🔥🌟 MDCAT/ECAT/FSC/XII Grand Chemistry MCQs Interactive Quiz # 4 🧪💡 | Learn & Practice!

 

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1. The number of isomers of hexane are

a) 3

b) 5

c) 9

d) 18

✅ Correct Answer: (b) 5
👉 Hexane (C₆H₁₄) is an alkane having 5 structural isomers due to different branching patterns i.e n Hexane, 2 Methylpentane, 3 Methylpentane, 2,2 Dimethylbutane and 2,3 Dimethylbutane

2. Cycloalkenes have the general formula:

a) CₙH₂ₙ

b) CₙH₂ₙ₊₂

c) CₙH₂ₙ₋₂

d) CₙH₂ₙ₊₄

✅ Correct Answer: (c) CₙH₂ₙ₋₂
👉 Cycloalkenes have one double bond in the ring, reducing 2 hydrogens from CₙH₂ₙ (cycloalkanes) making its general formula CₙH₂ₙ₋₂. (Cycloalkanes → CₙH₂ₙ (like alkanes but cyclic).

3. The functional group in RCN is

a) Thioalcohol

b) Ether

c) Nitrile

d) Thioether

✅ Correct Answer: (c) Nitrile
👉 RCN or R–C≡N (where carbon is triple bonded to nitrogen) is the general formula nitrile (alkanenitrile). The CN group is called the nitrile or cyano group.
👉 Thioalcohols (SH), ethers (-O-), and thioethers (-S-) involve oxygen/sulfur, not a C≡N bond.

4. Propanone and propanal are

a) Functional isomers

b) Positional isomers

c) Tautomers

d) Metamers

✅ Correct Answer: (a) Functional isomers
👉 Propanone (CH₃–CO–CH₃) → ketone. 👉 Propanal (CH₃–CH₂–CHO) → aldehyde. 👉 Same molecular formula (C₃H₆O) but different functional groups (aldehyde vs ketone). That makes them functional isomers.

5. The oxidation states of the elements of group VA is:

a) +1 and +3

b) −3 and −5

c) −3, +3 and +5

d) +1, −1 and +3

✅ Correct Answer: (c) −3, +3 and +5
👉 Group VA elements (N, P, As, Sb, Bi) exhibit −3, +3, and +5 oxidation states.
🔥 Example: Nitrogen in NH₃ → −3.
🔥Phosphorus in PCl₃ → +3.
🔥Phosphorus in PCl₅ → +5.

6. Which of the following s block elements form normal oxide when burnt in air?

a) Sr

b) Na

c) K

d) Li

✅ Correct Answer: (d) Li
👉 Lithium forms normal oxide (Li₂O) when burnt in air, unlike Na, K, or Sr which form peroxides/superoxides.

7. In the equilibrium of dichromate ion(Cr₂O₇²⁻), the colours of CrO₄²⁻ and Cr₂O₇²⁻ are respectively:

a) Orange and green

b) Green and yellow

c) Yellow and orange

d) Orange and yellow

✅ Correct Answer: (c)Yellow and orange
👉 Chromate ion (CrO₄²⁻) is yellow, while dichromate (Cr₂O₇²⁻) is orange. 👉 So the correct option is c

8. The coordination number of Pt in [Pt(en)₃]Cl₄ is:

a) 3

b) 4

c) 6

d) 7

✅ Correct Answer: (c) 6
👉 en = ethylenediamine, a bidentate ligand (each en donates 2 lone pairs).
👉 In the Complex: [Pt(en)₃]Cl₄ → Pt is bonded to 3 en ligands.
👉 Each en contributes 2 donor atoms → total = 3 × 2 = 6 coordination sites giving coordination number 6.

9. The knocking of internal combustion engine can be reduced by the following petroleum process:

a) Refining

b) Resonance

c) Distillation

d) Reforming

✅ Correct Answer: (d) Reforming
👉 Knocking is caused by premature ignition of fuel-air mixture.
👉 Reforming increases octane rating by converting straight chains into branched/aromatic hydrocarbons, reducing knocking thereby improving fuel quality. It converts straight-chain hydrocarbons into branched or aromatic ones → better combustion, less knocking.
👉 Refining and distillation separate components but don’t improve anti-knock quality.
👉 Resonance is unrelated to petroleum processing.

10. How many optical isomers are possible for HOOC(Br)CH–CH(OH)COOH?

a) 2

b) 3

c) 4

d) 3

✅ Correct Answer: (c) 4
👉 The molecule has two chiral centers, giving 2² = 4 optical isomers, with no meso form. Details
👉 The molecule is HOOC–CH(Br)–CH(OH)–COOH.
👉 It has two central carbons: one bonded to Br, the other bonded to OH. There are two chiral centers.
👉 C1 (CHBr): Attached to –COOH, –H, –Br, and –CH(OH)COOH → all different → chiral.
👉 C2 (CHOH): Attached to –COOH, –H, –OH, and –CH(Br)COOH → all different → chiral.
👉 Meso occurs if the molecule has an internal plane of symmetry.
👉 In this case, substituents (Br vs OH) are different → no meso form possible.
👉 General formula for Number of stereoisomers = = 2ⁿ, where n = number of chiral centers.
👉 Here, 2² = 4.

11. Which of the following pairs of compounds represent functional group isomerism?

a) Propanone and propanal

b) Ethanol and dimethyl ether

c) Acetic acid & methyl formate

d) All of these

✅ Correct Answer: (d) All of these
👉 Each pair has the same molecular formula but different functional groups, so all represent functional group isomerism.
👉 Hence the correct choice is All of these.
🧠 Stepwise Reasoning
Functional group isomers → same molecular formula but different functional groups.
Propanone (CH₃–CO–CH₃) vs Propanal (CH₃–CH₂–CHO):
Both C₃H₆O, one is ketone, other aldehyde → functional group isomers.
Ethanol (C₂H₅OH) vs Dimethyl ether (CH₃–O–CH₃):
Both C₂H₆O, one is alcohol, other ether → functional group isomers.
Acetic acid (CH₃COOH) vs Methyl formate (HCOOCH₃):
Both C₂H₄O₂, one is carboxylic acid, other ester → functional group isomers.
👉 Hence all three pairs are functional group isomers.

12. Which of the following pairs of compounds represent positional isomerism?

a) 1-butene and 2-butene

b) Iso-propyl alcohol and 1-propanol

c) n-butane and iso-butane

d) Both a and b

✅ Correct Answer: (d) Both a and b
🔎 1-butene/2-butene and iso propyl alcohol/1-propanol differ by position of double bond or OH group, so both represent positional isomerism.
🔎 🧠 Stepwise Reasoning
👉 Positional isomers → same carbon skeleton and functional group, but the position of the functional group or multiple bond differs.
👉 1-butene vs 2-butene:
Both are alkenes (C₄H₈).
Double bond position differs (C1 vs C2). ✅ Positional isomers.
👉 Iso-propyl alcohol (CH₃–CHOH–CH₃) vs 1-propanol (CH₃–CH₂–CH₂OH):
Both are alcohols (C₃H₈O).
OH group position differs (terminal vs middle carbon). ✅ Positional isomers.
👉 n-butane vs iso-butane:
Both are alkanes (C₄H₁₀).
Difference is in branching, not position of functional group. ❌ These are chain isomers, not positional.
✅ Correct Answer (d) Both a and b

13. Lucas Test is used to distinguish between:

a) Alcohol and phenol

b) Three types of alcohols

c) Alcohols and amines

d) None of these

✅ Correct Answer: (b) Three types of alcohols
🔎 Lucas test differentiates primary, secondary, and tertiary alcohols based on their reactivity with ZnCl₂/HCl.
🔎 🧠 Stepwise Reasoning
👉 Lucas reagent = ZnCl₂ + concentrated HCl.
👉 It tests the reactivity of alcohols with HCl to form alkyl chlorides.
🔎 Observation:
👉 Tertiary alcohols → react immediately (cloudy solution).
👉 Secondary alcohols → react within 5–10 minutes.
👉 Primary alcohols → no reaction at room temperature.
👉 Therefore, Lucas test is specifically used to distinguish primary, secondary, and tertiary alcohols.

14. General formula CₙH₂ₙ₋₂ represents:

a) Alkynes

b) Cycloalkenes

c) Alkadienes

d) All of these

✅ Correct Answer: (d) All of these
🔎 The formula CₙH₂ₙ₋₂ applies to alkynes, cycloalkenes, and alkadienes, so all are correct.
🔎 🧠 Stepwise Reasoning
👉 Alkynes: General formula = CₙH₂ₙ₋₂ (for one triple bond). ✅
👉 Cycloalkenes: A cycloalkene has one ring + one double bond → loses 2 hydrogens compared to CₙH₂ₙ. Formula = CₙH₂ₙ₋₂.✅
👉 Alkadienes: For conjugated/cumulative dienes with two double bonds, formula also = CₙH₂ₙ₋₂. ✅
👉 Therefore, all three classes of compounds can be represented by CₙH₂ₙ₋₂.

15. The kinetics and molecularity of elimination bimolecular (E₂) reaction are respectively:

a) Second order and 2

b) First order and 1

c) Zero order and 2

d) None of these

✅ Correct Answer: (a) Second order and 2
🔎 👉E₂ elimination is bimolecular: both substrate and base participate in the rate determining step. Hence, the reaction shows second order kinetics and has molecularity = 2.
🔎🧠 Stepwise Reasoning
📌 E₂ reaction = Elimination, bimolecular.
📌 In E₂, the rate determining step involves two species simultaneously:
⚡The substrate (alkyl halide)
⚡The base
📌 Therefore, the rate law depends on both concentrations: Rate = k [substrate][base]
📌 This makes the reaction second order kinetics.
📌 Since two species are involved in the slow step, the molecularity = 2.

16. The region of sphere which extends from 11 km to 50 km from Earth is known as:

a) Thermosphere

b) Mesosphere

c) Troposphere

d) Stratosphere

✅ Correct Answer: (d) Stratosphere
👉 The stratosphere lies just above the troposphere, spanning from about 11 km to 50 km altitude. It contains the ozone layer, which absorbs harmful UV radiation.
🔎🧠 Stepwise Reasoning
📌Troposphere: Extends from surface up to ~11 km.
📌Stratosphere: Extends from ~11 km to ~50 km.
📌Mesosphere: Extends from ~50 km to ~85 km.
📌Thermosphere: Extends above ~85 km up to 600 km+.
👉 The region between 11 km and 50 km is the Stratosphere.

17. Clemmensen reduction is the conversion of aldehydes and ketones into:

a) Alkyl halides

b) Alkanes

c) Alcohols

d) Alkenes

✅ Correct Answer: (b) Alkanes
👉 Clemmensen reduction removes the carbonyl oxygen and converts aldehydes/ketones into hydrocarbons (alkanes) using Zn(Hg)/HCl.
🔎🧠 Stepwise Reasoning
📌 Clemmensen reduction uses Zn(Hg) + concentrated HCl.
📌 It reduces the carbonyl group (C=O) of aldehydes and ketones.
📌 The oxygen is removed and the carbonyl carbon is converted into a –CH₂– group (for aldehydes) or –CH– group (for ketones).
📌 Final product = hydrocarbon (alkane).
👉 Example: CH₃COCH₃ (acetone) → CH₃–CH₂–CH₃ (propane).

18. Acetic acid is naturally found in:

a) Vinegar

b) Valerian root

c) Butter

d) Bee’s sting

✅ Correct Answer: (a) Vinegar
👉 Acetic acid is the sour tasting acid naturally present in vinegar, whereas butter has butyric acid and bee stings have formic acid.
🔎🧠 Stepwise Reasoning
📌 Acetic acid (CH₃COOH) is the main component of vinegar, giving it its sour taste and pungent smell.
📌 Vinegar typically contains 4–8% acetic acid in water.
📌 Valerian root contains other organic acids and compounds, not acetic acid.
📌 Butter contains butyric acid (C₄H₈O₂), not acetic acid.
📌 Bee’s sting venom contains formic acid (HCOOH), not acetic acid.

19. An example of quaternary structure of protein is:

a) Globulin

b) Albumin

c) Hemoglobin

d) Myoglobin

✅ Correct Answer: (c) Hemoglobin
🔎 Hemoglobin exhibits quaternary structure because it consists of multiple polypeptide subunits (α₂ β₂) working together, unlike myoglobin or albumin which are single chain proteins.
🔎 🧠 Stepwise Reasoning
🔎 Protein structures:
📌 Primary: sequence of amino acids.
📌 Secondary: α helix, β sheet.
📌 Tertiary: 3D folding of a single polypeptide chain.
📌 Quaternary: association of two or more polypeptide chains into a functional protein.
🔎 Myoglobin: single polypeptide chain → tertiary structure only.
🔎 Albumin: also a single chain → tertiary structure only.
🔎 Globulin: general class of proteins, many are single chains.
🔎 👉 Hemoglobin: composed of 4 polypeptide chains (2α + 2β) → classic example of quaternary structure.

20. It is a three membered cyclic ether:

a) Diazonium salt

b) Epoxide

c) Benzoquinone

d) Ozonide

✅ Correct Answer: (b) Epoxide
👉 Epoxides are three membered cyclic ethers with significant ring strain, making them highly reactive in organic chemistry.
🔎 🧠 Stepwise Reasoning
📌Diazonium salt: Not an ether, it’s an –N₂⁺ group attached to an aromatic ring.
📌Benzoquinone: A cyclic diketone, not an ether.
📌Ozonide: A five membered ring containing oxygen atoms, formed in ozonolysis.
📌Epoxide: A three membered cyclic ether (triangle shaped ring with one oxygen atom).
👉 Therefore, the correct option is Epoxide.

21. The coordination number of Cr in [Cr(H₂O)₄Cl₂]⁺ is:

a) 2

b) 4

c) 6

d) 8

✅ Correct Answer: (c) 6
👉 Four H₂O + two Cl⁻ ligands, all monodentate → CN = 6.

22. The number of unpaired electrons in gaseous species of Mn³⁺ and Co³⁺ respectively are:

a) 4, 4

b) 3, 3

c) 4, 3

d) 3, 4

✅ Correct Answer: (c) 4, 3
👉 Mn = atomic number 25 → Mn³⁺ = [Ar] 3d⁴ → 4 unpaired electrons
👉 Co = atomic number 27 → Co³⁺ = [Ar] 3d⁶ → 3 unpaired electrons (high spin)

23. Which type of glycosidic linkage is present in maltose molecule?

a) α-linkage

b) α-β linkage

c) β-linkage

d) γ-linkage

✅ Correct Answer: (a) α-linkage
👉 Maltose = glucose + glucose via α(1→4) glycosidic bond
👉 Both glucose units are in α-form → α-linkage

24. A zwitterion is:

a) An ion that is positively charged in solution

b) An ion that is negatively charged in solution

c) An ion electrically neutral

d) A carbohydrate with an electrical charge

✅ Correct Answer: (c) An ion electrically neutral
👉 Zwitterion has both positive and negative charges but overall neutral
👉 Common in amino acids at isoelectric point

25. Aspirin is a pain reliever, its chemical name is:

a) Ascorbic acid

b) Nicotinic acid

c) Acetylsalicylic acid

d) Benzoic acid

✅ Correct Answer: (c) Acetylsalicylic acid
👉 Aspirin = ester of salicylic acid and acetic acid → acetylsalicylic acid
👉 Ascorbic acid = Vitamin C
👉 Nicotinic acid = Vitamin B₃

26. Troposphere extends from planetary surface to bottom of ________?

a) Stratosphere

b) Mesosphere

c) Troposphere

d) Thermosphere

✅ Correct Answer: (a) Stratosphere
👉 Troposphere is the lowest layer of atmosphere; it extends up to the base of the stratosphere (~12 km).

27. The atomic number of an element belonging to group IVA and 3rd period is:

a) 14

b) 13

c) 15

d) 23

✅ Correct Answer: (a) 14
👉 Group IVA = carbon family; 3rd period element is silicon, atomic number = 14.

28. A mixture of ZnCl₂ + conc. HCl is called

a) Lucas reagent

b) Grignard’s reagent

c) Tollen’s reagent

d) Benedict’s reagent

✅ Correct Answer: (a) Lucas reagent
👉 Lucas reagent = A mixture of ZnCl₂ + conc. HCl, used to distinguish alcohols based on reactivity.

29. Select the incorrect statement regarding alkanes: (M)?

a) It is known as paraffin

b) It is an acyclic saturated hydrocarbon

c) In alkanes, carbon hybridization is sp²

d) Alkane have the general formula CₙH₂ₙ₊₂

✅ Correct Answer: (c) In alkanes, carbon hybridization is Sp²
👉 Alkanes have sp³ hybridized carbon atoms, not sp² (which is for alkenes).

30. Fractional distillation of coal gives coke, coal tar, coal gas and:

a) Petroleum

b) Natural gas

c) Naphtha

d) Ammonia liquor

✅ Correct Answer: (d) Ammonia liquor
👉 Ammonia liquor is a by-product of coal distillation, along with coke, coal tar, and coal gas.

31. An isomer of acetone is:

a) Formaldehyde

b) Acetaldehyde

c) Diethyl ether

d) Propionaldehyde

✅ Correct Answer: (b) Acetaldehyde
👉 Acetone = CH₃–CO–CH₃ (ketone)
👉 Acetaldehyde = CH₃–CHO (aldehyde)
👉 Both have formula C₃H₆O → functional isomers

32. Down the group atomic size increases due to:

a) Shielding effect

b) Photoelectric effect

c) Increase in nuclear force of attraction

d) Oxidation effect

✅ Correct Answer: (a) Shielding effect
👉 As you go down a group, more electron shells → inner electrons shield outer ones → atomic size increases

33. The most reactive molecule towards nucleophilic addition in the following is:

a) Formaldehyde

b) Acetaldehyde

c) diethyl ketone

d) Acetophenone

✅ Correct Answer: (a) Formaldehyde
👉 Formaldehyde has no alkyl groups → least steric hindrance and strongest partial positive on carbonyl carbon → Most reactive to nucleophilic attack

34. In flame test, Apple green (Pale green) is given by

a) Barium

b) Beryllium

c) Boron

d) Calcium

✅ Correct Answer: (a) Barium
👉 In flame tests, barium salts produce a pale apple green flame, which is diagnostic for barium.
🔎 🧠 Stepwise Reasoning
📌 Flame test colors are due to excitation of electrons in metal ions.
📌 Barium salts give a characteristic apple green (pale green) flame.
📌 Beryllium does not impart a distinct flame color (too small, high ionization energy).
📌 Boron gives a bright green flame, but not the apple green shade typical of barium.
📌 Calcium gives a brick red flame, not green.

35. When an alkylbenzene reacts with acidified potassium permanganate, the alkyl group is oxidized to:

a) –CHO group

b) –COO⁻ group

c) –COOH group

d) –CONH₂ group

✅ Correct Answer: (c) –COOH
👉 Acidified KMnO₄ oxidizes the alkyl side chain of benzene completely to a carboxylic acid group (–COOH), regardless of chain length.
🔎 🧠 Stepwise Reasoning
📌 Alkylbenzenes (Ar–CH₃, Ar–CH₂R, etc.) undergo strong oxidation with acidified KMnO₄.
📌 Regardless of the length of the alkyl chain, the benzylic carbon (the carbon directly attached to the benzene ring) is oxidized completely.
📌 The final product is always benzoic acid (Ar–COOH) or substituted benzoic acids.
📌 Example: C₆H₅–CH₃ (KMnO₄/ H⁺) → C₆H₅–COOH
🔎 –CHO group (aldehyde) may appear as an intermediate but is further oxidized to –COOH.
🔎 –COO⁻ is the conjugate base of –COOH, but the product is written as the acid.
🔎 –CONH₂ (amide) is not formed in this reaction.

36. The distinct aroma and flavour of dairy products is due to:

a) Propionic acid

b) Butyric acid

c) Acetic acid

d) Formic acid

✅ Correct Answer: (b) Butyric acid
👉 The characteristic flavour and aroma of dairy products, especially butter, comes from butyric acid.
🔎 🧠 Stepwise Reasoning
📌 Butyric acid (CH₃CH₂CH₂COOH):
📌 Found in butter and other dairy fats.
📌 Responsible for the characteristic aroma and flavour of dairy products.
📌 Also gives the rancid smell when butter spoils.
📌 Propionic acid: contributes to flavour in Swiss cheese but not the general dairy aroma.
📌 Acetic acid: gives vinegar its smell, not typical dairy flavour.
📌 Formic acid: pungent, found in ant venom, not related to dairy aroma.

37. Citric acid is used as a/an:

a) Preservative

b) Flavour enhancer

c) Acidity regulator

d) All of these

✅ Correct Answer: (d) All of these
👉 Citric acid functions as a preservative, flavour enhancer, and acidity regulator, making option All of these correct.
🔎 🧠 Stepwise Reasoning
📌 Citric acid is a weak organic acid naturally found in citrus fruits. It has multiple uses in food and beverages:
📌 Preservative: Prevents microbial growth by lowering pH.
📌 Flavour enhancer: Adds a tangy, sour taste to foods and drinks.
📌 Acidity regulator: Maintains the desired pH in processed foods and beverages.
👉 Since citric acid serves all these roles, the most comprehensive answer is All of these.

38. The IUPAC name of salicylic acid is:

a) 2-hydroxybenzoic acid

b) 3-hydroxybenzoic acid

c) 4-hydroxybenzoic acid

d) None of these

✅ Correct Answer: (a) 2-hydroxybenzoic acid
👉 Salicylic acid is benzoic acid with an –OH group at the ortho (2) position, hence 2-hydroxybenzoic acid.
🔎 🧠 Stepwise Reasoning
🔎 Salicylic acid has a benzene ring with:
📌 A –COOH group (carboxylic acid) at position 1.
📌 A –OH group (hydroxyl) at position 2 (ortho to the carboxyl group).
🔎 Therefore, its systematic IUPAC name is 2-hydroxybenzoic acid.
📌 If the –OH were at position 3, it would be 3-hydroxybenzoic acid (meta).
📌 If the –OH were at position 4, it would be 4-hydroxybenzoic acid (para).
📌 But salicylic acid specifically has the ortho (2 position) hydroxyl group.

39. The IUPAC name of sorbic acid is:

a) Hexa 2,4-dienoic acid

b) Hexa 2,3-dienoic acid

c) Hex-2-enoic acid

d) None of these

✅ Correct Answer: (a) Hexa-2,4-dienoic acid
👉 Sorbic acid has a six carbon chain with double bonds at positions 2 and 4, hence Hexa 2,4-dienoic acid.
🔎 🧠 Stepwise Reasoning
📌 Sorbic acid is a naturally occurring unsaturated fatty acid used as a preservative.
📌 Its structure: CH₃–CH=CH–CH=CH–COOH.
📌 This is a six carbon chain (hex ) with two double bonds at positions 2 and 4.
📌 Therefore, the correct systematic name is Hexa 2,4-dienoic acid.
🔎 Option (b) Hexa 2,3-dienoic acid would mean double bonds at carbons 2 and 3, which is incorrect.
🔎 Option (c) Hex-2-enoic acid has only one double bond, not two.

40. Methylbenzene (toluene) is oxidized by acidified potassium permanganate into:

a) Acetophenone

b) Benzaldehyde

c) Benzdioic acid

d) Benzoic acid

✅ Correct Answer: (d) Benzoic acid
👉 Acidified KMnO₄ oxidizes the methyl group of methylbenzene completely to a carboxylic acid (benzoic acid).
🔎 🧠 Stepwise Reasoning
📌 Methylbenzene (C₆H₅–CH₃) has a benzylic –CH₃ group.
📌 Acidified KMnO₄ is a strong oxidizing agent.
📌 It oxidizes the entire alkyl side chain at the benzylic position, regardless of length, into a –COOH group.
📌 Thus, methylbenzene is converted into benzoic acid (C₆H₅–COOH).
🔎 Benzdioic acid (terephthalic acid) is not formed here.
🔎 Acetophenone (C₆H₅–CO–CH₃) is a ketone, not produced by oxidation of methylbenzene.
🔎 Benzaldehyde (C₆H₅–CHO) may appear as an intermediate but is further oxidized to benzoic acid.

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صحنِ خیالِ یار میں کی نہ بسر شبِ فراقجب سے وہ چاندنہ گیا، جب سے وہ چاندنی گئی

اس کے بدن کو دی نمود ہم نے سخن میں اور پھراس کے بدن کے واسطے ایک قبا بھی سی گئی

🌙 ـــــــــــــــــــــــــــــــــــــــــــــ 🌙

نشئہ شوقِ رنگ میں، تجھ سے جدائی کی گئیایک لکیر خون کی، بیچ میں کھینچ دی گئی

تھی جو کبھی سرِسخن ، میری وہ خامشی گئیہائے! کہن سنن کی بات، ہائے وہ بات ہی گئی

شوق کی ایک عمر میں کیسے بدل سکے گا دلنبضِ جنون ہی تو تھی ، شہر میں ڈوبتی گئی

اُس کی گلی سے اُٹھ کے میں ،آن پڑا تھا اپنے گھرایک گلی کی بات تھی ، اور گلی گلی گئی

اُس کی اُمید ِناز کا مجھ سے یہ مان تھا کہ آپعمر گزار دیجیے ، عمر گزار دی گئی

دور به دور ،دل به دل، درد به درد ،دم به دمتیرے یہاں رعایت ِحال نہیں رکھی گئی

جونؔ! جنوب ِزرد کے خاک بسر، یہ دُکھ اُٹھاموجِ شمالِ سبز جاں، آئی تھی اور چلی گئی

کیا وہ گماں نہیں رہا؟ ہاں وہ گماں نہیں رہاکیا وہ اُمید بھی گئی؟ ہاں وہ اُمید بھی گئی

🌟 نوری نستعلیق میں رنگین غزل، مرکزی اور جَسٹفائیڈ متن 🌟

🔥 جونؔ ایلیا --- غزل 🔥

💜 حالتِ حال کے سبب، حالتِ حال ہی گئی 💙 شوق میں کچھ نہیں گیا، شوق کی زندگی گئی

💚 ایک ہی حادثہ تو ہے اور وہ یہ کہ آج تک 💛 بات نہیں کہی گئی ، بات نہیں سنی گئی

🧡 بعد بھی تیرے جانِ جاں ، دل میں رہا عجب سماں 💖 یاد رہی تری یہاں، پھر تری یاد بھی گئی

💗 اس کی امیدِ ناز کا ہم سے یہ مان تھا کہ آپ 💘 عمر گزار دیجئے، عمر گزار دی گئی

💞 اس کے وصال کے لئے، اپنے کمال کے لئے 💕 حالتِ دل، کہ تھی خراب ،اور خراب کی گئی

💓 تیرا فراق جانِ جاں! عیش تھا کیا میرے لئے 💔 یعنی ترے فراق میں خوب شراب پی گئی

💬 اس کی گلی سے اٹھ کے میں آن پڑا تھا اپنے گھر 📿 ایک گلی کی بات تھی اور گلی گلی گئی

🌙 صحنِ خیالِ یار میں کی نہ بسر شبِ فراق ✨ جب سے وہ چاندنہ گیا، جب سے وہ چاندنی گئی

🎨 اس کے بدن کو دی نمود ہم نے سخن میں اور پھر 🧵 اس کے بدن کے واسطے ایک قبا بھی سی گئی

🔥 جونؔ ایلیا --- غزل 🔥

💜 نشئہ شوقِ رنگ میں، تجھ سے جدائی کی گئی 💙 ایک لکیر خون کی، بیچ میں کھینچ دی گئی

💚 تھی جو کبھی سرِسخن ، میری وہ خامشی گئی 💛 ہائے! کہن سنن کی بات، ہائے وہ بات ہی گئی

🧡 شوق کی ایک عمر میں کیسے بدل سکے گا دل 💖 نبضِ جنون ہی تو تھی ، شہر میں ڈوبتی گئی

💗 اُس کی گلی سے اُٹھ کے میں ،آن پڑا تھا اپنے گھر 💘 ایک گلی کی بات تھی ، اور گلی گلی گئی

💞 اُس کی اُمید ِ ناز کا مجھ سے یہ مان تھا کہ آپ 💕 عمر گزار دیجیے ، عمر گزار دی گئی

💓 دور به دور ،دل به دل، درد به درد ،دم به دم 💔 تیرے یہاں رعایت ِ حال نہیں رکھی گئی

💬 جونؔ! جنوب ِ زرد کے خاک بسر، یہ دُکھ اُٹھا 📿 موجِ شمالِ سبز جاں، آئی تھی اور چلی گئی

🌙 کیا وہ گماں نہیں رہا؟ ہاں وہ گماں نہیں رہا ✨ کیا وہ اُمید بھی گئی؟ ہاں وہ اُمید بھی گئی