at the same external
pressure due to;
(a) Amounts of
liquids
(b) Intermolecular
forces
(c) Surface
area
(d) Viscosities
2.
Explanation (Answer; b)
The boiling points of
liquids depend upon the strength of intermolecular forces. Greater the strength
of IMF, greater is the boiling point. Since different liquids have different
strength of IMF, therefore, different liquids have different boiling points at
the same external pressure.
3. Which
of the following possesses weakest London dispersion forces:
(a) Cl2
(b) F2
(c) Br2
(d) I2
3.
Explanation (Answer; b)
The strength of London
dispersion forces depends upon the size of atom or molecules. Large size atom
or molecules have greater number of electrons therefore more distortion of
electrons is possible, which increases the strength of London forces. In this questions
given options belongs to halogens of group VIIA in which atomic size increases
down the group. Hence iodine has the largest size and has the strongest
dispersion forces while fluorine has the smallest size and hence has the
weakest dispersion forces.
4. A
non-polar molecule with bigger size will experience:
(a) London
forces
(b) dipole-dipole
interaction
(c) Hydrogen
bonding
(d) All of these
4.
Explanation (Answer; a)
Non-polar molecules have
only London dispersion forces. The strength of London dispersion forces depends
upon the Size of atom or molecules. Large size atom or molecules have greater
number of electrons therefore more distortion of electrons is possible, which
increases the strength of London forces.
5. Which
of the following liquids show maximum surface tension?
(a)
Water
(b)
Mercury
(c) Ethyl
alcohol
(d) Gasoline
5.
Explanation (Answer; b)
Surface tension of
liquids depends upon the strength of IMF. Greater is the strength of IMF,
greater is the surface tension. Mercury has strong interatomic forces in the
form strong metallic bond. Hence mercury has the highest surface tension among
all liquids.
6.The
boiling point of water (H2O) is 100oC where as that of
hydrogen sulphide (H2S) is 42oC, This can be
attributed to:
(a) Smaller bond angle of
H2S than H2O
(b) Smaller radii of
oxygen than sulphur
(c) High I.P of oxygen than sulphur
(d) Tendency of water to
form hydrogen bond
6.
Explanation (Answer; d)
The boiling points of
liquids depend upon the strength of intermolecular forces. Greater the strength
of IMF, greater is the boiling point. Water being a polar liquid forms strong
hydrogen bonds with neighbouring water molecules while hydrogen sulphide being
non-polar cannot do so. Hence due to hydrogen bonding water shows high boiling
point than hydrogen sulphide.
7.
Liquids can form convex meniscus in a narrow glass tube when:
(a) Cohesive forces are
stronger than adhesive
forces
(b) Adhesive forces are
stronger than cohesive forces
(c) Cohesive and adhesive
forces are equal in
strength
(d) None of these
7.
Explanation (Answer; a)
The formation of meniscus
by liquids in a narrow tube depends the strength of cohesive and adhesive
forces in that liquids. Liquids with strong cohesive forces than adhesive
forces form convex meniscus while liquids with strong adhesive forces than
cohesive forces forms convex meniscus.
8. Which
of the following statement is incorrect?
(a) Viscosity is
the resistance against flow of liquid
(b) Liquid possesses
definite volume
(c) One feels sense
of cooling after bath due to condensation
(d) Sublimation is
endothermic process
8.
Explanation (Answer; d)
Cooling sensation after taking bath is due to evaporation not due to condensation. Evaporation is a cooling process.
9. Which
statement is incorrect about evaporation?
(a) It is an exothermic process
(b) It is a reverse
process of condensation
(c) It occurs at all temperatures and pressure
(d) It causes cooling
effect
9.
Explanation (Answer; a)
Evaporation is an endothermic process.
10.
Cooking time is reduced in a pressure cooker because:
(a) Boiling point of water increases
(b) Boiling point of
water decreases
(c) Vapor pressure of liquid is reduced
(d) Heat is uniformly
distributed
10.
Explanation (Answer; a)
In pressure cooker, boiling point of water increases which reduces cooking time.
🔎💥XI MCQs Chapter 6 - Solid State
🔎💥XI MCQs Chapter 7 -Chemical Equilibrium
1.
In the equilibrium system of PCl5(g) ⇌ PCl3(g)+ Cl2(g),
the correct relationship between Kc and Kp is:
(a) Kp> Kc
(b) Kp< Kc
(c) Kp =
Kc
(d) Kp/Kc=1
Explanation; (Answer, a)
2. In
the reaction A(g)+ B(g) ⇌2C(g), the
equilibrium constants Kp=Kc because:
(a) ∆n >
1
(b) ∆n <
1
(c) ∆n =
1
(d) ∆n = 0
Explanation;
(Answer, d)
3. What
happens to the value of Kc when a catalyst is added to a
chemical system at equilibrium?
(a) It
decreases
(b) it increases
(c) It becomes zero
(d) it remains unchanged
Explanation;
(Answer, d)
4. The
correct Ksp expression of a sparingly soluble salt Li2C2O4 among
the following is
(a) Ksp=[Li+][C2O42−]
(b) Ksp=[Li+]2[C2O42−]
(c) Ksp=[2Li+][
C2O42−]
(d) Ksp=[2Li+]2[C2O42−]2
Explanation;
(Answer, b)
5. If
the equilibrium expression of a reversible reaction is
Kc = [C]2/[A][B]
The
balanced equilibrium equation should be:
(a) C ⇌
A+B
(b) A+B ⇌
C
(c) 2C ⇌ A+
B
(d) A+B ⇌ 2C
Explanation;
(Answer, d)
6. The
term active mass use in law of mass action means:
(a) No. of
mole
(b) No. of molecules
(c) mole per dm³
(d) gram per dm³
Explanation;
(Answer, c)
7. The
equilibrium of N2(g) +O2(g) ⇌ 2NO(g) (∆H
= +ve) is affected by change in:
(a) Temperature
only
(b) Pressure only
(c) Both temperature
& pressure
(d) Neither temperature
nor pressure
Explanation;
(Answer, a)
8. The
equilibrium of which of the following reaction would not be affected by an
increase in pressure:
(a) PCl5(g) ⇌
PCl3(g)+ Cl2(g)
(b) 2NO(g) +
Cl2(g) ⇌ 2NOCl(g)
(c) N2(g) +
O2(g) ⇌ 2NO(g)
(d) 2SO2(g) +
O2(g) ⇌ 2SO3(g)
Explanation;
(Answer, b)
9. NaCl
when added to an aqueous silver chloride solution:
(a) Decreases the
solubility of AgCI
(b) Increases the
solubility of
AgCl
(c) Forms a clear
solution
(d) Does not effect
Explanation;
(Answer, a)
10. Some
reactions are nearly to completion in the forward direction and identified by
their:
(a) Very high value of Kc
(b) Very low value of Kc
(c) Very high value of
∆H
(d) Very low value of ∆H
Explanation; (Answer, a)
🔎💥XI MCQs Chapter 8 - Acids, bases and Salts
1. H2SO4 is
stronger acid than CH3COOH because:
(a) It gives two H+ ion
per
molecule
(b) Its boiling point is
high
(c) Its degree of
ionization is
high
(d) It is highly
corrosive
Explanation
(Answer; c)
The strength of an acid
depends upon its degree of ionization and acid ionization constant (Ka).
Greater the value of degree of ionization (equal or greater than 30%) and Ka
(more than 10−3), stronger is the acid. The degree of ionization of
H2SO4 is quite high about 60% (first step) and its
Ka is also quite high about 103 (Ka2 for HSO4− =
1.3 x 10−2) showing that it is a very strong acid. Contrarily, The
degree of ionization of CH3COOH is quite low about 1.4% and its Ka
is also quite low about 1.7 x 10−5 showing that it is a weaker
acid than H2SO4.
2. Which
of the following statements is not correct about the bases?
(a) They have bitter tastes
(b) They have high pH value
(c) They react with acids to form salts
(d) They turn blue litmus red
Explanation
(Answer; d)
Bases turns red litmus
paper blue.
3. Al2O3 is
amphoteric oxide because it reacts with:
(a)
Acids
(b) neither acid nor
base
(c) Both acids and
base
(d)
Base
Explanation
(Answer; c)
amphoteric oxides have
dual characteristics of acid and base and hence they react both with acid and
base through neutralization forming salt and water.
The oxides of amphoteric
metals like Be, Al, Sn, Pb, Zn, Cr etc are amphoteric . These oxides react with
both acids and base via neutralization to form salt and water. Al2O3,
ZnO, BeO, SnO, SnO2, PbO, PbO2, Cr2O3 are
amphoteric oxides.
4. Which
of the following is not a buffer solution?
(i) Na2CO3 /NaHCO3
(b) CH3COOH/CH3COONa
(c) NH4OH/NH4Cl
(d) NaOH/HCl
Explanation
(Answer; d)
Buffer solution is a
mixture of weak acid and its salt with strong base or weak base and its salt
with strong base. In option ‘d’, NaOH and HCl are both base and acid.
5. Which
oxide is amphoteric in nature:
(a) K2O
(b) CO2
(c) CaO
(d) Al2O3
Explanation
(Answer; d)
The oxides of amphoteric
metals like Be, Al, Sn, Pb, Zn, Cr etc are amphoteric . These oxides react with
both acids and base via neutralization to form salt and water. Al2O3,
ZnO, BeO, SnO, SnO2, PbO, PbO2, Cr2O3 are
amphoteric oxides.
The oxides of metals like
K2O, Li2O, Na2O, MgO, CaO, BaO etc. are basic.
The oxides of non-metals
like CO2, SO2, SO3, NO2, N2O5,
P2O3, P2O5, Cl2O, Cl2O7 etc.
are acidic. (Some non-metallic oxides like H2O, N2O, NO
and CO are neutral).
6. Which
of the following does not alter the pH of a solution?
(a) NH4Cl
(b) Na2CO3
(c)
NaCl
(d) Mg(OH)Cl
Explanation
(Answer; c)
Neutral salts formed from
strong acids and strong bases like NaCl, KBr, KCl, BaCl2, Na2SO4,
KNO3 etc. do not change the pH of solution.
7. Conjugated
acids of NH3 is:
(a) NH4+
(b) NH2-
(c) NH2
(d) NH
Explanation
(Answer; a)
To form conjugate acid
from a compound or ion, just add H into it while to form conjugate base of a
compound or ion just remove H from it. Don’t forget to balance charge either
adding or subtracting H. Addition of H increases one unit + charge while removing
of H decreases one + charge.
NH3 (adding
add H into it will introduce + charge) …………… NH4+ (conjugate
acid)
NH3 (removing
H from it will introduce – charge) ……………. NH2− (conjugate
base)
8. Salt
which is formed by the neutralization of weak acid and strong base is:
(a) NaNO3
(b) NH4Cl
(c) Na2CO3
(d) NH4CN
Explanation
(Answer; c)
Strong bases contain Na+, K+, Ba2+ as cation while weak acids contains anions other than Cl−, Br−, I−, NO3−, SO4−, ClO4−, and ClO3−. A salt formed from weak acid and strong base via neutralization contains any one of the cation like Na+, K+, Ba2+ and anion other than Cl−, Br−, I−, NO3−, SO4−, ClO4−, and ClO3−. Sodium carbonate (Na2CO3) is a salt derived from the neutralization of weak acid (H2CO3 containing CO32− anion) and strong base (NaOH).
9. A
conjugate acid base pair has the difference of only:
(a) One
electron
(b) One proton
(c) One electron
pair
(d) One proton pair
Explanation
(Answer; b)
conjugate acid-base pair
differs from one another only by a single proton (H+). To identify a
conjugate acid base pair, we look at the specie that donates a proton and the
specie that is formed from it by the loss of proton. In the given reaction, NH3 accepts
a proton acting as a Bronsted base to form NH4+ which
is its conjugate acid. Similarly, HCl donates a proton acting as a Bronsted
acid to form Cl− which is its conjugate base.
10. Salt
formed by neutralization of weak acid and weak base is:
(a) NH4Cl
(b) Na2CO3
(c) NH4CN
(d) Na2SO4
Explanation
(Answer; c)
Strong bases contain Na+, K+, Ba2+ as cation while weak acids contains anions other than Cl−, Br−, I−, NO3−, SO4−, ClO4−, and ClO3−. A salt formed from weak acid and weak base via neutralization contains any cation other Na+, K+, Ba2+ and any anion other than Cl−, Br−, I−, NO3−, SO4−, ClO4−, and ClO3−. Ammonium cyanide (NH4CN) is a salt derived from the neutralization of weak acid (HCN containing CN− anion) and weak base (NH4OH).
🔎💥XI MCQs Chapter 9 - Chemical Kinetics
1. For a
hypothetical reaction x+y → z, if the conc. of x is double, the rate increases
by square and if the conc. of y is doubled the rate increases by twice. The
experimental rate law of this reaction is:
(a) R=K [x]1 [y]1
(b) R= K
[x][y]
(c) R= K [x]2 [y]1
(d) R=K [x][y]
Explanation
(Answer; c)
For second order
reaction, doubling the concentration of reactant (x) quadruples the rate (i.e.
increases by square).
For first order reaction,
doubling the concentration of reactant (y) doubles the rate (i.e. increases by
twice).
Hence for given reaction,
order of reaction with respect to x is 2 and that of y is 1. The experimental
rate law of this reaction is R= K [x]2 [y]1
2.The
unit of rate constant (K) for the first order reaction is:
(a) s−1
(b) conc. s−1
(c) conc−1.
s
(d) conc−1. s−1
Explanation
(Answer; a)
3. Rate
constant of a reaction is affected by:
(a)Conc. of
reactants
(b) Conc. of
products
(c)
Temperature
(d) Reaction time
Explanation
(Answer; c)
Rate constant of a
reaction is affected by temperature only.
4. The
purpose of using of catalyst in a chemical reaction is to change:
(a) Equilibrium
constant
(b) Enthalpy of reaction
(c) Activation
energy
(d) Nature of reaction
Explanation
(Answer; c)
The main function of
catalyst is to change the activation energy.
5. The
overall order of reaction to which the rate law is R= K:
(a) Zero
order
(b) First
order
(c) Second order
(d) Third order
Explanation
(Answer; a)
For zero order reaction,
concentration of reactants are taken as 1, so for such reactions the rate law
becomes R = K
Rate of reaction = K [A]x[B]y ⇒
Rate of reaction = K [A]0[B]0 ⇒ Rate of reaction = K
x 1 x 1⇒ Rate=K
6. Ionic
reactions are classified into
(a) Reversible
reaction
(b) Moderate reactions
(c) Fast reactions
(d) Slow reaction
Explanation
(Answer; c)
Ionic reactions are
classified into fast reactions as they have low activation energy due to
involvement of ions.
7. The
decomposition of H2O2 is inhibited by:
(a) 2%
ethanol
(b)
Glycerin
(c)
MnO₂
(d) V₂Os
Explanation
(Answer; b)
Prevention of
decomposition of H2O2 is done by using glycerin as a
negative catalyst.
8. The
best alternative term for the velocity of reaction is:
(a) Rate of
appearance
(b) Rate of disappearance
(c) Average rate
(d) Instantaneous rate
Explanation
(Answer; d)
The velocity of reaction
is called Instantaneous rate when time interval is approached to zero.
9. Order
of reaction is the power to which concentration of reactant is:
(a)
Lowered
(b)
Raised
(c) Stopped In the
reaction
(d) Constant
Explanation
(Answer; b)
Order of reaction is the
power to which concentration of reactant is raised.
10. In
the reaction A→B, the rate of disappearance is written as:
(a) dA/dt
(b) -dA/dt
(c) dB/dt
(d) -dB/dt
Explanation
(Answer; b)
The rate of disappearance of reactant is written as -dA/dt
🔎💥XI MCQs Chapter 10 - Solution and Colloids
1.Only
one pair of liquid in the following set does not obey Raoult’s law, identify
it:
(a) Methanol and
Ethanol
(b) Benzene and toluene
(c) n-Hexane and
n-heptane
(d) Ethanol and
Acetone
Explanation
(Answer; d)
The solutions which obey
Raoult’s law over the entire range of concentration are known as ideal
solutions. Some examples of ideal solutions are methanol and ethanol, benzene,
and toluene, n-hexane and n-heptane, bromoethane and chloroethane etc.
Methanol and ethanol are
polar in nature with similar intermolecular forces, hence they form an ideal
solution.
Both benzene and toluene
and n-hexane and n-heptane are non-polar in nature with similar intermolecular
forces. Hence, they form an ideal solution.
Ethanol and Acetone
When Acetone and ethanol are mixed, the weakening of hydrogen bonds in ethanol takes place. Hence, an ideal solution is not formed.
2.
Identify the incorrect statement about colloidal solution:
(a) It shows Tyndall
effect
(b) Its particles
movement is Brownian type
(c) Its particle size is less than 1 nm
(d) Its physical
appearance is translucent.
Explanation
(Answer; c)
In colloidal solution,
the particles is in between 1 nm to 1000 nm.
4. Which
is not a colligative property?
(a) Lowering in vapours pressure
(b) Elevation in boiling
point
(c) Depression in freezing point
(d) Atmospheric pressure
Explanation
(Answer; d)
Atmospheric pressure is a colligative property.
3. Effect
of pressure change play significant role in the solubility of:
(a) Solid into
liquid
(b) Liquid into
liquid
(c) Gas into
liquid
(d) All of
them
Explanation
(Answer; c)
The solubility of gas into liquid is significantly changed by the change in pressure obeying Henry’s law accordingly the solubility of gas in liquid solvent is directly proportional to the pressure.
5.
According to Raoult law the relative lowering of vapour pressure is equal to:
(a)
Molality
(b) Mole fraction of
solute
(c) Mole fraction of
solvent
(d) Molarity
Explanation
(Answer; b)
According to the third
form of Raoult’s law, the relative lowering of vapour pressure is equal to the
mole fraction of solute.
∆P/Po = X2---------------- [Here ∆P/Po is referred as relative lowering in vapours pressure]
6.
The sum of mole fractions of components of a solution is equal to:
(a)
0.0
(b)
1.0
(c)
10
(d) 100
Explanation
(Answer; b)
The sum of mole fractions of components of a solution is equal to 1.0.
7. How
many mole of NaOH are present in 2dm3 of 1 molar aqueous
solution of it.
(a) 0.5
mole
(b) 1
mole
(c) 1.5
mole
(d) 2 mole
Explanation
(Answer; d)
1 molar aqueous solution contains 1 mole of solute per 1 dm3. Hence 1 molar aqueous solution contains 2 mole per 2 dm3.
8. A
colloidal solution of liquid into liquid is known as:
(a)
Gel
(b)
Foam
(c)
Sol
(d) Emulsion
Explanation
(Answer; d)
A colloidal solution of liquid into liquid is known as Emulsion
9. An
example of completely immiscible liquid pair is:
(a) Benzene to
toluene
(b) Water and
phenol
(c) Water and
Benzene
(d) Water & methanol
Explanation
(Answer; b)
Water and phenol is an example of completely immiscible liquid pair
10. A 15%
W/W KOH solution can be prepared by mixing 15g KOH in:
(a) 15g
water
(b) 85g
water
(c) 100g
water
(d) 115g
water
Explanation
(Answer; b)
The numerical value of percent concentration indicates amount of solute. The amount of solvent can be calculated by subtracting amount of solute from 100 (only in case of W/W % or V/V%). Hence 15% W/W KOH solution can be prepared by mixing 15g KOH in 85 g (100 – 15) water.
🔎💥XI MCQs Chapter 11 - Thermochemistry
1. Least
entropy found in which of the following state of water:
(a) Steam at
100°C
(b) Liquid water at
25°C
(c) Liquid water at
4°C
(d) Ice at 0°C
Explanation
(Answer; d)
Entropy, S, is a
thermodynamic quantity and it is defined as the measure of the disorder or
randomness in a system. The more “messed up” or “disorderly” or “messy” a
system becomes, we say the entropy of the system increases.
The entropy of substances
with different states are different since the arrangement of the molecules are
different. Solids have the lowest entropy value among the other states of
matter since the molecules in solids are stationary. These molecules only
vibrate but do not move due to the strong interactions. Their molecules are
therefore more ordered than those in other states of matter. Thus among three
states of matter, the randomness and also entropy follows the order:
Gas > liquid >
solid
Sgas >
Sliquid > Ssolid
2. The
energy corresponds to the given thermochemical process is labeled as:
Li+(g) +
Cl−(g) →LiCl(s)
(a) Ionization energy
(b) Enthalpy of
atomization
(c) Enthalpy of
combustion
(d) Lattice energy
Explanation
(Answer; d)
The given equation for
thermochemical process represents the lattice formation of ionic solid. The
energy released during the formation of 1 mole solid lattice of ionic compound
from its consistent gaseous ions under standard conditions is called lattice
energy. It is negative.
3. Which
of the following change is not an endothermic reaction
(a) Cracking of alkanes
(b) Decomposition of
lime
(c) Combustion of
butane
(d) Photosynthesis
Explanation
(Answer; c)
Combustion is always
exothermic releasing heat called heat of combustion.
4. Heat
transfer cannot be feasible across the boundary of a
(a) Open system
(b) Thermo permeable
system
(c) Isolated system
(d) Close system
Explanation
(Answer; c)
Isolated system is an
ideal thermodynamic system that does not allow either matter or heat transfer
across the boundary.
5. In a
thermochemical process, no work is done if the system is kept at:
(a) Constant pressure
(b) Constant
temperature
(c) Constant
volume
(d) Constant mass
Explanation
(Answer; c)
When a reaction is
carried out at constant volume (ΔV = 0), no work is done on or by the system.
6.
Standard enthalpy of formation of all of the following elements at 25°C and 1
atm pressure are zero except:
(a) C(diamond))
(b) C(graphite)
(c)
O₂
(d) N₂
Explanation
(Answer; a)
Conventionally ΔHf°
of most the elements (H2, O2, N2, Na,
graphite, rhombic sulphur etc.) are taken as zero because no further change is
needed to bring them to standard state conditions. However, for those elements
which exists in more than one allotropic forms, the most stable allotropic form
is conventionally assumed to have ΔHf° = 0.
For
example
(i) ΔHf° of
graphite is taken as zero as it is thermodynamically more stable allotrope of
carbon than diamond at 25°C & 1 atm
(ii) O2 is
more stable allotropic form than O3 (ozone) and therefore, ΔHf°
of O2 is assumed to be zero.
(iii) Rhombic sulphur has
ΔHf° = 0 as it is the most stable allotrope of sulphur.
7. In the
equation of First law of thermodynamics (∆E=q + w), the property(s) which
depends upon initial and final state is (are):
(a)
ΔE
(b)
q
(c)
W
(d) Both q and W
Explanation
(Answer; a)
Only ΔE is a state
function which depends upon initial and final state. The other two variables of
the first law of thermodynamics i.e. q and W both are path functions that
do not depend upon initial and final state.
8. Volume
is a:
(a) State
function
(b) Colligative
property
(c) Intensive
properties
(d) Path function
Explanation
(Answer; a)
Volume is a state
function as it depends upon initial and final states.
9. Which
statement is incorrect?
(a) For constant pressure process, ∆H = ∆E + P∆V
(b) For constant volume
process, ∆E = q
(c) For exothermic
reactions, ∆H > 0
(d) For Hess law Σ∆H°(cycle) =
0
Explanation
(Answer; c)
For exothermic changes, ∆H < 0
10. Which
of the following enthalpy change is always negative:
(a) Enthalpy of formation
(b) Enthalpy of
decomposition
(c) Enthalpy of
combustion
(d) Enthalpy of reaction
Explanation
(Answer; c)
Enthalpy of decomposition
is always negative as decomposition is always endothermic.
Enthalpy of formation and Enthalpy of reaction may be positive or negative. Enthalpy of combustion is always negative.
🔎💥XI MCQs Chapter 12 - Electrochemistry
1. The
outer body of dry cell serves as anode, it is made up of
(a)
Copper
(b)
Zinc
(c)
Lead
(d) Iron
1.
Explanation (Answer; b)
Composition
and Construction of Dry Cell
1. Zinc vessel (Outer
body………….……; Acts as anode
2. carbon or graphite rod
……………...; acts as cathode (Centrally placed)
3. manganese dioxide
(40%) ………….; Acts as cathode, helps to depolarize electrodes to produce steady
current
4. Ammonium Chloride(8%)
and Zinc chloride (8%) ………….; Acts as electrolyte giving necessary ions for
conduction
5. Glycerin or Starch
(4%) ……………..; serves to retains moisture
6. Carbon powder (40%)
……………….; serves to increase the surface area of graphite cathode
7. copper cap
………………………………..; for conduction of electricity
8. Zinc container
internally lined with porous paper works as a separator.
Electrolyte
The region between cathode
and anode is filled a paste of MnO2, carbon, NH4Cl and
ZnCl2.
2. The
conduction of electricity through an electrolytic solution is due to the flow
of
(a)
Electrons
(b)
ions
(c)
Atoms
(d) Molecules
2.
Explanation (Answer; b)
The conduction of
electricity through an electrolytic solution is called electrolytic conduction
which is due to the movement of free floating ions of electrolyte. In contrast,
the electronic or metallic conduction of electricity through metal or graphite is
due to free moving valence electrons.
3. During
electrolysis, the reaction that takes place at anode is
(a) Oxidation
(b) Simultaneous
oxidation and reduction
(c)
Hydrolysis
(d) Reduction
3.
Explanation (Answer; a)
In electrochemical cells
(whether electrolytic or Galvanic cell), oxidation always takes place at anode
while reduction occurs at cathode.
4.The
strongest oxidizing agent in the electro chemical series is:
(a)
Li
(b)
H₂
(c)
Cu
(d) F
4.
Explanation (Answer; d)
The strongest oxidizing
must have highest reduction potential. In ECS, F has the highest reduction
potential, therefore fluorine is the strongest oxidizing agent.
5.
Galvanized rod of iron is coated with:
(a)
Nickel
(b)
Zinc
(c)
Chromium
(d) Carbon
5.
Explanation (Answer; b)
Galvanized iron is coated
with zinc. Galvanizing is the type of electroplating involving coating of zinc
on baser metals like iron.
6.
Oxidation number of Cr in Na2Cr2O7, is:
(a)
+3
(b)+6
(c)
+8
(d) +12
6.
Explanation (Answer; b)
Oxidation number of Na =
+1
Oxidation number of O =
−2
In a neutral compound,
sum of oxidation numbers of all atoms is equal to zero.
2(Na) + 2Cr + 7(O) = 0
2(+1) + 2Cr + 7(−2) = 0
2 + 2Cr + (−14) = 0
2 + 2Cr −14 = 0
2Cr −12 = 0
2Cr = +12
Cr = +12/2 = +6
7. Which
of the following half cell reaction show oxidation?
(a) Fe3+ →
Fe2+
(b) Cl₂→ 2Cl−
(c) SO42− →
SO32−
(d) Zn → Zn2+
7. Explanation (Answer;
d)
Oxidation means increase
in oxidation number of atom during a reaction. The conversion of Zn (ON =0) to
ion (ON = +2) involves loss of 2 electrons which is manifested by the 2 units
increase in oxidation number.
In reaction (a) where
ferric ion (ON = +3) changes into ferrous ion (ON = +2) is reduction as
oxidation number of Fe has been decreased.
In reaction (b) where
chlorine (ON = 0) changes into chloride ion (ON = −1) is reduction as oxidation
number of Cl has been decreased.
In reaction (c) where
sulphate ion (ON = ++6) changes into sulphite ion (ON = +4) is reduction as
oxidation number of S has been decreased.
8. Fuel
cell is a typical Galvanic cell which is based on the reaction between:
(a) Nitrogen and oxygen
(b) Hydrogen and
oxygen
(c) Methane and
oxygen
(d) Hydrogen & zinc
6.
Explanation (Answer; b)
A fuel cell is an electrochemical cell that generates electrical energy from fuel via an electrochemical reaction. A hydrogen oxygen fuel cell is a unique type of Galvanic cell based on the reaction between hydrogen and oxygen to produce water and the heat released in the reaction is used to produced electricity.
Cathode ………….. Platinum
coated porous graphite
Anode …………….
Platinum coated porous graphite
Electrolyte ……...
Concentrated alkaline electrolyte like KOH
Hydrogen gas …. fed to
anode side (where it reacts with four OH- to produce 4 water molecules)
Oxygen gas ……… Fed to the
cathode side (where it reacts with water to produced four OH- ions)
9. In Zn-SHE
voltaic cell, the half reaction occurs at anode is
(a) Zn2+ +
2e− →
Zn
(b) Zn → Zn2+ +2e−
(c) 2H+ +
2e− →
H₂
(d) H₂ →2H2+ +
2e−
6.
Explanation (Answer; b)
In Zn-SHE voltaic cell,
zinc oxidizes to zinc ion (Zn2+) at anode (while SHE or hydrogen ion
reduces to H2 gas at cathode).
10. This
statement is not correct for lead storage battery:
(a) It can be recharged
(b) It is a primary
battery
(c) Anode is made up of
lead
(d) Cathode is made up of
lead oxide
10.
Explanation (Answer; b)
Lead storage battery is a secondary cell.
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🌙 یعنی میرے بعد بھی یعنی سانس لیے جاتے ہوں گے
💥 غزل ۔۔۔۔ جونؔ ایلیا 💥
💭 ہر لمحہ جی رہے ہیں مگر خیریت سے ہیں
🌌 دشتِ گماں کے خاک بسر خیریت سے ہیں
🕊️ اللہ اور تمام بشر خیریت سے ہیں
💧 مژگانِ خشک و دامنِ تر خیریت سے ہیں
🌟 یعنی تمام اہلِ نظر خیریت سے ہیں
🔥 سودائیانِ حال کے سر خیریت سے ہیں
💭 شکوے کی بات ہے، وہ اگر خیریت سے ہیں
🏚️ خاک اڑ رہی ہے اور کھنڈر خیریت سے ہیں
✨ جونؔ! ایک معجزہ ہے اگر خیریت سے ہیں
📜 اور اپنے صاحبانِ ہنر خیریت سے ہیں
⚔️ برگستوان و تیغ و تبر خیریت سے ہیں
🚪 بس در ہے اور بندئہ در خیریت سے ہیں
📖 ورنہ تمام جوشؔ و جگرؔ خیریت سے ہیں
🌙 باقی جو ہیں وہ شام و سحر خیریت سے ہیں
💥 غزل ۔۔۔۔ جونؔ ایلیا 💥
☀️ دھوپ آنگن میں پھیل جاتی ہے
🌆 شہر کوچوں میں خاک اڑاتی ہے
🕰️ میز پر گرد جمتی جاتی ہے
🌙 اب کسے رات بھر جگاتی ہے
💔 بے دلی بھی تو لب ہلاتی ہے
🌸 زندگی خواب کیوں دکھاتی ہے
💭 خواہشِ غیر کیوں ستاتی ہے
😮 ہمنشیں! سانس پھول جاتی ہے
👀 غور کرنے پہ یاد آتی ہے
💔 روز ایک چیز ٹوٹ جاتی ہے
💥 غزل ۔۔۔۔ جونؔ ایلیا 💥
💭 یہ دل کے خواب کی صورت نہ رائیگاں جائے
🌌 یہ شہر شہر کی محنت نہ رائیگاں جائے
💡 یہ خود سے اپنی رفاقت نہ رائیگاں جائے
🌙 کہیں یہ حسنِ طبیعت نہ رائیگاں جائے
💔 ہمارا عہدِ محبت نہ رائیگاں جائے
✨ یہ اجتماع یہ صحبت نہ رائیگاں جائے
🌟 رہے خیال یہ مہلت نہ رائیگاں جائے
🔥 تیرے جنون کی حالت نہ رائیگاں جائے
💥 غزل ۔۔۔۔ جونؔ ایلیا 💥
💭 شوق اس کا کمال ہے، تاحال
😔 جی ہمارا نڈھال ہے، تاحال
⚡ شوقِ بحث و جدال ہے، تاحال
❓ ہر جواب اک سوال ہے، تاحال
💔 دل میں زخمِ کمال ہے، تاحال
🌟 ذہن میں اک مثال ہے، تاحال
🌿 ہوسِ اندمال ہے، تاحال
🌸 آپ اپنی مثال ہے، تاحال
💔 بے امیدِ وصال ہے، تاحال
🌙 وہ جو تھا اک ملال ہے، تاحال
🎨 رنگ بے خدوخال ہے، تاحال
🦌 تو غزل کا غزال ہے، تاحال
💭 تجھ کو پانا محال ہے، تاحال
😔 پر وہی میرا حال ہے، تاحال
💥 غزل ۔۔۔۔ جونؔ ایلیا 💥
🌙 ہم ہیں حیران اپنی حیرت کے
💔 تم نہیں تھے مری طبیعت کے
🌟 کیا عجب عیش تھے شکایت کے
🎁 یہ عطیے ہیں دل کی عادت کے
⚖️ ہم ہی مفتی ہیں اہلسنت کے
🛠️ نہیں خوگر کسی مشقت کے
🌙 ہیں یہ لمحے تمام ہجرت کے
📖 ہیں عجب معجزے حکایت کے