✅ Correct Answer: (b)
Short reason:
Sodium (Group 1) loses one electron → M⁺, oxygen is O²⁻ → formula is M₂O.
Detailed reasoning:
Given electronic configuration: 1s² 2s² 2p⁶ 3s¹
🔎 The outermost electron is in 3s¹ → this means it belongs to Group 1 (alkali metals) → This is sodium (Na, Z = 11)
🔎 Group = IA (alkali metal)
Oxidation State of M = +1 (M⁺)
Oxygen forms oxide ion O²⁻
🔎 Oxide formula: To balance charges: 2M⁺ + O²⁻ → M₂O
👉🏼 Alkali metals form oxides with general formula M₂O.
🔎 Checking options:
(a) MO → for Group IIA metals like MgO, not Na
(b) M₂O → correct for Na
(c) M₂O₂ → sodium peroxide Na₂O₂ (exists but formed on burning oxygen with Na)
(d) MO₂ → not for Na 👈
✅ Correct Answer: (b)
Short reason:
Lithium is the first element of Group IA (alkali metals), shows a diagonal relationship in the periodic table with magnesium (Mg) from Group 2 showing strong resemblance with magnesium due to similar sizes and charge densities.
✅ Correct Answer: (c)
Short reason:
Given configuration of M: 1s² 2s² → Beryllium (Be), Z = 4 (an alkaline earth metal, Group 2, Period 2).
Diagonal relationship: Be (Group 2) ↔ Aluminium (Al, Group 13, Period 3).
Electronic configuration of Al (Z = 13) → 1s² 2s² 2p⁶ 3s² 3p¹
✅ Correct Answer: (a)
Short reason:
Li → forms normal oxide (Li₂O) when burned in air. (Na forms peroxide, K/Rb/Cs form superoxides)
✅ Correct Answer: (c)
Short reason:
Alkali metals (Group IA) have oxidation state +1, so they form M⁺ ions.
Nitrogen forms the nitride ion N³⁻.
To balance charges: 3(M⁺) + (N³⁻) → M₃N
So, alkali metals form nitrides of the type M₃N (e.g., Li₃N). The general formula is M₃N because alkali metals have +1 oxidation state and nitrogen has −3, so three M⁺ are needed to balance N³⁻.
Alkali metals (Group 1) typically form ionic nitrides with the formula M₃N, where M is Li, Na, etc., though in practice, only lithium nitride (Li₃N) is stable at room temperature. Other alkali metals form nitrides only under extreme conditions and they are unstable.
Details:
To form a neutral compound, we need total positive charge = total negative charge.
(a) MN₃ → M⁺ and 3N³⁻ → charges: +1 vs. −9 ❌ not possible.
(b) M₂N₃ → 2M⁺ and 3N³⁻ → charges: +2 vs. −9 ❌ not possible.
(c) M₃N → 3M⁺ and N³⁻ → charges: +3+ vs. −3 ✅ perfectly balanced.
(d) MN → M⁺ and N³⁻ → charges: +1 vs. −3 ❌ not possible.
✅ Correct Answer: (d)
Short reason:
Group IIIA have oxidation state +3, so they form M³⁺ ions.
Nitrogen forms the nitride ion N³⁻.
To balance charges: M³⁺ + N³⁻ → MN (charges are already balanced)
So, group IIIA elements form nitrides of the type MN
✅ Correct Answer: (c)
Short reason:
The general valence shell electronic configuration is (n –1)d¹⁻¹⁰, ns¹⁻² (where n ranges 4-7) or (n –1)d¹⁻¹⁰, ns⁰⁻².
✅ Correct Answer: (c)
Short reason:
The general valence shell electronic configuration is (n–2)f¹⁻¹⁴, (n–1)d⁰⁻¹, ns²
✅ Correct Answer: (b)
Short reason:
Transition elements in the 4d series have their outermost electrons in the 4d orbitals. These are located in period 5 of the periodic table.
4d series elements are: Yttrium (Y), Zr, Nb, Mo, Tc, Ru, Rh, Pd, Ag, Cd
✅ Correct Answer: (d)
Short reason:
Titanium → In making artificial joints, bone plates, screws and dental implants.
Vanadium → Use in batteries, as a catalyst and a pigment in glass making.
Iron → Building and bridge construction and tool making.
Copper → In making copper wires, alloys and sanitary works.
Zinc → Galvanizing, alloying and also use in batteries
Platinum → In making jewelry and also serves as a catalyst.
Mercury → Use in thermometers, B.P apparatus and amalgam formation
✅ Correct Answer: (d)
Short reason:
In the 3d series (Sc to Zn), atomic size generally decreases from Sc to Ni due to increasing nuclear charge without much shielding by d-electrons, then slightly increases for Cu and Zn due to electron-electron repulsion in the more filled 3d subshell. The smallest atomic radius in the 3d series is for Ni. 👈
Sc (largest) → Ti → Zn → Cu → V → Cr → Mn → Fe → Co → Ni (smallest)
✅ Correct Answer: (b)
Short reason:
In Ostwald’s process, ammonia (NH₃) is oxidized to nitric oxide (NO) using platinum (Pt) or platinum–rhodium (Pt–Rh) alloy as a catalyst. This is the first step in the industrial production of nitric acid (HNO₃).
4NH₃ + 5O₂ (Pt or Pt-Rh) → 4NO + 6H₂O
✅ Correct Answer: (a)
Short reason:
Ruthenium (Ru) catalysts are being explored for water splitting reactions, which involve the conversion of water into hydrogen and oxygen through electrolysis. These catalysts play a vital role in renewable energy technologies like hydrogen production and fuel cells.
✅ Correct Answer: (d)
Short reason:
Paramagnetism strength ∝ number of unpaired electrons (spin only).
Mn²⁺ → [Ar] 3d⁵ → 5 unpaired electrons (half-filled d-shell, maximum stability and paramagnetism).
Fe³⁺ → [Ar] 3d⁵ → 5 unpaired electrons (also half-filled).
Both have maximum spin = strongest paramagnetism.
Cr³⁺ → [Ar] 3d³ → only 3 unpaired.
Sc³⁺ → [Ar] 3d⁰ → no unpaired (diamagnetic).
Zn²⁺ → [Ar] 3d¹⁰ → no unpaired (diamagnetic).
Cr²⁺ → [Ar] 3d⁴ → 4 unpaired, but less than 5.
Mn³⁺ → [Ar] 3d⁴ → 4 unpaired, but less than 5.
✅ Correct Answer: (d)
Short reason:
👉 Amide (–CONH₂), imine (–C=NH or –C=NR), and oxime(–C=NOH) all have nitrogen atoms, but thiocarbonyl (C=S) does not.
✅ Correct Answer: (a)
Short reason:
👉 Terephthalic acid officially called Benzene-1,4-dicarboxylic acid is an aromatic dicarboxylic acids with formula C₆H₄(CO₂H)₂. It is principally as a precursor or a key monomer in the production of a high-performance polymer known as polyethylene terephthalate (PET, used to make clothing and plastic bottles.
✅ Correct Answer: (a)
Short reason:
👉 The production of the anti-inflammatory drug, ibuprofen from a compound called cumene is an example of partial synthesis. During the partial synthesis, cumene undergoes several chemical transformations including oxidation and rearrangement to yield ibuprofen.
The process of synthesizing a complex target organic molecule directly from a partially synthesized intermediate compound or readily available natural product, rather than through stepwise reaction is called as partial synthesis. It involves starting with a simple molecule and modifying it through a series of chemical reactions to create a more complex target molecule. Partial synthesis is frequently used when synthesizing the target molecule using total synthesis is extremely difficult either due to lack of availability of starting material or complexity of the multistep route.
✅ Correct Answer: (b)
Short reason:
👉 The combustion of carbon monoxide does not give soot as it always gives gaseous carbon dioxide.
✅ Correct Answer: (c)
Short reason:
👉 Cellulose is a naturally occurring polymer of glucose units found in plant cell walls made by living organisms, hence a biopolymer, while the others (Nylon-6, Polyester and Bakelite) are synthetic polymers.
✅ Correct Answer: (c)
Short reason:
C₆₀ (Buckminsterfullerene) is a spherical molecule made of 60 carbon atoms arranged in pentagons and hexagons (like a soccer ball).
👉 Each carbon atom is bonded to three other carbons. That means each carbon uses sp² hybridization.
Each carbon forms 3 sigma bonds with neighbors and has 1 delocalized π-electron, similar to graphite using sp² hybridized orbitals.
✅ Correct Answer: (c)
Short reason:
👉 Hydroxylic group (–OH) contains oxygen.
👉 Amino group (–NH₂) contains nitrogen, making this pair not exclusively oxygen-containing.
Let’s analyze each pair:
(a) Aldehyde (–CHO), Ketone (–CO–) → Both contain oxygen
(b) Carboxyl (–COOH), Ester (–COOR) → Both contain oxygen
(c) Hydroxylic (–OH), Amino (–NH₂) → Amino group contains nitrogen, not oxygen
(d) Amide (–CONH₂), Acid halide (–COCl) → Both contain oxygen in the carbonyl group
✅ Correct Answer: (a)
Short reason:
👉 Pentyl group (C₅H₁₁–) has the following 8 isomers, 4 are primary, 3 are secondary and 1 is tertiary:
⚡Four Primary isomers: n-pentyl (pentan 1 yl), isopentyl (3 methylbutan 1 yl), neopentyl (2,2 dimethylpropan 1 yl), 3-methylbutyl (2 methylbutan 1 yl.)
⚡Three Secondary isomers: sec pentyl (pentan 2 yl), 3-pentyl (pentan 3 yl,), 2-methylbutyl (3 methylbutan 2 yl.)
⚡One Tertiary isomers: tert-pentyl (1 methyl 1 butyl).
👉 So the counts are primary = 4, secondary = 3, tertiary = 1 → (a) 4, 3, 1
🔎 All 8 Isomeric Pentyl Groups with their Classification
✨1. n-pentyl → –CH₂CH₂CH₂CH₂CH₃ → primary
✨2. isopentyl → –CH₂CH(CH₃)CH₂CH₃ → primary
✨3. neopentyl → –CH₂C(CH₃)₃ → primary
✨4. sec-pentyl → –CH(CH₂CH₂CH₃)CH₃ → secondary
✨5. tert-pentyl → –C(CH₃)₂CH₂CH₃ → tertiary
✨6. 3-pentyl → –CH(CH₂CH₃)CH₂CH₃ → secondary
✨7. 2-methylbutyl → –CH(CH₃)CH₂CH₂CH₃ → secondary
✨8. 3-methylbutyl → –CH₂CH(CH₃)CH₂CH₃ → primary
✅ Correct Answer: (d)
Short reason:
👉 The tertiary butyl group is also known as the neo-butyl group.
⚡The trivial prefix "neo" is used when all but two carbons in a molecule form a continuous chain, and the remaining two carbons are part of a terminal tert-butyl group.
⚡It is used as a STANDARD for different types of higher branched radicals.
⚡Its full systematic name is 1,1-dimethylethyl or 2-methylpropan-2-yl.
✅ Correct Answer: (b)
Short reason:
👉 Acid amides contain acid amide functional group (−CONH₂) which is the combination of carbonyl group and amino group. Hence the general formula of acid amides is CₙH₂ₙ₊₁₋₋₋₋CONH₂
✅ Correct Answer: (b)
Short reason:
👉 Two secondary carbons (each CH₂) contribute 2 hydrogens each → total 4 secondary H atoms.
Detailed Explantion:
🔎 Step 1: Expand the structure
⚡ (CH₃)₂CH = isopropyl group → central carbon bonded to two CH₃ groups and one H.
⚡Then attached to –CH₂– (methylene).
⚡Then attached to –C₂H₅ (ethyl group).
⚡So the full chain is: CH₃–CH(CH₃)–CH₂–CH₂–CH₃
🔎 Step 2: Identify carbon types
⚡CH₃ at left end → primary carbon (attached to one other C).
⚡CH(CH₃) → central carbon attached to 3 carbons → tertiary carbon.
⚡CH₂ next to it → attached to 2 carbons → secondary carbon.
⚡CH₂ in ethyl group → attached to 2 carbons → secondary carbon.
⚡CH₃ at right end → primary carbon.
🔎 Step 3: Count secondary hydrogens
⚡Secondary carbons = 2 (each CH₂). 👈
⚡Each CH₂ has 2 hydrogens. 👈
🔎 Total secondary hydrogens = 2 × 2 = 4. 👈
✅ Correct Answer: (b)
Short reason:
👉 A free radical is an atom, molecule, or ion that has at least one unpaired valence electron and highly chemically reactive. 👉 Carbon free radicals are paramagnetic due to the presence of unpaired electron.
✅ Correct Answer: (d)
Short reason:
👉 Alkynes and alkadienes have the same general formula CₙH₂ₙ₋₂.
🔎 For example: Molecular formula C₅H₈
⚡CH≡C−CH₂−CH₂−CH₃ (1-Pentyne),
⚡CH₂=CH−CH₂−CH=CH₂ (1,4-pentadiene)
✅ Correct Answer: (d)
Short reason:
👉 C₄H₈ does not fit alkane formula (C₄H₁₀). It is saturated only if cyclic → cyclobutane.Detailed reasoning:
🔎 Saturated acyclic alkanes follow CₙH₂ₙ₊₂ → for n = 4, that’s C₄H₁₀ (butane).🔎 If a hydrocarbon has CₙH₂ₙ, it's either a cycloalkane (saturated) or an alkene (unsaturated). C₄H₈ is either an alkene (unsaturated) or a cycloalkane; since it’s saturated, it must be cyclobutane.
🔎 Details
⚡General formula for alkanes (saturated hydrocarbons) is CₙH₂ₙ₊₂ → For 4 carbons: C₄H₁₀ would be saturated
⚡But C₄H₈ has two fewer hydrogens, meaning one degree of unsaturation.
⚡Step 2: What causes one degree of unsaturation? Either one double bond → alkene Or one ring → cycloalkane
⚡So C₄H₈ could be Butene (unsaturated due to double bond) or Cyclobutane (saturated but cyclic)
👉 Key Point: The question asks for a saturated hydrocarbon. Only cyclobutane fits that bill.
✅ Correct Answer: (c)
Short reason:
👉 1,3-pentadiene has 5 carbons (pent-) with two double bonds at positions C-1 and C-3 structurally formulated as CH₂=CH–CH=CH–CH₃
🔎 (a) has 6 carbons (hexadiene).
🔎 (b) is 1,4-pentadiene.
🔎 (d) is a cumulated diene (allenic) containing a cumulene structure (C=C=C), not typical for simple dienes like 1,3-pentadiene.
✅ Correct Answer: (b)
Short reason:
👉 Cyclopropane carbons are sp³ hybridized, which corresponds to tetrahedral geometry, though distorted due to ring strain.
👉 Hybridization basis: sp³ → tetrahedral
👉 Actual ring shape: strained triangle due to ring constraints
👉 (answer; a in book but should be b)
Detailed reasoning:
🔎 Cyclopropane is the smallest and simplest cycloalkanes with a ring composed of three carbon atoms and six hydrogen atoms. The carbon atoms in cyclopropane are sp³-hybridized arranged in a trigonal geometry with an internal angel of 60°, which is quite smaller than the bond angle (109.5°) exists in alkanes.🔎 In cyclopropane, each carbon atom is sp³-hybridized, which typically corresponds to a tetrahedral geometry with bond angles of approximately 109.5°. Despite the tetrahedral hybridization, the three-membered ring forces the carbon atoms into a triangle, with internal bond angles of 60°. This leads to angle strain because the ideal tetrahedral angle is much larger than 60°. So while the electronic geometry is tetrahedral due to sp³ hybridization, the molecular shape is distorted into a planar triangle.
✅ Correct Answer: (b)
Short reason:
👉 The high s character of sp carbon stabilizes the conjugate base, making ethyne much more acidic than typical hydrocarbons, with a pKₐ around 25. ).
The pKₐ value of ethyne is 25 which shows that its acidic nature. Ethyne is a terminal alkyne. The hydrogen attached to the sp-hybridized carbon is relatively acidic compared to alkenes and alkanes. This is due to the high s-character (50%) of the sp orbital, which stabilizes the conjugate base (acetylide anion).
Detailed reasoning:
🔎 Cyclopropane is the smallest and simplest cycloalkanes with a ring composed of three carbon atoms and six hydrogen atoms. The carbon atoms in cyclopropane are sp³-hybridized arranged in a trigonal geometry with an internal angel of 60°, which is quite smaller than the bond angle (109.5°) exists in alkanes.🔎 In cyclopropane, each carbon atom is sp³-hybridized, which typically corresponds to a tetrahedral geometry with bond angles of approximately 109.5°. Despite the tetrahedral hybridization, the three-membered ring forces the carbon atoms into a triangle, with internal bond angles of 60°. This leads to angle strain because the ideal tetrahedral angle is much larger than 60°. So while the electronic geometry is tetrahedral due to sp³ hybridization, the molecular shape is distorted into a planar triangle.
🔎 pKa Comparison Table
⚡Ethyne; HC≡CH (alkyne) → Hybridization of C–H sp →pKa ~25 →Most acidic of hydrocarbons
⚡Ethene; CH₂=CH₂ (alkene) → Hybridization of C–H sp² →pKa ~44 →Less acidic
⚡Ethane; CH₃CH₃ (alkane) → Hybridization of C–H sp³ →pKa ~50 →Least acidic
⚡Phenol→ Hybridization of C–H — uninvolved in acidity →pKa ~10 →Much more acidic due to resonance
✅ Correct Answer: (b)
Short reason:
👉 Decarboxylation of sodium salts of carboxylic acids gives an alkane with one carbon less than the acid. Propanoic acid (3 carbons) → ethane (2 carbons).
Detailed reasoning:
🔎 Sodium propanoate = CH₃–CH₂–COONa.🔎 Decarboxylation means removal of the –COONa group as CO₂.
🔎 Reaction: CH₃–CH₂–COONa (NaOH / CaO; heat) → CH₃–CH₃ + CO₂
🔎 The product is ethane (CH₃–CH₃).
👉 Decarboxylation of sodium propanoate gives ethane.
👉 Decarboxylation of sodium ethanoate gives methane.
🔎 The decarboxylation of sodium alkanoate or sodium carboxylate on heating with soda lime gives lower previous alkane containing one carbon atom less than the corresponding acid salt. The new hydrogen atom in alkane is derived from the NaOH of soda lime.
✅ Correct Answer: (c)
Short reason:
👉 The molecular formula of benzene is C₆H₆. Benzene has 12 sigma bonds and 3 pi bonds. Benzene molecule contains 12 σ and 3 π bonds, giving a ratio (12:3) of 4:1 (4)
✅ Correct Answer: (d)
Short reason:
👉 The halogenation of benzene in presence of sunlight gives an addition product called hexachloro cyclohexane which is also called lindane or gammaxene or Gammallin formulated C₆H₆Cl₆. It is erroneously called Benzene hexachloride (BHC).
✅ Correct Answer: (d)
Short reason:
👉 Toluene contains methyl group which is the strongest activating group.
✅ Correct Answer: (d)
Short reason:
👉 All three compounds—toluene, ethylbenzene, and n-propylbenzene—contain a benzylic hydrogen, which is essential for oxidation by acidified KMnO₄ or K₂Cr₂O₇. These oxidizing agents convert the benzylic carbon (the carbon directly attached to the benzene ring) of the alkyl side chain (regardless of its length) into a carboxylic acid group. The final product is benzoic acid (C₆H₅COOH).
C₆H₅–CH₃ (toluene) → C₆H₅–COOH
C₆H₅–CH₂CH₃ (ethylbenzene) → C₆H₅–COOH
C₆H₅–CH₂CH₂CH₃ (n-propylbenzene) → C₆H₅–COOH
✅ Correct Answer: (a)
Short reason:
👉 Ortho/para directors: Electron-donating groups (EDGs), like −OH, −NH₂, −OCH₃, −R, −N(CH₃)₂
👈
👉 Meta directors: Electron-withdrawing groups (EWGs), like −NO₂, −CN, −CHO, −COOH, −COR
🔎 (a) −N(CH₃)₂ (EDG) , −NH₂ (EDG) → Both Ortho/Para ✅ 👈
🔎 (b) −OH (EDG) , RCO− (EWG) → Ortho/para + Meta ❌
🔎 (c) −NR₃ (EDG) , −CN (EWG) → Ortho/para + Meta ❌
🔎 (d) −OCH₃ (EDG) , −CHO (EWG) → Ortho/para + Meta ❌
✅ Correct Answer: (b)
Short reason:
👉 In Friedel–Crafts alkylation, primary alkyl halides like n-propyl chloride form (or behave like) primary carbocations, which readily rearrange to the more stable secondary carbocation (isopropyl). The benzene ring then attacks this rearranged electrophile → cumene.
In Friedel–Crafts alkylation, carbocation rearrangement occurs. The unstable n propyl cation rearranges to the more stable isopropyl cation, giving cumene as the major product.
Detailed reasoning
🔎 Reactants: Benzene + n-propyl chloride (CH₃CH₂CH₂Cl)
🔎 Catalyst: AlCl₃ (Lewis acid)
🔎 Goal: Electrophilic aromatic substitution to form an alkylbenzene
🔎 Key Mechanistic Insight:
🔥 AlCl₃ abstracts Cl⁻ from n-propyl chloride → forms n-propyl carbocation (CH₃CH₂CH₂⁺)
🔥 BUT: n-propyl carbocation is primary, and unstable 👈
🔥 So it undergoes a 1,2-hydride shift → forms isopropyl (secondary) carbocation (CH₃–CH⁺–CH₃), which is secondary and more stable. Benzene attacks the isopropyl carbocation, not the original n-propyl forming Iso-propylbenzene (cumene) 👈
✅ Correct Answer: (c)
Short reason:
👉 In Friedel–Crafts acylation of toluene, the –CH₃ group in spite of its ortho/para-directing nature exclusively gives only the para product. Steric hindrance makes the para acylated product the major one 👈 (a smaller amount of the ortho is also formed; meta is not).
Detailed reasoning
In acylation, though, virtually all the substitution happens in the 4-position.
This is due to the steric hindrance created by newly entrant bulky acyl group.
The reason that you get virtually none of the 2-isomer in this instance is because of the size of the incoming acyl group 👈.
Everything gets too cluttered (and therefore less stable) if you try to put the acyl group next door to the methyl group.👈
✅ Correct Answer: (b)
Short reason:
👉 Resonance in benzene makes all C–C bonds identical with bond order ≈ 1.5, lying between single and double. 👈
⚡Benzene (C₆H₆) has a resonance structure with alternating double bonds.
⚡In reality, all six C–C bonds are equivalent due to delocalization of π electrons.
⚡Each bond is not a pure single bond (bond order = 1) nor a pure double bond (bond order = 2).
⚡Instead, the bond order is the average:
👉 Bond order = (3 double bonds × 2) + (3 single bonds × 1)/6 = 9/6 = 1.5 👈
👉 So, each C–C bond in benzene has a bond order between one and two. 👈
Detailed reasoning
⚡Benzene (C₆H₆) is a resonance hybrid of two Kekulé structures, where the π-electrons are delocalized over all six carbon ⚡atoms. This delocalization leads to:
⚡Equal bond lengths for all C–C bonds (~1.39 Å)
⚡Intermediate bond character between a single (1.54 Å) and double bond (1.34 Å)
🔎 Bond Order Calculation
⚡Each C–C bond in benzene shares:
⚡1 σ-bond (from the framework)
⚡½ π-bond (from delocalized electrons)
👉 Bond Order = 1 + ½ = 1.5 👈
Final Score
💥 جونؔ ایلیا 🎯 💥
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⚡ کیا ستم ہے کہ ہم لوگ مر جائیں گے
🕊️ بے خبر آئے ہیں بے خبر جائیں گے
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🔥 بس گزرنے کو ہے موسمِ ہاؤ و ہُو، تم کہاں جاؤ گے ہم کہاں جائیں گے؟
💥 غزل ۔۔۔۔ جونؔ ایلیا 💥
💭 جانے کیسے لوگ وہ ہوں گے جو اس کو بھاتے ہوں گے
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💔 آنے والوں سے کیا مطلب آتے ہیں آتے ہوں گے
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💔 وہ جو سمٹتے ہوں گے ان میں وہ تو مر جاتے ہوں گے
🌙 یعنی میرے بعد بھی یعنی سانس لیے جاتے ہوں گے
💥 غزل ۔۔۔۔ جونؔ ایلیا 💥
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💧 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
💥 غزل ۔۔۔۔ جونؔ ایلیا 💥
💔 خود کو ہلاک کر لیا، خود کو فدا نہیں کیا
😔 تو نے تو ہم سے آج تک کوئی گلہ نہیں کیا
📜 میں نے بھی ایک شخص کا قرض ادا نہیں کیا
🌟 آپ بہت شریف ہیں، آپ نے کیا نہیں کیا
🙏 ہم نے نہیں کیا وہ کام، ہاں باخُدا نہیں کیا
💡 اُس نے تو کارِ جہل بھی بے علما نہیں کیا
💥 غزل ۔۔۔۔ جونؔ ایلیا 💥
🌙 تھا تو اک شہر خاکساروں کا
😔 اب پتہ کیا ہے دل فگاروں کا
🛏️ بسترا اب کہاں ہے یاروں کا
🕊️ کون پُرساں ہے یادگاروں کا
🌌 مجھ سے کیا ذکر رہ گزاروں کا
🌟 عیش مت پوچھ دعویداروں کا
🐎 نہ پیادوں کا نہ سواروں کا
📚 دہر ہے صرف استعاروں کا
🤝 کیا ہوا جانے جانثاروں کا
🌪️ ایک جلسہ تھا شعلہ خواروں کا
💥 حضرتِ جونؔ ایلیا 💥 🎨 خاتم الشعرا و شاعرِ بے مثل و بے بدل 🎨
🌙 اور ہم ہیں کہ نام کر رہے ہیں
✨ ہم جو یہ اہمتمام کررہے ہیں
🔥 آپ تو قتلِ عام کررہے ہیں
💭 ہم تو خود سے کلام کررہے ہیں
🤝 ہر کسی کو سلام کررہے ہیں
💔 اپنا ہونا حرام کررہے ہیں
📢 ہم یہ اعلانِ عام کررہے ہیں
🥂 ناف پیالے کو جام کررہے ہیں
🙏 اور وہ احترام کررہے ہیں
🌌 کوئے دل میں خرام کررہے ہیں
🕊️ بات ہی ہم تمام کررہے ہیں
🥁 بے سبب دھوم دھام کررہے ہیں
🗡️ تیغ کو بے نیام کررہے ہیں
⏳ دم کو بس دوام کررہے ہیں
🏛️ ہم ابد میں قیام کررہے ہیں
💥 حضرتِ جونؔ ایلیا 💥 🎨 خاتم الشعرا و شاعرِ بے مثل و بے بدل 🎨
💔 بچھڑنا ہے تو جھگڑا کیوں کریں ہم
🔥 کوئی ہنگامہ برپا کیوں کریں ہم
❌ پر اب اس باب کو وا کیوں کریں ہم
⚖️ بھلا گھاٹے کا سودا کیوں کریں ہم
🌟 تمہاری ہی تمنّا کیوں کریں ہم
📜 تو ساری عمر ایفا کیوں کریں ہم
🚶 فقط کمروں میں ٹہلا کیوں کریں ہم
💡 وہ سرمایہ اکٹھا کیوں کریں ہم
😔 تو پھر دنیا کی پروا کیوں کریں ہم
👀 بھلا اندھوں سے پردہ کیوں کریں ہم
🤝 سو خود پر بھی بھروسا کیوں کریں ہم
🍽️ تمہیں راتب مہیا کیوں کریں ہم
🌪️ زمیں کا بوجھ ہلکا کیوں کریں ہم
🕊️ یہاں کارِ مسیحا کیوں کریں ہم