Chemical Bonding and Molecular Structure: Class 11 Chemistry Chapter 4 Hinglish mein

Chemistry Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

105 min
48 practice questions
33 MCQs

Is chapter mein aap kya seekhenge

Molecules aur ions ke Lewis symbols aur Lewis structures banao, formal charges nikalo aur octet rule ke exceptions pehchano
Ionic compounds ka banna ionisation enthalpy, electron gain enthalpy aur lattice enthalpy ki madad se samjhao
Bonds ko unki length, angle, enthalpy aur order se compare karo, aur unhe resonance aur dipole moments se describe karo
VSEPR theory se molecules aur ions ke shapes predict karo, lone pairs ke effect ke saath
Covalent bonds, σ aur π bonds aur molecular shapes ko valence bond theory aur sp, sp², sp³, sp³d aur sp³d² hybridisation se samjhao
Homonuclear diatomic molecules ke molecular orbital diagrams bharo aur unse bond order aur magnetic behaviour nikalo
Hydrogen bonding, uske do types aur physical properties par uska effect samjhao

NCERT Chemistry Class 11, Chapter 4, pp. 100–135 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

Table salt par chot maaro to woh toot kar bikhar jaata hai, aur sirf 800 °C se upar pighalta hai. Sugar stove par rakhe pan mein pighal jaati hai, aur cooking cylinder ki gas room temperature se kaafi neeche hi boil ho jaati hai. Diamond, jo pure carbon hai, ab tak ka sabse hard natural material hai. In sab differences ki jad ek hi cheez hai: har substance ke particles aapas mein kaise jude rehte hain.

Aas-paas dekho, tumhe oxygen, iron ya carbon ke akele atoms kahin nahi milenge. Sirf noble gases hi nature mein single atoms ke roop mein milti hain; baaki har element ke atoms jude hue milte hain, molecules mein ya ions ke bahut bade arrays mein, jinhe chemical bonds (रासायनिक आबंध) pakad kar rakhte hain. Chemical bond woh attraction hai jo kisi substance mein atoms ya ions ko saath rakhta hai. Isse kuch tez sawaal uthte hain. Atoms bond banate hi kyun hain? Hydrogen $H_{2}$ ke roop mein kyun rehta hai jabki helium akela rehta hai? Water molecule bent kyun hai lekin carbon dioxide molecule seedha?

Tum chaar models se miloge, aur har model pichhle model ki kami poori karta hai. Pehle Kössel–Lewis electron counting aati hai, phir shapes ke liye VSEPR theory, directed bonds ke liye valence bond theory aur magnetism ke liye molecular orbital theory. Unit hydrogen bond ke saath khatam hoti hai.

4.1 Chemical Bonding ka Kössel–Lewis Approach

1916 mein do chemists ne alag-alag kaam karte hue, Germany mein Walther Kössel aur United States mein Gilbert N. Lewis ne, valence ki pehli explanation electrons ke terms mein di. Dono ne ek hairaan karne wale fact se shuruaat ki: noble gases mushkil se hi react karti hain, isliye unka outer electron arrangement khaas taur par stable hona chahiye.

Lewis ke model ne har atom ko do parts mein baanta. Kernel, yaani nucleus aur andar wale electrons milakar, positive charge rakhta hai aur reactions mein badalta nahi. Uske chaaron taraf outer shell hota hai, jise unhone ek cube ki tarah banaya, jiske aathon corners par ek-ek electron ki jagah hai. Noble gas ke saare corners bhare hote hain. Lewis ke hisaab se koi bhi doosra atom apna cube bharne ke liye bond banata hai, ya to electrons de kar ya unhe share karke.

Chemistry atom ki surface par hoti hai. Sirf sabse bahari shell ke electrons, yaani valence electrons, hi move ya share hote hain; andar wale electrons isse door rehte hain. Lewis ke shorthand mein element ka symbol likha jaata hai aur har valence electron ke liye ek dot lagaya jaata hai. Aisi drawing ko Lewis symbol (लूइस प्रतीक) kehte hain; Fig. 4.1 mein period 3 ke Lewis symbols diye hain.

Fig. 4.1Dots pehle chaaron sides par akele lagte hain, phir pair bante hain. Chaar dots tak group valence dots ki ginti hoti hai, uske baad aath minus dots.NCERT p. 101

Symbols Na, Mg, Al, Si, P, S, Cl aur Ar do rows mein, har ek ke saath har valence electron ke liye ek dot, sodium ke ek dot se lekar argon ke aath dots tak. Pair banne se pehle dots chaaron sides par akele lagte hain, aur har symbol ke neeche electron count likha hai.

Dots ginkar pata chalta hai ki koi element aam taur par kitne bonds banata hai, yaani uski group valence. Ek se chaar dots hon to atom saare dots use karta hai (Mg do bonds banata hai, Al teen, Si chaar). Paanch ya zyada dots hon to missing electrons ikattha karna aasaan hota hai, isliye valence aath minus dots hoti hai (P 3, S 2, Cl 1).

Kössel ka starting point periodic table ka layout tha. Har noble gas ke ek taraf halogen hai, jo aasani se electrons le leta hai, aur doosri taraf alkali metal, jo aasani se electrons de deta hai. Metal atom se ek electron halogen atom par le jao, to dono ka electron count unke noble-gas padosi jaisa ho jaata hai: outer $n s^{2} n p^{6}$ octet, ya $L i^{+}$ ke case mein helium ka $1 s^{2}$ pair. Products opposite charge wale ions hote hain, aur electrostatic attraction unhe saath pakad kar rakhta hai.

Potassium ( $[A r] 4 s^{1}$ ) apna 4s electron kho deta hai aur uske paas argon ka configuration bach jaata hai. Bromine ( $[A r] 3 d^{10} 4 s^{2} 4 p^{5}$ ) ek electron leta hai aur krypton ke configuration tak pahunch jaata hai.

\begin{aligned} K & \overset{}{\to} K^{+} + e^{-} \\ B r + e^{-} & \overset{}{\to} B r^{-} \\ K^{+} + B r^{-} & \overset{}{\to} K B r \end{aligned}

Magnesium ke paas do valence electrons hain, isliye woh do chlorine atoms mein se har ek ko ek electron deta hai: $M g \overset{}{\to} M g^{2 +} + 2 e^{-}$ , aur har $C l + e^{-} \overset{}{\to} C l^{-}$ . Compound $M g C l_{2}$ banta hai (Fig. 4.2). Chemists is ion-to-ion attraction ko electrovalent bond kehte the; aaj hum aam taur par ionic bond (आयनिक आबंध) kehte hain. Kisi element ke ion par jitne unit charges hote hain, woh uski electrovalence hai: yahan magnesium ke liye 2 (positive) aur chlorine ke liye 1 (negative). Ionic crystals ki hamari picture aaj bhi isi scheme par tiki hai, lekin jaisa Kössel ne khud maana, yeh carbon compounds aur kai doosre compounds ke baare mein kuch kaam ki baat nahi batati.

3/3 $M g^{2 +}$ ke paas neon ka shell hai; har $C l^{-}$ ke paas argon ka octet.

Fig. 4.2Electron transfer se har ion ko noble-gas configuration mil jaata hai.NCERT p. 101

Do electrons (crosses ke roop mein) wala magnesium atom, saat-saat dots wale do chlorine atoms ke beech hai. Step 1 neutral atoms dikhata hai. Step 2 mein curved arrows hain jo magnesium ka ek-ek electron har chlorine tak le jaate hain. Step 3 mein bina outer electrons ka Mg 2+ hai, aur har chloride ion brackets mein aath electrons aur minus charge ke saath hai.

4.1.1 Octet Rule

Definition

Octet rule

Atoms valence electrons kho kar, le kar ya share karke aapas mein judte hain, taaki har atom ke valence shell mein aath electrons ho jaayein (hydrogen ke liye do), bilkul sabse paas wali noble gas ki tarah.

ExampleMagnesium do electrons khota hai, chlorine ek leta hai, aur $C H_{4}$ mein carbon chaar pairs share karta hai.

Kössel aur Lewis ne chemical combination ki yeh electronic theory 1916 mein di thi. Yeh ek kaafi kaam ka rule of thumb hai, nature ka law nahi: Section 4.1.5 mein bataya hai ki yeh kahan fail hota hai.

4.1.2 Covalent Bond

Transfer se $F_{2}$ samajh nahi aata: do ek jaise atoms ke paas electron ki adla-badli karne ki koi wajah nahi hai. 1919 mein Irving Langmuir ne Lewis ka sharing wala idea aage badhaya, rigid cube ko chhod diya, aur result ko covalent bond (सहसंयोजी आबंध) naam diya: aisa bond jismein do atoms electrons ko milkar share karte hain.

Fluorine atom ( $[H e] 2 s^{2} 2 p^{5}$ ) ke paas saat valence electrons hain aur use aath chahiye. Agar do fluorine atoms ek common pair mein ek-ek electron de dein, to har atom us pair ko apne total mein gin leta hai, aur dono aath tak pahunch jaate hain. Hum $F - F$ likhte hain, jismein line shared pair ko dikhati hai. Har fluorine ke paas teen aise pairs bhi hote hain jinhe woh share nahi karta: yeh lone pairs sirf ek atom ke hote hain.

Aisi drawings Lewis dot structures kehlati hain, aur atoms ek jaise hon ya alag, yahi logic kaam karta hai. Ek line hamesha ek shared pair hoti hai; aam taur par har partner usmein ek electron daalta hai; aur poori drawing mein har atom ke paas noble-gas count hota hai, aath electrons, ya hydrogen ke liye do.

Atoms ek se zyada pair bhi share kar sakte hain. Methanal (formaldehyde, $H C H O$ ) mein carbon aur oxygen do pairs share karte hain, yeh double bond hai. Hydrogen cyanide, $H C N$ , mein carbon aur nitrogen teen pairs share karte hain, yeh triple bond hai, aur $N_{2}$ ke dono atoms bhi yahi karte hain. Ek shared pair single bond hota hai.

4.1.3 Simple Molecules ka Lewis Representation (Lewis Structures)

Lewis structure shape nahi dikhata, lekin yeh dikhata hai ki kaun se atoms jude hain, kitne pairs se, aur lone pairs kahan hain. Isse chemistry ka bahut bada hissa samajh aata hai, isliye inhe sahi tarike se banana seekhna faayde ka hai. Yeh recipe use karo:

Saare atoms ke valence electrons gino. Anion ke liye har negative charge par ek electron jodo; cation ke liye har positive charge par ek electron ghatao. ( $S O_{4}^{2 -}$ : $6 + 4 \times 6 + 2 = 32$ ; $H_{3} O^{+}$ : $3 + 6 - 1 = 8$ .)
Skeleton sketch karo. Sabse kam electronegative atom, lekin kabhi hydrogen nahi, aam taur par centre mein jaata hai: $S O_{4}^{2 -}$ mein S, $C O C l_{2}$ mein C.
Har outer atom ko centre se single bond se jodo, har bond mein do electrons.
Outer atoms ke octets lone pairs se poore karo, phir jo electrons bachein unhe central atom par rakho.
Centre ko theek karo. Agar central atom ka octet abhi bhi adhoora hai, to padosi atom ke ek lone pair ko shared pair bana do, jisse double ya triple bond ban jaaye.

4/4O ka ek lone pair shared pair ban jaata hai: C=O. Saare octets poore.

Fig. 4.3Har electron apni jagah par hai aur har atom ka octet poora hai. Aisi ginti kisi bhi molecule ya ion ke liye kaam karti hai.NCERT Table 4.1 · p. 103

Phosgene ka skeleton: C centre mein, O upar, do Cl neeche. Step 1 mein sirf symbols aur tally 4 + 6 + 14 = 24 electrons hain. Step 2 mein teen single bonds jud jaate hain, 18 electrons bachte hain. Step 3 mein har outer atom par teen lone pairs lagte hain aur note hai ki carbon ke paas sirf 6 hain. Step 4 mein oxygen ka ek lone pair doosra C–O bond ban jaata hai, to har atom ke paas aath ho jaate hain.

Example 4.1NCERT Problem 4.1 · p. 104

Hydrogen cyanide ka Lewis structure

Hydrogen cyanide, $H C N$ , ka Lewis structure banao.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 4.2NCERT Problem 4.2 · p. 104

Nitronium ion ka Lewis structure

Nitronium ion, $N O_{2}^{+}$ , ka Lewis structure banao.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

4.1.4 Formal Charge

$N O_{2}^{+}$ ka charge asal mein kahan baitha hai? Poora ion yeh charge rakhta hai, lekin ek simple bookkeeping rule se hum har atom ko ek formal charge (औपचारिक आवेश) de sakte hain. Yeh rule har bond ko bilkul barabar sharing maanta hai: atom ko uske bonds ke aadhe electrons aur uske apne lone pairs ke saare electrons milte hain. Is hisaab ko atom ke valence electrons se compare karne par milta hai:

Formal charge

F.C. = V - L - \frac{1}{2} S

(4.1)

jahan

$V$: free atom ke valence electrons
$L$: atom par lone pairs mein electrons
$S$: atom se jude bonds mein electrons (shared electrons)

$N O_{2}^{+}$ (Example 4.2) mein nitrogen ke liye $V = 5$ , $L = 0$ aur $S = 8$ hai, isliye uska formal charge $5 - 0 - 4 = + 1$ hai. Har oxygen ka $6 - 4 - 2 = 0$ hai. Formal charges ka total +1 hai, jo ion ka charge hai: kisi bhi structure ko check karne ka yeh ek kaam ka tarika hai.

Formal charges ko hisaab-kitaab samjho, measurement nahi. $N O_{2}^{+}$ mein oxygen atoms zyada electronegative hone ki wajah se, zero formal charge ke bawajood, nitrogen se zyada electron density rakhte hain. Formal charges tab kaam aate hain jab tumhe kai possible drawings mein se ek chunni ho: woh structure chuno jismein sabse kam aur sabse chhote formal charges hon, aur negative charge (agar ho) zyada electronegative atom par ho.

Example

Dinitrogen oxide ke liye structure chunna

Dinitrogen oxide, $N_{2} O$ (laughing gas), mein 16 valence electrons hain. Teen octet structures possible hain: (a) $N = N = O$ , (b) $N \equiv N - O$ aur (c) $N = O = N$ . Formal charges ki madad se decide karo ki kaun se structures molecule ko sabse achchhe se describe karte hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

4.1.5 Octet Rule ki Limitations

Zyada tar organic molecules aur second period ke elements ke lagbhag saare compounds octet rule follow karte hain. Is comfort zone ke bahar yeh rule teen tarah se toot jaata hai.

Compare

Octet rule tootne ke teen tarike

Incomplete octet

Central atom ke paas aath se kam electrons hote hain.
Chaar se kam valence electrons wale atoms mein yeh aam hai: Be, B, Al.
Jaise: $B e C l_{2}$ gas (Be par 4 electrons); $B F_{3}$ aur $B (C H_{3})_{3}$ (B par 6).

Odd-electron molecules

Electrons ka total odd ho to kam se kam ek atom ka octet poora nahi ho sakta.
Jaise: $N O$ (11 valence electrons), $N O_{2}$ (17), $C l O_{2}$ (19).

Expanded octet

Period 3 ya usse aage ka central atom aath se zyada electrons rakhta hai.
Jaise: $P C l_{5}$ (P ke chaaron taraf 10), $S F_{6}$ (S ke chaaron taraf 12), $I F_{7}$ (I ke chaaron taraf 14).

Pehle expanded octets ko period-3 atoms ke khaali d orbitals se samjhaya jaata tha; yeh description tum Section 4.6.3 mein dekhoge. Wahi element rule follow bhi kar sakta hai: $H_{2} S$ mein sulfur ka octet hai lekin $S F_{6}$ mein uske chaaron taraf baarah electrons hain.

Is rule mein aur gehri kamiyan bhi hain. Iski buniyaad hi, yaani 'unreactive' noble gas, poori tarah sach nahi hai: xenon $X e F_{4}$ , $X e F_{6}$ aur $X e O_{3}$ banata hai, aur krypton $K r F_{2}$ banata hai. Yeh molecule ke shape ke baare mein koi clue nahi deta. Aur kyunki yeh energy ka koi zikr kiye bina sirf electrons ginta hai, yeh molecules ki stability ya bonds ki strength compare nahi kar sakta.

4.2 Ionic ya Electrovalent Bond

Kössel ki picture kehti hai ki ionic compound do stages mein banta hai: pehle atoms ions bante hain, phir ions ek crystal mein pack ho jaate hain. Compound banega ya nahi, yeh is par depend karta hai ki ions kitni aasani se bante hain aur crystal kitni energy wapas deta hai. Ek metal M aur ek non-metal X ke liye steps yeh hain:

\begin{aligned} M (g) & \overset{}{\to} M^{+} (g) + e^{-} \\ X (g) + e^{-} & \overset{}{\to} X^{-} (g) \\ M^{+} (g) + X^{-} (g) & \overset{}{\to} M X (s) \end{aligned}

Pehle step mein ionisation enthalpy lagti hai, jo hamesha positive hoti hai. Doosre step mein electron gain enthalpy, $Δ_{e g} H$ , aati hai: ground state wale ek isolated gaseous atom mein ek electron jodne par enthalpy change, jo kuch elements ke liye negative (energy release hoti hai) aur kuch ke liye positive hota hai. Electron affinity isi change ko ulte sign ke saath batati hai. Isliye ionic bonds sabse aasani se ek low ionisation enthalpy wale element aur ek large negative electron gain enthalpy wale element ke beech bante hain: yaani ek metal aur ek non-metal ke beech.

To ek typical ionic compound mein cation metal se aata hai aur anion non-metal se. Ek mashhoor exception ammonium ion, $N H_{4}^{+}$ , hai: iske saare atoms non-metals hain, phir bhi ammonium chloride aur ammonium nitrate jaise salts mein yeh metal ion ka role nibhata hai.

Ab potassium chloride ke numbers dekho (Fig. 4.4). Potassium ko ionise karne mein $419 k J m o l^{- 1}$ lagte hain; chlorine mein electron jodne par $349 k J m o l^{- 1}$ release hote hain. Isliye gaseous ions ka pair banane mein ab bhi $+ 70 k J m o l^{- 1}$ lagte hain: octets tak pahunchna uphill hai. Crystal balance badal deta hai. Ions ko paas laakar solid banane par lagbhag $718 k J m o l^{- 1}$ release hote hain, isliye overall change kaafi downhill hai.

4/4Lattice lagbhag $718 k J m o l^{- 1}$ release karti hai: overall lagbhag $648 k J m o l^{- 1}$ downhill.

Fig. 4.4Ions banane mein energy lagti hai; lattice use kai guna wapas kar deti hai.NCERT p. 106

K(g) + Cl(g) se shuru hone wali ek energy ladder. Step 1 sirf starting level dikhata hai. Step 2 mein potassium ko ionise karne ke liye 419 kJ/mol ka upar jaata arrow hai. Step 3 mein K+ aur Cl− ions tak 349 kJ/mol ka neeche aata arrow hai, jo abhi bhi start se upar hai. Step 4 mein KCl(s) tak 718 kJ/mol ka neeche aata arrow hai, jo overall start se 648 kJ/mol neeche hai.

Solid mein har ion ek regular three-dimensional lattice mein opposite charge wale ions se ghira hota hai. KCl (rock-salt arrangement) mein har $K^{+}$ chhe $C l^{-}$ ions ko chhoota hai aur har $C l^{-}$ chhe $K^{+}$ ko; koi salt kaun sa arrangement apnata hai, yeh uske ions ke size par depend karta hai. Isliye ionic compound ki stability uski lattice enthalpy se judge hoti hai, sirf octets se nahi.

4.2.1 Lattice Enthalpy

Definition

Lattice enthalpy

Ek mole ionic solid ko alag karke aise gaseous ions banane ke liye zaroori energy jo aapas mein interact nahi karte.

Example $K C l (s) \overset{}{\to} K^{+} (g) + C l^{-} (g)$ ke liye lagbhag $718 k J m o l^{- 1}$ chahiye.

Crystal mein ek akele ion ko har opposite charge wala ion kheenchta hai aur har same charge wala ion dhakelta hai, paas wale bhi aur door wale bhi, ek aise pattern mein jo crystal structure tay karta hai. Isliye lattice enthalpy calculate karne ka matlab hai poori lattice par in sab interactions ko jodna; ek ion pair kaafi nahi hai. Do trends yaad rakhne layak hain. Ions par charge badhne se lattice enthalpy tezi se badhti hai aur ions bade hone par ghatti hai. Magnesium oxide, jismein $M g^{2 +}$ aur $O^{2 -}$ ions hain, ki lattice enthalpy lagbhag $3800 k J m o l^{- 1}$ hai, jo KCl se paanch guna se bhi zyada hai.

4.3 Bond Parameters

Kisi bond ko kuch measurable quantities se describe kiya ja sakta hai: uski length, uske banaye angles, use todne ke liye zaroori energy aur uska order. Inhi par do aur ideas tike hain: resonance aur polarity.

4.3.1 Bond Length

Bond length woh doori hai jis par do bonded atoms ke nuclei rehte hain jab molecule apni lowest-energy state mein ho; ise spectroscopy aur X-ray ya electron diffraction se naapa jaata hai. Bond length ko do hisson mein baanta ja sakta hai, har atom se ek: yeh atoms ke covalent radii hain. Kisi element ka covalent radius uske apne do atoms ke beech single bond ki length ko aadha karke milta hai (Cl–Cl distance ka aadha Cl ka radius deta hai). Phir atoms A aur B ke liye:

Covalent radii se bond length

R = r_{A} + r_{B}

(4.2)

jahan

$R$: bond length
$r_{A}, r_{B}$: A aur B ke covalent radii

Bromine monochloride par try karo: Br ( $114 p m$ ) aur Cl ( $99 p m$ ) ke covalent radii jodkar $213 p m$ milta hai, jo measured $214 p m$ ke kaafi kareeb hai. Jo atoms bonded nahi hain, jaise solid mein padosi molecules ke atoms, woh kaafi door rehte hain, kyunki unke poore electron clouds aapas mein takrate hain. Ek jaise atoms ke beech is contact distance ka aadha van der Waals radius hai, jo non-bonded atom ke size ka measure hai. Bromine ka van der Waals radius lagbhag $185 p m$ hai, jo uske covalent radius se kaafi bada hai (Fig. 4.5).

Fig. 4.5Covalent radius ek bonded pair se aata hai; van der Waals radius padosi molecules ke un atoms se jo bas ek doosre ko chhoo rahe hain.NCERT Fig. 4.2 · p. 107

Do Br2 molecules ek ke baad ek, har atom ek chhote core circle ke roop mein bana hai jo ek bade van der Waals circle ke andar hai. Bonded pair ke upar ek bracket 228 pm dikhata hai, jo covalent radius ka double hai; padosi molecules ke chhoote hue atoms ke beech ek bracket 370 pm dikhata hai, jo van der Waals radius ka double hai.

4.3.2 Bond Angle

Jahan do bonds kisi atom par milte hain, wahan woh ek bond angle banate hain; sahi maayne mein, yeh kisi molecule ya complex ion ke central atom par do bonding pairs wale orbitals ke beech ka angle hai. Bond angles spectroscopic measurements se milte hain aur degrees mein likhe jaate hain. Bond lengths ke saath milkar yeh atoms ki positions tay karte hain, aur molecule ke shape ka matlab yahi hai: ammonia mein H–N–H 107° hai, aur hydrogen sulfide mein H–S–H sirf lagbhag 92°.

4.3.3 Bond Enthalpy

Definition

Bond enthalpy

Do atoms ke beech kisi khaas tarah ke bond ko todne ke liye per mole zaroori energy, jab har species gas phase mein ho. Unit: $k J m o l^{- 1}$ .

ExampleEk mole Cl–Cl bonds todne ke liye $243 k J m o l^{- 1}$ chahiye.

\begin{aligned} C l_{2} (g) \overset{}{\to} 2 C l (g) \\ Δ_{a} H = 243 k J m o l^{- 1} \\ H B r (g) \overset{}{\to} H (g) + B r (g) \\ Δ_{a} H = 366 k J m o l^{- 1} \end{aligned}

Bond enthalpy jitni zyada, bond utna strong. Multiple bonds khaas taur par strong hote hain: carbon monoxide ke triple bond ke liye lagbhag $1076 k J m o l^{- 1}$ chahiye, jo kisi bhi diatomic molecule mein sabse zyada hai.

Kai identical bonds wale molecule mein har agla bond todne ke liye alag energy lagti hai, kyunki bache hue bonds ka mahaul badal jaata hai. Methane ke pehle C–H bond ke liye lagbhag $439 k J m o l^{- 1}$ chahiye, aakhri ke liye kaafi kam. Isliye hum average (mean) bond enthalpy use karte hain: us tarah ke saare bonds todne ki total energy, unki ginti se divide karke. Methane ko atomise karne, $C H_{4} (g) \overset{}{\to} C (g) + 4 H (g)$ , ke liye lagbhag $1664 k J m o l^{- 1}$ chahiye, isliye average C–H bond enthalpy $416 k J m o l^{- 1}$ hai. Tables mein aise hi averages diye jaate hain.

4.3.4 Bond Order

Definition

Bond order

Lewis picture mein, do atoms ke beech shared electron pairs ki ginti: single bond ke liye 1, double bond ke liye 2, triple bond ke liye 3.

To $C l_{2}$ ka bond order 1 hai, $O_{2}$ ka 2 aur $N_{2}$ ka 3; carbon monoxide ka bhi, teen shared pairs ke saath, 3 hai. Isoelectronic species (electrons ki same ginti, milta-julta structure) ka bond order same hota hai: $C N^{-}$ , $C_{2}^{2 -}$ (acetylide ion) aur $N O^{+}$ , $N_{2}$ ke saath isoelectronic hain aur sabka bond order 3 hai.

Ek correlation poori chemistry mein chalta hai: bond order badhne par bond enthalpy badhti hai aur bond length ghatti hai. Table 4.1 ise carbon–carbon aur nitrogen–nitrogen bonds ke liye dikhata hai.

Table 4.1Single, double aur triple bonds ki typical lengths aur average enthalpies

Bond	Bond order	Length / pm	Enthalpy / kJ mol⁻¹
$C - C$	1	154	348
$C = C$	2	134	614
$C \equiv C$	3	120	839
$N - N$	1	145	163
$N = N$	2	125	418
$N \equiv N$	3	110	946

Quick check

3 mein se 1

Covalent radii: C

77 p m

, Cl

99 p m

C C l_{4}

mein C–Cl bond length ka estimate pm mein lagao.

Answer dekho

Answer 176 pm

R = r_{C} + r_{C l} = 77 + 99 = 176 p m

, jo measured value ke kareeb hai. Dono radii jodne ki jagah ek radius ko double karne se galat answer aata hai.

4.3.5 Resonance Structures

Kuch species ko ek Lewis structure se banaya hi nahi ja sakta. Formate ion, $H C O O^{-}$ , ko lo, jo formic acid (cheenti ke dank wala acid) ka anion hai. Har octet structure mein ek C=O aur ek C–O bond hota hai, jinki lengths alag honi chahiye: lagbhag $122 p m$ aur $143 p m$ . Experiment mein dono carbon–oxygen bonds identical milte hain, lagbhag $125 p m$ , dono ke beech mein.

Iska hal hai resonance. Jab ek single Lewis structure kisi species ko describe nahi kar paata, to hum do ya zyada structures likhte hain jinmein har nucleus same jagah rehta hai aur fark sirf is baat ka hota hai ki electrons kahan banaye gaye hain. Yeh canonical ya resonance structures hain, jo double-headed arrow $⟷$ se jude hote hain. Species ka asal mein ek hi structure hota hai, resonance hybrid (अनुनाद संकर), jo in sabki features ko mila deta hai. Formate mein har C–O bond lagbhag dedh bond hai, aur negative charge dono oxygens par barabar baanta hota hai (Fig. 4.6).

3/3Asli ion: lagbhag $125 p m$ ke do barabar C–O bonds, har O par aadha negative charge.

Fig. 4.6Koi bhi canonical structure exist nahi karta; hybrid mein do identical C–O bonds hain.NCERT Fig. 4.3 · p. 109

Formate ion: C, H aur do O atoms se bonded hai. Step 1, structure I: upar wale O tak double bond, neeche wale O tak single bond, jo minus charge aur teen lone pairs rakhta hai. Step 2, structure II: same nuclei, lekin double bond aur charge ki jagah badli hui. Step 3, hybrid: har C–O ek solid aur ek dashed line se bana hai, 125 pm, aur har O par aadha negative charge.

Definition

Resonance hybrid

Kisi species ka woh ek asli structure jise describe karne ke liye do ya zyada canonical Lewis structures chahiye. Yeh kisi bhi canonical structure se kam energy wala hota hai.

Resonance ke do measurable effects hain. Asli species kisi bhi single canonical structure se kam energy wali, aur isliye zyada stable, hoti hai. Aur jo properties ek structure mein bonds ke beech alag hoti, jaise length, woh identical nikalti hain. Ek mixed colour ke baare mein socho: green paint aadhe time blue aur baaki time yellow nahi hota, aur woh kabhi dono ka ek-ek dibba bhi nahi tha. Resonance hybrid ke saath bhi yahi baat hai:

Iska hamesha ek hi structure hota hai; kuch bhi aage-peeche flip nahi hota.
Canonical structures sirf drawings hain jinki zaroorat isliye hai kyunki Lewis structure partial bond nahi dikha sakta; inmein se koi bhi asli molecule nahi hai.
Resonance tautomerism nahi hai. Keto aur enol tautomers sach mein ek equilibrium mein ek doosre mein badalte hain kyunki ek hydrogen atom move karta hai; canonical structures mein fark sirf is baat ka hai ki electrons kahan banaye gaye hain.

Example 4.3NCERT Problem 4.3 · p. 110

Nitrate ion mein resonance

Nitrate ion, $N O_{3}^{-}$ , ke teenon N–O bonds barabar lambe hain. Ise resonance se samjhao.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 4.4NCERT Problem 4.4 · p. 110

Benzene mein bond lengths

Benzene, $C_{6} H_{6}$ , mein chhe carbon atoms ek flat ring banate hain aur har carbon–carbon bond $139 p m$ lamba hai. Ek typical C–C bond $154 p m$ ka hota hai aur C=C bond $134 p m$ ka. Yeh data benzene ke structure ke baare mein kya batata hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

4.3.6 Bonds ki Polarity

Pure ionic aur pure covalent bonds limiting cases hain; asli bonds kahin beech mein hote hain. Agar dono atoms same hon, jaise $H_{2}$ , $N_{2}$ ya $C l_{2}$ mein, to rassakashi mein koi nucleus nahi jeetta: shared pair beech mein rehta hai aur bond non-polar hota hai. Agar ek atom zyada electronegative ho, to woh pair ko apni taraf kheench leta hai. Hydrogen chloride mein density chlorine ki taraf shift hoti hai, isliye chlorine thoda negative ( $δ -$ ) aur hydrogen thoda positive ( $δ +$ ) hota hai: yeh polar covalent bond hai.

Polar bond charge ko alag karta hai, isliye uska ek dipole moment (द्विध्रुव आघूर्ण) hota hai, jise $μ$ likhte hain. Yeh alag hue charge aur positive aur negative charge ke centres ke beech ki doori ka product hai.

Dipole moment

μ = Q \times r

(4.3)

jahan

$Q$: har centre par charge ka magnitude
$r$: centres ke beech ki doori

Dipole moments aam taur par debye (D) mein diye jaate hain: $1 D = 3.336 \times 10^{- 30} C m$ . Dipole moment ek vector hai. Physicists ise negative centre se positive centre ki taraf banate hain. Chemists ek crossed arrow banate hain, jiska cross positive end par aur head negative end par hota hai, jo dikhata hai ki electron density kidhar shift hui hai. Dono conventions ulti directions mein point karti hain.

Kai polar bonds wale molecule mein har bond ka apna bond dipole hota hai, aur molecule ka dipole moment inka vector sum hota hai. Isliye shape bhi utna hi matter karta hai jitni polarity (Fig. 4.7). Water mein do O–H bond dipoles 104.5° par milte hain; woh cancel nahi hote, aur unka resultant water ko 1.85 D ka dipole moment deta hai. Carbon dioxide mein do C=O dipoles bilkul ulti directions mein point karte hain aur cancel ho jaate hain, isliye har bond polar hone ke bawajood $C O_{2}$ ka dipole moment zero hai. Boron trifluoride mein ek plane mein 120° par teen B–F dipoles bhi cancel ho jaate hain.

Ek puzzle dekho. Fluorine nitrogen se zyada electronegative hai aur hydrogen kam, isliye N–F bonds N–H bonds ke ulte sense mein polar hain. Phir bhi $N H_{3}$ ka dipole moment 1.47 D hai aur $N F_{3}$ ka sirf 0.23 D. Dono molecules pyramids hain jinmein nitrogen par ek lone pair hai, aur yeh lone pair nitrogen ke doosri taraf negative charge ka ek region hai. Ab vectors jodo. Ammonia mein bonds density ko nitrogen ki taraf kheenchte hain aur lone pair usse aage hai, isliye dono contributions ek hi taraf point karte hain aur jud jaate hain. Nitrogen trifluoride mein bonds density ko neeche fluorines ki taraf kheenchte hain jabki lone pair upar point karta hai, isliye yeh kaafi had tak cancel ho jaate hain.

4/4NH₃ mein lone pair bond dipoles ko support karta hai lekin NF₃ mein unke khilaaf jaata hai.

Fig. 4.7Shape decide karta hai ki bond dipoles judenge ya cancel honge. Arrows chemists ka crossed-arrow convention use karte hain.NCERT p. 111

Crossed arrows bond dipoles dikhate hain. Step 1: H–Cl, H par δ+, Cl par δ− aur electron density Cl ki taraf shift. Step 2: bent water, do O–H dipoles O ki taraf point karte hue aur 1.85 D ka resultant. Step 3: linear O=C=O, barabar aur ulte dipoles ke saath aur net dipole zero. Step 4: NH3 aur NF3 pyramids; NH3 mein bond aur lone-pair arrows ek hi taraf point karte hain (1.47 D), NF3 mein ulti taraf (0.23 D).

Table 4.2Gas phase mein kuch molecules ke dipole moments

Molecule	Shape	μ / D
$H_{2} O$	bent	1.85
$H_{2} S$	bent	0.97
$S O_{2}$	bent	1.63
$C O_{2}$	linear	0
$N H_{3}$	trigonal pyramidal	1.47
$N F_{3}$	trigonal pyramidal	0.23
$B F_{3}$	trigonal planar	0
$C H_{3} C l$	tetrahedral	1.89
$C H_{2} C l_{2}$	tetrahedral	1.60
$C H C l_{3}$	tetrahedral	1.04
$C C l_{4}$	tetrahedral	0

Chloromethanes par dhyaan do. Chaaron lagbhag tetrahedral hain, phir bhi jaise-jaise chlorine atoms hydrogens ki jagah lete hain, dipole moment badalta hai, aur bilkul symmetric $C C l_{4}$ mein yeh zero ho jaata hai.

Example

H–Cl bond kitna ionic hai?

HCl ki bond length $127 p m$ hai aur dipole moment $1.11 D$ . Agar bond poori tarah ionic hota, dono nuclei par $\pm e$ charges ke saath, to uska dipole moment kitna hota? Asli value kitne percent ionic character dikhati hai? ( $e = 1.602 \times 10^{- 19} C$ .)

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Jaise covalent bonds mein kuch ionic character hota hai, waise hi ionic bonds mein kuch covalent character hota hai. Chhota ya zyada charge wala cation paas ke anion ke electron cloud ko apni taraf kheench kar use bigaad deta hai (polarise karta hai). Tab do nuclei ke beech electron density jama ho jaati hai, jaise covalent bond mein. Kasimir Fajans ne bataya ki yeh sabse zyada kab matter karta hai:

Chhota cation, bada anion: zyada covalent. $L i I$ , $L i F$ se zyada covalent hai, kyunki bada iodide ion aasani se distort ho jaata hai.
Cation par zyada charge: zyada covalent. $A l^{3 +}$ wala $A l C l_{3}$ , $N a C l$ se kaafi zyada covalent hai: yeh lagbhag 180 °C par sublime ho jaata hai, jabki NaCl sirf 801 °C par pighalta hai.
Same size aur charge: d-electron shell, $(n - 1) d^{n} n s^{0}$ , wala cation noble-gas shell, $n s^{2} n p^{6}$ , wale cation se zyada polarise karta hai. $H g^{2 +}$ aur $C a^{2 +}$ lagbhag same size ke hain, phir bhi $H g C l_{2}$ 276 °C par pighalta hai aur $C a C l_{2}$ 772 °C par.

Cation ki polarising power, anion ki polarisability aur usse hone wala distortion milkar decide karte hain ki ionic bond mein kitna covalent character hai.

4.4 Valence Shell Electron Pair Repulsion (VSEPR) Theory

Lewis structures shape ke baare mein kuch nahi batate, jabki shape hi molecule ki polarity, uske react karne ka tarika aur yahan tak ki uski smell bhi decide karta hai. Valence shell electron pair repulsion (VSEPR) theory sirf electron pairs ki ginti se shapes predict kar deti hai. Nevil Sidgwick aur Herbert Powell ne ise 1940 mein diya; Ronald Nyholm aur Ronald Gillespie ne 1957 mein ise behtar banaya. Yeh kuch assumptions par tiki hai:

Sirf central atom ke valence shell ke electron pairs matter karte hain, chahe bonding pairs hon ya lone pairs.
Negative charge ke clouds hone ki wajah se pairs ek doosre ko door dhakelte hain aur wahan tik jaate hain jahan dhakka sabse kam ho: ek doosre se jitna ho sake utna door, maano ek sphere ki surface par phaile hon.
Double ya triple bond ke saare electrons unhi do atoms ke beech hote hain, isliye multiple bond ek baar gina jaata hai, ek single 'super pair' ki tarah.
Resonance structures wali species ke liye koi bhi ek structure same prediction deta hai.
Saare pairs barabar zor se nahi dhakelte: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair.

Yeh aakhri ranking Nyholm aur Gillespie ka contribution thi. Bonding pair do nuclei ke beech pakda rehta hai aur bond ke saath khincha hota hai; lone pair sirf ek nucleus se pakda rehta hai, isliye woh central atom ke paas phail jaata hai aur apne padosiyon ko daba deta hai. Isliye lone pairs wale shapes distorted hote hain, aur unke bond angles ideal values se kuch degrees chhote hote hain.

Hum kisi species ko ${A B}_{x} E_{y}$ likhte hain: A central atom hai, B bonded atom aur E, A par ek lone pair. Total $x + y$ electron pairs ka arrangement tay karta hai (Fig. 4.8); shape sirf yeh batata hai ki atoms kahan hain. Pehle hum bina lone pairs wale central atoms dekhenge, phir lone pairs wale.

3/5Tetrahedral: 109.5°.

Fig. 4.8Electron pairs jitna ho sake utna door phailte hain: yeh arrangements sabse zyada possible doori dete hain.NCERT Fig. 4.6 · p. 113

Ek central atom A, outer atoms ke saath; tumhari taraf aane wale bonds wedges se aur door jaane wale bonds dashes se bane hain. Step 1: do pairs, linear, 180°. Step 2: teen, trigonal planar, 120°. Step 3: chaar, tetrahedral, 109.5°. Step 4: paanch, trigonal bipyramidal, axis tak 90° aur equator mein 120°. Step 5: chhe, octahedral, 90°.

Koi lone pair nahi ( ${A B}_{x}$ ). Shape wahi hai jo arrangement hai: $H g C l_{2}$ linear hai, $B C l_{3}$ trigonal planar, $C C l_{4}$ tetrahedral, $P F_{5}$ trigonal bipyramidal aur $S F_{6}$ octahedral. Multiple bonds se kuch nahi badalta: do C=O 'super pairs' wala $C O_{2}$ linear hai, aur $S O_{3}$ trigonal planar hai.

Ek ya zyada lone pairs ( ${A B}_{x} E_{y}$ ). Pairs ka arrangement same rehta hai, lekin shape sirf atoms ko naam deta hai. Methane, ammonia aur water, teenon mein chaar pairs tetrahedral tarike se arranged hain (Fig. 4.9). Ammonia mein ek lone pair hai, isliye uska shape trigonal pyramid hai, aur zyada strong lone pair–bond pair repulsion H–N–H angle ko 109.5° se 107° tak kar deta hai. Water mein do lone pairs hain, isliye woh bent hai, aur usse bhi strong lone pair–lone pair repulsion H–O–H ko 104.5° tak sikod deta hai.

2/3Ek lone pair: trigonal pyramid, 107°.

Fig. 4.9Har lone pair bond pair se zyada jagah leta hai aur bond angles ko daba deta hai.NCERT Table 4.8 · p. 116

Ek tetrahedral arrangement teen baar bana hai. Step 1: methane, chaar C–H bonds, 109.5°. Step 2: ammonia, sabse upar ki position par lone-pair lobe hai, shape pyramidal, angle 107°. Step 3: water, do lone-pair lobes aur do O–H bonds, shape bent, angle 104.5°.

Trigonal bipyramid mein paanchon positions ek jaisi nahi hoti. Teen equatorial positions 120° ke faasle par hain; har axial position teenon equatorial positions ke saath 90° banati hai. Axial position par rakha lone pair teen bond pairs se 90° par milega; equatorial position par sirf do se (Fig. 4.10). Kyunki 90° wale repulsions hi mehenge padte hain, lone pairs equatorial jagah lete hain. Isliye $S e F_{4}$ ( $A B_{4} E$ ) see-saw hai, $B r F_{3}$ ( $A B_{3} E_{2}$ ) T-shaped hai, aur teen equatorial lone pairs ke saath $X e F_{2}$ ( $A B_{2} E_{3}$ ) linear hai. Octahedron mein do lone pairs ek doosre ke aamne-saamne baithte hain, jisse $A B_{4} E_{2}$ species square planar ban jaati hain.

Fig. 4.10Equatorial lone pair ke 90° wale padosi kam hote hain, isliye AB₄E molecules see-saw shape ke hote hain.NCERT Table 4.8 · p. 116

Chaar atoms aur ek lone-pair lobe wale do trigonal bipyramids. Left: lone pair axial hai aur dashed lines use 90° par teen bond pairs se jodti hain, label less stable. Right: lone pair equatorial hai aur 90° ke sirf do dashed contacts hain, label more stable.

Table 4.3VSEPR arrangements aur shapes

Type	Bond + lone pairs	Arrangement	Shape	Examples
$A B_{2}$	2 + 0	linear	linear	$C O_{2}$ , $H g C l_{2}$
$A B_{3}$	3 + 0	trigonal planar	trigonal planar	$S O_{3}$ , $B C l_{3}$
$A B_{2} E$	2 + 1	trigonal planar	bent	$S n C l_{2}$ (gas), $N O_{2}^{-}$
$A B_{4}$	4 + 0	tetrahedral	tetrahedral	$C C l_{4}$ , $N H_{4}^{+}$
$A B_{3} E$	3 + 1	tetrahedral	trigonal pyramidal	$P C l_{3}$ , $H_{3} O^{+}$
$A B_{2} E_{2}$	2 + 2	tetrahedral	bent	$O F_{2}$ , $H_{2} S$
$A B_{5}$	5 + 0	trigonal bipyramidal	trigonal bipyramidal	$P F_{5}$ , $A s F_{5}$
$A B_{4} E$	4 + 1	trigonal bipyramidal	see-saw	$S e F_{4}$ , $T e C l_{4}$
$A B_{3} E_{2}$	3 + 2	trigonal bipyramidal	T-shaped	$B r F_{3}$
$A B_{2} E_{3}$	2 + 3	trigonal bipyramidal	linear	$X e F_{2}$ , $I_{3}^{-}$
$A B_{6}$	6 + 0	octahedral	octahedral	$S F_{6}$ , $P F_{6}^{-}$
$A B_{5} E$	5 + 1	octahedral	square pyramidal	$I F_{5}$
$A B_{4} E_{2}$	4 + 2	octahedral	square planar	$X e F_{4}$ , $I C l_{4}^{-}$

Practice mein VSEPR zyada tar p-block molecules ke shapes sahi batata hai, close cases mein bhi. Yaad rakho ki yeh ek working rule hai, derivation nahi: electron-pair repulsion shape ko itni achchhi tarah kyun control karta hai, is par aaj bhi bahas hoti hai. Har combination Fig. 4.11 mein explore karo.

Fig. 4.11Bond pairs aur lone pairs ke numbers chuno, phir model ko ghumane ke liye drag karo.

Example

Step by step shapes predict karna

$C l O_{3}^{-}$ , $I C l_{2}^{-}$ , $I F_{4}^{+}$ aur $B r F_{4}^{-}$ ke shapes predict karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Method: central atom ke valence electrons se shuru karo, har negative charge par ek jodo ya har positive charge par ek ghatao, aur bonding mein use hone wale electrons ghatao: har monovalent atom (H, F, Cl, Br, I) ke liye 1 aur har oxygen ke liye 2, kyunki oxygen ko apna octet poora karne ke liye do electrons chahiye. Jo bache uska aadha lone pairs deta hai; usmein bonded atoms jodo to total milta hai.

Try karke dekho

Electron pairs ke balloon models

Tumhe chahiye

Ek hi size ke 6 gol balloons
1 bada balloon
ek marker pen

Pehle guess karo

Tum chaar barabar balloons ko unki gaanthon par ek saath marodte ho. Woh kaise set honge?

ek flat square mein, 90° ke faasle partetrahedron ke corners ki taraf point karte huesab ek taraf ikatthe

Chhe gol balloons ko same size tak phulao aur baandh do.
Do balloons ko unki gaanthon par ek saath marodo aur chhod do. Unke beech ka angle note karo.
Centre mein teesra balloon marod kar jodo, phir chautha, paanchva aur chhatha. Har baar shape ka sketch banao.
Chaar balloons ka aisa guchha banao jismein ek balloon saaf taur par bada ho, jo lone pair ko dikhaye, aur baaki teen ko dekho.

Quick check

4 mein se 1

Iodine trifluoride,

I F_{3}

, mein iodine par teen bond pairs aur do lone pairs hain. Iska shape kya hai?

trigonal planartrigonal pyramidalT-shapedsee-saw

Answer dekho

Answer (C) T-shaped

Paanch pairs trigonal bipyramid banate hain; dono lone pairs equatorial positions lete hain, jisse teen F atoms T shape mein bach jaate hain. Trigonal planar kehna lone pairs ko ignore karna hoga.

4.5 Valence Bond Theory

Lewis structures aur VSEPR bookkeeping aur geometry hain; dono mein se koi yeh nahi samjhata ki shared pair do atoms ko saath kyun rakhta hai. Yeh kuch zaroori differences bhi miss kar dete hain. Fluorine aur chlorine molecules dono mein ek-ek shared pair hai, phir bhi F–F bond ( $142 p m$ , $158 k J m o l^{- 1}$ ), Cl–Cl bond ( $199 p m$ , $243 k J m o l^{- 1}$ ) se kamzor hai, jabki woh chhota hai. Quantum mechanics par bani do theories aise sawaalon ko handle karti hain: valence bond (VB) theory aur molecular orbital (MO) theory.

VB theory 1927 se hai, jab Walter Heitler aur Fritz London ne hydrogen molecule par quantum mechanics apply kiya; phir Linus Pauling aur doosron ne ise aage badhaya. Iska poora form mathematical hai, isliye yahan hum sirf ideas tak rahenge: Unit 2 se atomic orbitals aur configurations, aur saath mein overlap aur hybridisation, jo aage introduce honge.

Socho do hydrogen atoms, A aur B, dheere-dheere paas aa rahe hain (Fig. 4.12). Bahut door hone par kuch nahi hota. Jaise-jaise woh paas aate hain, har nucleus apne electron ke saath-saath doosre atom ke electron ko bhi kheenchne lagta hai, jabki dono electrons ek doosre ko repel karte hain aur dono nuclei bhi. Kul milakar extra attraction zyada hota hai, isliye pair ki energy gir jaati hai. Yeh girna $74 p m$ par rukta hai, jahan kheench aur dhakka balance ho jaate hain aur energy sabse kam hoti hai: yahi separation H–H bond length hai. Nuclei ko aur paas dabao to unka repulsion haavi ho jaata hai, aur energy tezi se upar chali jaati hai.

Is energy well ke bottom par baitha molecule do free atoms se kam energy wala hota hai, aur yahan 'stable' ka matlab yahi hai. Well ki gehraai bond enthalpy hai: ek mole $H_{2}$ molecules ko alag karne ke liye tumhe $436 k J$ dene padte hain, isliye $H_{2} (g) \overset{}{\to} 2 H (g)$ ke liye $Δ H = + 436 k J m o l^{- 1}$ hai.

3/4 $74 p m$ par minimum: alag atoms se $436 k J m o l^{- 1}$ neeche.

Fig. 4.12Bond length wahan hai jahan energy sabse kam hai; well ki gehraai bond enthalpy hai.NCERT Fig. 4.8 · p. 118

Do H nuclei ke beech doori ke saath potential energy: kam doori par steep, 74 pm aur −436 kJ/mol par minimum, phir zero ki taraf badhti hui. Ek marker aur atoms ka ek pair har stage dikhate hain. Step 1: bahut door, energy zero. Step 2: paas, attraction arrows, energy girti hui. Step 3: minimum par, overlap karte atoms. Step 4: bahut paas, repulsion arrows, energy tezi se badhti hui.

4.5.1 Orbital Overlap ka Concept

Well ke bottom par dono 1s orbitals kuch same space gherte hain: woh overlap karte hain. Shared region mein dono electrons, jinke spins opposite hone chahiye, pair ban jaate hain. Covalent bond ki VB theory wali picture yahi hai: do atoms ke half-filled valence orbitals overlap karte hain, aur unke electrons opposite spins ke saath pair bante hain. Zyada overlap ka matlab zyada strong bond.

4.5.2 Bonds ki Directional Properties

Spherical 1s orbitals kisi bhi direction se barabar achchhe se overlap karte hain, isliye $H_{2}$ ke liye direction ka sawaal hi nahi uthta. Bade molecules ke liye uthta hai: methane ke chaar bonds bilkul 109.5° par hain, ammonia ek neecha pyramid hai aur water bent hai. Ek poori theory ko yeh angles dene chahiye, aur VB theory overlap ke saath hybridisation jodkar yeh karti hai (Section 4.6).

4.5.3 Atomic Orbitals ka Overlapping

Orbital ke har lobe ka ek sign hota hai, + ya −: wahan wave function ka sign, kuch-kuch paani ki lehar ke crest aur trough jaisa. Iska electric charge se koi lena-dena nahi hai. Jab same sign ke do lobes milte hain, to waves ek doosre ko mazboot karti hain: overlap positive hota hai aur bond ban sakta hai. Opposite sign ke lobes cancel karte hain, jisse negative overlap milta hai. Aur agar ek orbital doosre ke + aur − hisson se barabar milta hai, jaise s orbital p orbital ke paas side se aata hai, to effects bilkul cancel ho jaate hain aur overlap zero hota hai (Fig. 4.13). Bonds ke liye positive overlap chahiye.

Ab ise methane par test karo. Maan lo carbon pehle ek 2s electron ko khaali 2p orbital mein bhejta hai, jisse $2 s^{1} 2 p_{x}^{1} 2 p_{y}^{1} 2 p_{z}^{1}$ aur chaar unpaired electrons milte hain; yeh kharcha do ki jagah chaar bonds banakar wasool ho jaata hai. Agar yeh chaar pure orbitals hydrogen ke 1s orbitals se overlap karte, to teen C–H bonds x, y aur z axes ke saath ek doosre se 90° par hote, aur chautha, spherical 2s orbital se, kisi khaas direction mein point nahi karta. Yeh methane nahi hai, jiske chaaron bonds identical hain aur 109.5° ke faasle par hain. Ammonia aur water ka haal bhi yahi hai: pure p orbitals 90° predict karte hain, 107° aur 104.5° nahi.

4/5Axis ke saath overlap: σ bond.

Fig. 4.13Signs wave function ke phases hain, charges nahi. Matching phases bonding overlap dete hain.NCERT Fig. 4.9 · p. 119

Ek horizontal axis par + ya − mark kiye orbital lobes. Step 1: same sign ke do s orbitals overlap karte hain, positive overlap. Step 2: opposite sign ke lobes end to end milte hain, negative overlap. Step 3: ek s orbital, axis ke across pade p orbital se milta hai, zero overlap. Step 4: do p orbitals same sign ke saath head-on overlap karte hain, σ bond. Step 5: do parallel p orbitals axis ke upar aur neeche overlap karte hain, π bond.

4.5.4 Overlapping ke Types aur Covalent Bonds ka Nature

Overlap do geometries mein hota hai. Sigma (σ) bond mein orbitals internuclear axis ke saath head-on milte hain, ise axial overlap kehte hain. Yeh do s orbitals ke beech ho sakta hai (jaise $H_{2}$ mein), ek s aur ek p orbital ke beech (jaise $H F$ mein), ya ek doosre ki taraf point karte do p orbitals ke beech (jaise $F_{2}$ mein). σ bond ki electron density axis ke chaaron taraf symmetrical hoti hai.

Pi (π) bond ke liye do p orbitals ko agal-bagal khada hona chahiye, ek doosre ke parallel aur axis se right angle par. Woh axis ke upar aur neeche edge to edge overlap karte hain, isliye bond ki electron density nuclei ko jodne wali line ke dono taraf do regions mein hoti hai, aur line par bilkul nahi hoti.

4.5.5 Sigma aur pi Bonds ki Strength

Numbers compare karo: C–C single bond todne ke liye $348 k J m o l^{- 1}$ chahiye, C=C double bond ke liye $614 k J m o l^{- 1}$ , jo double se kaafi kam hai. Doosra bond π bond hai, aur sideways overlap head-on overlap se kam hota hai, isliye π bond σ bond se kamzor hota hai. Do atoms ke beech pehla bond hamesha σ hota hai, aur baaki extra bonds π: double bond σ + π hai, triple bond σ + 2π.

4.6 Hybridisation

VB theory ko methane wali problem se bachane ke liye (Section 4.5.3), Linus Pauling ne atom ko bond banane se pehle apne valence orbitals mix karne diye. Hybridisation (संकरण) mein ek atom ke milti-julti energy wale orbitals milkar utne hi naye orbitals banate hain, jo shape aur energy mein ek jaise hote hain, aur phir bonds in hybrid orbitals se bante hain. Jaise, carbon ka ek 2s aur teen 2p orbitals chaar $s p^{3}$ hybrids dete hain.

Definition

Hybridisation

Ek atom par milti-julti energy wale atomic orbitals ka mixing, jisse utne hi equivalent hybrid orbitals bante hain, jo phir bonding mein use hote hain.

Hybrid orbitals ke baare mein chaar facts yaad rakhne layak hain:

Jitne atomic orbitals daaloge, utne hi hybrids niklenge.
Ek set ke andar hybrids identical hote hain, sirf har ek ki direction alag hoti hai.
Har hybrid ka ek mota lobe hota hai jo pure orbital se aage tak jaata hai, isliye woh behtar overlap karta hai aur zyada strong bond banata hai.
Ek set aisi directions mein point karta hai jo electron pairs ko jitna ho sake utna door rakhein, isliye hybridisation ka naam lena hi geometry ka naam lena hai.

Aur chaar conditions:

Sirf valence shell ke orbitals hi hissa lete hain.
Unki energies paas-paas honi chahiye: 1s orbital, jo kaafi neeche hai, kabhi 2s aur 2p ke saath mix nahi hota.
Kisi electron ko pehle promote karna zaroori nahi; promotion sirf bookkeeping hai.
Lone pairs wale orbitals bhi shaamil ho sakte hain, sirf half-filled wale nahi.

4.6.1 Hybridisation ke Types

sp hybridisation. Ek s aur ek p orbital do sp hybrids dete hain, har ek character mein aadha s aur aadha p. Dono bilkul ulti directions mein, 180° ke faasle par, point karte hain, har ek ka aage bada lobe aur peeche chhota lobe hota hai; is arrangement ko diagonal hybridisation bhi kehte hain. Isliye do partners wala sp atom linear molecule deta hai. Gaseous beryllium fluoride yeh dikhata hai. Beryllium ( $1 s^{2} 2 s^{2}$ ) mein tab tak koi unpaired electron nahi hota jab tak ek 2s electron 2p mein nahi jaata; phir 2s aur woh 2p orbital milkar do sp hybrids banate hain, jinmein se har ek fluorine ke ek singly occupied 2p orbital se end-on overlap karta hai, isliye F–Be–F ek seedhi line hai.

sp² hybridisation. Ek s aur do p orbitals teen hybrids banate hain, har ek mein ek-tihaai s, jo ek plane mein ek doosre se 120° par phaile hote hain. Boron trifluoride mein boron ( $2 s^{2} 2 p^{1}$ ) apne teen hybrids teen B–F σ bonds ke liye use karta hai, isliye molecule 120° angles wala ek flat triangle hai. Boron ka ek 2p orbital mixing se bahar reh jaata hai; woh khaali rehta hai aur plane ke upar aur neeche nikla rehta hai.

sp³ hybridisation. Teenon p orbitals s orbital ke saath mil jaate hain, jisse chaar hybrids bante hain jo ek-chauthai s aur teen-chauthai p hote hain. Yeh tetrahedron ke corners ki taraf, 109.5° ke faasle par, point karte hain. Methane mein har hybrid ek hydrogen atom ke 1s orbital se overlap karta hai, jisse chaar identical C–H σ bonds bante hain (Fig. 4.14).

Fig. 4.14Promotion se chaar half-filled orbitals milte hain; mixing unhe identical bana deti hai.NCERT Fig. 4.12 · p. 122

Carbon ke orbital boxes ki teen rows. Ground state: 2s paired, teen 2p boxes mein se do mein ek-ek electron. Excited state: 2s mein aur har 2p box mein ek-ek electron. sp3: chaar barabar boxes, har ek mein ek electron, chaar C–H σ bonds banane ke liye taiyaar.

3/3Chaar hybrids, 109.5°: tetrahedral (CH₄).

Fig. 4.15Zyada p character ka matlab zyada hybrids aur unke beech chhote angles.NCERT Fig. 4.10 · p. 121

Hybrid orbitals ek nucleus ke chaaron taraf bade lobes aur chhote back lobes ke roop mein bane hain. Step 1: ek line mein do sp lobes, 180°, 50 % s character. Step 2: ek plane mein teen sp2 lobes, 120°, 33 % s. Step 3: tetrahedron ke corners ki taraf chaar sp3 lobes, 109.5°, ek peeche dashed bana hai, 25 % s.

Jab kuch hybrids lone pairs rakhte hain, to chaar ka yahi set ammonia aur water ko bhi samjha deta hai. Nitrogen ke paas paanch valence electrons hain: do ek sp³ hybrid ko lone pair ke roop mein bharte hain aur teen baaki hybrids mein akele baithte hain, teen hydrogens se bond banane ko taiyaar. Oxygen ke paas chhe hain: do lone-pair hybrids aur do singly occupied, isliye woh do hydrogens se bond banata hai. Phir lone pairs bonds ko bilkul waise hi dabate hain jaise VSEPR kehta hai, pyramidal $N H_{3}$ mein 109.5° se 107° tak aur bent (V-shaped) $H_{2} O$ mein 104.5° tak.

4.6.2 sp³, sp² aur sp Hybridisation ke Aur Examples

Ethane, $C_{2} H_{6}$ : dono carbons sp³. Carbons ek sp³–sp³ σ bond se judte hain, jo $154 p m$ lamba hai, aur har carbon apne baaki teen hybrids hydrogen se sp³–1s σ bonds ke liye use karta hai, jo $109 p m$ lambe hain.

Ethene, $C_{2} H_{4}$ : dono carbons sp². Har carbon ke teen sp² hybrids flat σ framework banate hain: ek C–C bond aur do C–H bonds ( $108 p m$ ). Har carbon par ek p orbital bach jaata hai jo molecular plane se seedha upar khada hai; dono sideways overlap karke ek π bond banate hain, jiski density plane ke upar aur neeche hoti hai. Isliye C=C double bond ( $134 p m$ ) ek σ bond plus ek π bond hai. Chhe ke chhe atoms ek hi plane mein hain, H–C–H lagbhag 117° aur H–C–C lagbhag 121°.

Ethyne, $C_{2} H_{2}$ : dono carbons sp. Har carbon ke do sp hybrids ulti directions mein point karte hain, ek doosre carbon ki taraf aur ek hydrogen ki taraf, isliye chaaron atoms ek line mein hote hain. Har carbon ke paas do p orbitals bachte hain jo ek doosre se aur axis se right angle par hain. Yeh apne partners ke saath sideways pair bankar do π bonds banate hain jo milkar axis ko chaaron taraf se gher lete hain. C≡C bond, σ + 2π, $120 p m$ lamba hai (Fig. 4.16).

2/4p orbitals sideways overlap karte hain: plane ke upar aur neeche ek π bond.

Fig. 4.16Hybrids σ framework banate hain; bache hue p orbitals sideways overlap karke π bonds banate hain.NCERT Fig. 4.15 · p. 123

Step 1: perspective mein ethene, sp2 hybrids se bane σ bonds aur har carbon par ek seedha khada p orbital. Step 2: woh p orbitals C–C axis ke upar aur neeche π clouds mein mil jaate hain. Step 3: linear ethyne, sp hybrids se bane σ bonds aur har carbon par right angles par do p orbitals. Step 4: yeh do pairs mein overlap karte hain, jisse axis ke chaaron taraf do π bonds bante hain.

Example

Acrylonitrile mein hybridisation aur bonds ki ginti

Acrylonitrile, $C H_{2} = C H - C \equiv N$ , acrylic fibres banane mein use hota hai. Har carbon ka hybridisation batao (left se unhe C1, C2, C3 kaho) aur σ aur π bonds gino.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

4.6.3 d Orbitals wale Elements ka Hybridisation

Period 2 ke baad ke atoms ke paas d orbitals pahunch mein hote hain. Period 3 mein khaali 3d level energy mein ek taraf 3s aur 3p ke kareeb hai aur doosri taraf 4s aur 4p ke. Isse do tarah ke d wale hybrids ban sakte hain: $n s$ , $n p$ aur $n d$ orbitals se bane, aur $(n - 1) d$ , $n s$ aur $n p$ se bane. 3p, 3d aur 4s ka mixture nahi ban sakta, kyunki 3p–4s energy gap bahut bada hai. Table 4.4 mein common types diye hain.

Table 4.4Hybridisation ki common schemes

Hybrid	Mix hue orbitals	Geometry	Angles	Examples
$s p$	s + p	linear	180°	$B e F_{2}$ (gas), $C_{2} H_{2}$
$s p^{2}$	s + 2p	trigonal planar	120°	$B F_{3}$ , $C_{2} H_{4}$
$s p^{3}$	s + 3p	tetrahedral	109.5°	$C H_{4}$ , $N H_{4}^{+}$
$d s p^{2}$	(n−1)d + s + 2p	square planar	90°	$[P t (N H_{3})_{4}]^{2 +}$
$s p^{3} d$	s + 3p + d	trigonal bipyramidal	90°, 120°	$P C l_{5}$ , $A s F_{5}$
$s p^{3} d^{2}$	s + 3p + 2d	octahedral	90°	$S F_{6}$ , $P F_{6}^{-}$
$d^{2} s p^{3}$	2(n−1)d + s + 3p	octahedral	90°	$[F e (C N)_{6}]^{3 -}$

$P C l_{5}$ (sp³d). Phosphorus ( $[N e] 3 s^{2} 3 p^{3}$ ) ke paas sirf teen unpaired electrons hain, lekin ek 3s electron ko khaali 3d orbital mein bhejne se paanch ho jaate hain. 3s, teen 3p aur ek 3d orbital milkar paanch sp³d hybrids banate hain jo trigonal bipyramid ke corners ki taraf point karte hain, aur har ek chlorine atom ke ek half-filled 3p orbital se overlap karta hai. Methane ke bonds ke ulat, yeh paanch bonds do tarah ke hain (Fig. 4.17). Teen equatorial bonds phosphorus ke chaaron taraf ek flat triangle banate hain, 120° ke faasle par; do axial bonds seedhe upar aur neeche point karte hain, us triangle se 90° par. Axial bond pair ke 90° par teen padosi hote hain, equatorial ke sirf do, isliye axial bonds thoda bahar dhakel diye jaate hain, lagbhag $212 p m$ tak, jabki equatorial $202 p m$ hain. Lambe hone ki wajah se woh kamzor hain, aur garam karne par $P C l_{5}$ aasani se chlorine chhod deta hai.Modern calculations dikhati hain ki aise bonds mein d orbitals ka role is model se kaafi kam hai, lekin sp³d aur sp³d² labels hybrids aur VSEPR shapes ke beech ek convenient link bane hue hain.

Fig. 4.17Axial bonds 90° par teen padosiyon se milte hain, equatorial bonds sirf do se, isliye axial bonds lambe hote hain.NCERT Fig. 4.17 · p. 125

Trigonal bipyramid ke centre mein phosphorus. Teen equatorial Cl atoms ek shaded plane mein ek doosre se 120° par hain, ek bond par label 202 pm. Do axial Cl atoms plane se 90° par upar aur neeche hain, upar wale bond par label 212 pm.

$S F_{6}$ (sp³d²). Sulfur ( $[N e] 3 s^{2} 3 p^{4}$ ) ke paas do unpaired electrons hain; ek 3s aur ek 3p electron ko 3d orbitals mein promote karne se chhe ho jaate hain. Chhe sp³d² hybrids regular octahedron ke corners ki taraf point karte hain, aur har ek fluorine ke 2p orbital ke saath σ bond banata hai. Chhe ke chhe S–F bonds equivalent hain, padosiyon ke beech 90° ke saath.

Quick check

3 mein se 1

C O_{2}

mein carbon ka hybridisation kya hai?

s p^{2}

s p^{3}

s p

s p^{3} d

Answer dekho

Answer (C) $s p$

Carbon do σ bonds aur do π bonds (do C=O) banata hai, isliye woh do sp hybrids aur do unhybridised p orbitals use karta hai; molecule linear hai. Chaar bonds gin kar

s p^{3}

chunna yeh bhool jaana hai ki π bonds unhybridised p orbitals use karte hain.

4.7 Molecular Orbital Theory

VB theory electrons ko do atoms ke beech khaas bonds se baandh deti hai. Molecular orbital (MO) theory, jise Friedrich Hund aur Robert S. Mulliken ne 1932 mein develop kiya, unhe poore molecule ka maanti hai. Iske main ideas yeh hain:

Energy mein milte-julte aur symmetry mein match karne wale atomic orbitals milkar molecular orbitals banate hain, jo do ya zyada nuclei par phaile hote hain (polycentric, monocentric atomic orbitals ke ulat).
Orbitals ki ginti bani rehti hai: do atomic orbitals hamesha do molecular orbitals dete hain, ek bonding, jo kam energy wala aur zyada stable hai, aur ek antibonding, jo zyada energy wala hai.
Har molecular orbital batata hai ki molecule ke nuclei ke aas-paas electron milne ki sambhavna kahan hai.
Molecule ke electrons in orbitals ko unhi rules se bharte hain jo atoms ke liye use hote hain: pehle sabse kam energy (aufbau), har orbital mein zyada se zyada do, opposite spins ke saath (Pauli), aur pairing se pehle barabar energy wale orbitals mein akele (Hund).

4.7.1 Molecular Orbitals ka Banna: Linear Combination of Atomic Orbitals (LCAO)

Ek exact molecular orbital molecule ke liye Schrödinger equation solve karke milta, lekin ek se zyada electron hone par is equation ka koi exact solution nahi hota. Isliye chemists atomic orbitals se approximate molecular orbitals banate hain, yaani wave functions $ψ$ jinki values electron wave ka amplitude deti hain. Linear combination of atomic orbitals (LCAO) method mein molecular orbital (आण्विक कक्षक) atomic wave functions ka sum ya difference hota hai. Do hydrogen atoms A aur B ke liye, jinke 1s wave functions $ψ_{A}$ aur $ψ_{B}$ hain:

Do 1s orbitals ke liye LCAO

\begin{aligned} σ & = ψ_{A} + ψ_{B} \\ σ^{*} & = ψ_{A} - ψ_{B} \end{aligned}

(4.4)

jahan

$σ$: bonding molecular orbital (addition)
$σ^{*}$: antibonding molecular orbital (subtraction)

Ek combination bonding aur doosra antibonding kyun hai? Waves ko jodne se nuclei ke beech ke region mein woh ek doosre ko mazboot karti hain (Fig. 4.18). Wahan electron density jama hoti hai aur dono nuclei ko beech ki taraf kheenchti hai, isliye σ mein electron ki energy kisi bhi 1s orbital se kam hoti hai. Ghatane par waves beech mein cancel ho jaati hain: antibonding orbital mein nuclei ke beech ek nodal plane hota hai, jahan density zero hai, aur iske electrons zyada tar nuclei ke peeche rehte hain, jisse nuclei ek doosre ko repel karne ke liye reh jaate hain. Isliye $σ^{*}$ mein electron ki energy 1s orbital se zyada hoti hai aur woh bond ko kamzor karta hai. Mote taur par, bonding orbital atomic level se utna hi neeche jaata hai jitna antibonding orbital upar.

4/4 $σ 1 s$ atomic orbitals se neeche hai, $σ^{*} 1 s$ unse upar.

Fig. 4.18Waves jodne se nuclei ke beech density jama hoti hai; ghatane se wahan node ban jaata hai.NCERT Fig. 4.19 · p. 126

Ek axis par do nuclei A aur B. Step 1: alag-alag 1s wave curves, ψA aur ψB. Step 2: unka sum, dono nuclei par ek curve aur unke beech density, σ. Step 3: unka difference, A par positive, B par negative, ek dashed nodal plane par zero ko cross karta hua, σ*. Step 4: energy diagram, jismein σ1s do 1s levels se neeche aur σ*1s upar hai.

Definition

Molecular orbital

Aisa orbital jo molecule ke do ya zyada nuclei par phaila ho aur atomic orbitals ko milakar bana ho. Bonding orbitals parent atomic orbitals se kam energy wale hote hain; antibonding orbitals zyada energy wale.

4.7.2 Atomic Orbitals ke Combination ki Conditions

Atomic orbitals tabhi achchhe se combine hote hain jab teen conditions poori hon:

Milti-julti energies. Do identical atoms ke liye 2s orbital doosre atom ke 2s se combine hota hai, kabhi uske bahut gehre 1s se nahi. (Bahut alag atoms ke beech matching thodi dheeli hoti hai.)
Bond axis (z maana gaya) ke around same symmetry. Ek $2 p_{x}$ orbital doosre atom ke $2 p_{x}$ se combine ho sakta hai, lekin uske $2 p_{y}$ ya $2 p_{z}$ se nahi, jabki teenon ki energy same hai.
Achchha overlap. Orbitals jitna zyada overlap karte hain, nuclei ke beech utni zyada electron density jama hoti hai.

4.7.3 Molecular Orbitals ke Types

Diatomic molecules ke molecular orbitals symmetry ke hisaab se classify hote hain aur σ, π, δ wagairah label kiye jaate hain. σ orbital ko bond axis ke around ghumao to woh har taraf se same dikhta hai. Do 1s orbitals $σ 1 s$ aur $σ^{*} 1 s$ ka pair dete hain; do $2 p_{z}$ orbitals, jo axis ke saath point karte hain, $σ 2 p_{z}$ aur $σ^{*} 2 p_{z}$ dete hain. π orbital is test mein fail hota hai, kyunki uska ek sign ka lobe axis ke upar aur doosre sign ka neeche hota hai. Agal-bagal ke $2 p_{x}$ orbitals $π 2 p_{x}$ aur $π^{*} 2 p_{x}$ dete hain, aur $2 p_{y}$ orbitals iska matching pair. Bonding π orbital mein density axis ke upar aur neeche hoti hai; antibonding $π^{*}$ orbital mein nuclei ke beech ek node bhi hota hai (Fig. 4.19).

Fig. 4.19σ orbitals axis ke chaaron taraf same dikhte hain; π orbitals ke lobes axis ke upar aur neeche hote hain.NCERT Fig. 4.20 · p. 128

Combine hote atomic orbitals ki teen rows, lobes par + aur − mark hain. Row a: do 1s orbitals dono nuclei ke chaaron taraf ek σ1s cloud dete hain aur beech mein node wale lobes ka ek σ*1s pair. Row b: do 2pz orbitals bade central lobe wala σ2pz aur beech mein node wala σ*2pz dete hain. Row c: do agal-bagal ke 2px orbitals axis ke upar aur neeche π2px lobes dete hain aur chaar lobes aur ek node wala π*2px.

4.7.4 Molecular Orbitals ka Energy Level Diagram

Do atoms ke 2s aur 2p orbitals, kul aath atomic orbitals, aath molecular orbitals dete hain. Bonding orbitals $σ 2 s$ , $σ 2 p_{z}$ , $π 2 p_{x}$ aur $π 2 p_{y}$ hain; antibonding orbitals $σ^{*} 2 s$ , $σ^{*} 2 p_{z}$ , $π^{*} 2 p_{x}$ aur $π^{*} 2 p_{y}$ hain. Inka order spectroscopy se tay hota hai. $O_{2}$ aur $F_{2}$ ke liye badhti energy ka order yeh hai:

O₂, F₂

\begin{aligned} σ 1 s < σ^{*} 1 s < σ 2 s \\ < σ^{*} 2 s < σ 2 p_{z} \\ < (π 2 p_{x} = π 2 p_{y}) \\ < (π^{*} 2 p_{x} = π^{*} 2 p_{y}) \\ < σ^{*} 2 p_{z} \end{aligned}

Halke molecules $L i_{2}$ , $B e_{2}$ , $B_{2}$ , $C_{2}$ aur $N_{2}$ ke liye experiment ek alag order dikhata hai:

Li₂ to N₂

\begin{aligned} σ 1 s < σ^{*} 1 s < σ 2 s \\ < σ^{*} 2 s \\ < (π 2 p_{x} = π 2 p_{y}) \\ < σ 2 p_{z} \\ < (π^{*} 2 p_{x} = π^{*} 2 p_{y}) \\ < σ^{*} 2 p_{z} \end{aligned}

Main fark yeh hai ki in halke molecules mein $σ 2 p_{z}$ , $π 2 p$ orbitals ke pair se upar hota hai (Fig. 4.20). Iski wajah, halke atoms mein energy mein paas-paas wale 2s aur 2p orbitals ka mixing, is book ke scope se bahar hai.

1/2 $π 2 p$ , $σ 2 p_{z}$ se neeche (2s–2p mixing ke saath).

Fig. 4.20Dono diagrams ke beech sirf

σ 2 p_{z}

aur

π 2 p

pair ki positions badalti hain.NCERT Fig. 4.21 · p. 129

Ek molecular orbital energy diagram, jismein dono taraf 2s aur 2p atomic levels hain aur beech mein molecular levels, σ2s se lekar σ*2pz tak. Step 1, B2, C2 aur N2 ke liye: colour mein bana σ2pz level π2p levels ke pair se upar hai. Step 2, O2 aur F2 ke liye: σ2pz level π2p pair se neeche aa gaya hai.

4.7.5 Electronic Configuration aur Molecular Behaviour

Molecular orbitals mein electrons ka distribution molecule ka electronic configuration hai. Isse kai properties predict hoti hain.

Stability. Bonding orbitals mein electrons, $N_{b}$ , aur antibonding orbitals mein electrons, $N_{a}$ , gino. Jab $N_{b}$ zyada ho to bonding effect haavi rehta hai aur molecule bana rehta hai; jab $N_{a}$ zyada ho to woh toot jaata hai.

MO theory mein bond order

Bond order = \frac{1}{2} (N_{b} - N_{a})

(4.5)

Bond order. Positive: stable molecule. Zero ya negative: koi stable molecule nahi.
Bond ka type. Bond orders 1, 2 aur 3 Lewis structures ke single, double aur triple bonds se match karte hain; MO theory aadhe values bhi allow karti hai, jaise $O_{2}^{+}$ ke liye 2.5.
Bond length. Bond order jitna zyada, bond utna chhota.
Magnetism. Jis molecule ke har orbital mein do electrons hon woh diamagnetic hai, magnetic field se halka sa bahar dhakela jaata hai. Ek ya zyada singly filled orbitals use paramagnetic bana dete hain, jo field ki taraf kheencha jaata hai.

Quick check

3 mein se 1

z ko molecular axis maano, to atomic orbitals ka kaun sa pair combine hokar molecular orbitals bana sakta hai?

2 p_{z}

aur

2 p_{x}

1 s

aur

2 p_{y}

2 p_{y}

aur

2 p_{z}

2 p_{x}

aur

2 p_{x}

Answer dekho

Answer (D) $2 p_{x}$ aur $2 p_{x}$

2 p_{x}

aur

2 p_{x}

ki axis ke around same symmetry hai aur yeh

π 2 p_{x}

aur

π^{*} 2 p_{x}

dete hain. Baaki pairs ki symmetries alag hain, isliye unka overlap zero hai.

4.8 Kuch Homonuclear Diatomic Molecules mein Bonding

Ab rules apply karo: electrons gino, orbitals ko sahi order mein bharo, phir bond order aur magnetism padh lo. Neeche KK ka matlab bhari hui inner shells hain, $(σ 1 s)^{2} (σ^{*} 1 s)^{2}$ , jo bonding mein kuch nahi jodti. Padhte samay har case ko Fig. 4.21 mein check karo.

InteractiveMolecular orbital filler

H2 se Ne2 tak diatomic species ke liye ek molecular orbital energy diagram, sides par atomic levels aur beech mein molecular levels. Buttons se element aur 2− se 2+ tak charge chuna jaata hai, ya ek-ek electron joda aur hataya jaata hai. Electrons aufbau principle, Pauli principle aur Hund's rule se bharte hain, zaroorat ke hisaab se Li2–N2 ya O2–Ne2 order mein. Readouts configuration, bonding aur antibonding electrons ki ginti, bond order, unpaired electrons aur magnetism batate hain.

Fig. 4.21Ek molecule ya ion chuno aur ek-ek karke electrons jodo, phir dekho bond order aur magnetism kaise badalte hain.

Hydrogen, $H_{2}$ . Do electrons $σ 1 s$ ko bharte hain, jisse $(σ 1 s)^{2}$ aur bond order $\frac{1}{2} (2 - 0) = 1$ milta hai, yaani single bond. Saare electrons paired hain, isliye $H_{2}$ diamagnetic hai. Iski bond length $74 p m$ hai aur bond enthalpy $436 k J m o l^{- 1}$ .

Helium, $H e_{2}$ . Chaar electrons $(σ 1 s)^{2} (σ^{*} 1 s)^{2}$ dete hain. Antibonding pair bonding pair ko cancel kar deta hai: bond order 0, isliye $H e_{2}$ stable molecule ke roop mein exist nahi karta. Beryllium, $B e_{2}$ , bhi isi tarah fail hota hai: $K K (σ 2 s)^{2} (σ^{*} 2 s)^{2}$ , bond order 0.

Lithium, $L i_{2}$ . Chhe electrons: $K K (σ 2 s)^{2}$ . Chaar electrons bonding orbitals mein aur do antibonding orbitals mein hain, isliye bond order $\frac{1}{2} (4 - 2) = 1$ hai. Isliye MO theory ek stable, diamagnetic molecule predict karti hai, aur spectroscopists ko yahi milta hai: garam lithium vapour mein $L i_{2}$ molecules milte hain.

Carbon, $C_{2}$ . Baarah electrons, halke molecules wale order mein bhare gaye:

\begin{aligned} C_{2} : & K K (σ 2 s)^{2} (σ^{*} 2 s)^{2} \\ (π 2 p_{x})^{2} (π 2 p_{y})^{2} \end{aligned}

Bond order $\frac{1}{2} (8 - 4) = 2$ , aur har electron paired, isliye $C_{2}$ diamagnetic hona chahiye; aise molecules high temperature par carbon vapour mein milte hain. Dekho bonding kahan se aati hai: $σ 2 s$ aur $σ^{*} 2 s$ pairs cancel ho jaate hain, aur saari net bonding do π orbitals ke chaar electrons se aati hai. Isliye $C_{2}$ ka double bond do π bonds hai, bina kisi σ bond ke, jo ethene ke σ + π double bond se alag hai. Nitrogen, $N_{2}$ , $σ 2 p_{z}$ mein ek pair aur jodta hai: bond order 3, diamagnetic.

Oxygen, $O_{2}$ . Solah electrons, bhaari molecules wale order mein bhare gaye:

\begin{aligned} O_{2} : & K K (σ 2 s)^{2} (σ^{*} 2 s)^{2} \\ (σ 2 p_{z})^{2} (π 2 p_{x})^{2} \\ (π 2 p_{y})^{2} (π^{*} 2 p_{x})^{1} \\ (π^{*} 2 p_{y})^{1} \end{aligned}

Inner shells ko milakar das electrons bonding orbitals mein aur chhe antibonding orbitals mein hain, isliye bond order $\frac{1}{2} (10 - 6) = 2$ hai, yaani double bond. Hund's rule ke hisaab se aakhri do electrons do $π^{*}$ orbitals mein akele, parallel spins ke saath jaate hain: $O_{2}$ mein do unpaired electrons hain aur yeh paramagnetic hai. Liquid oxygen sach mein ek strong magnet ke poles ke beech chipak jaati hai. Lewis structure O=O, jismein har electron paired hai, ise samjha nahi sakta; MO theory samjha sakti hai. Table 4.5 is method ko nitrogen aur oxygen ke ions tak le jaata hai.

Table 4.5Nitrogen aur oxygen species ke liye MO predictions

Species	Electrons	Bond order	Unpaired e⁻	Magnetic?
$N_{2}$	14	3	0	dia
$N_{2}^{+}$	13	2.5	1	para
$O_{2}^{+}$	15	2.5	1	para
$O_{2}$	16	2	2	para
$O_{2}^{-}$ (superoxide)	17	1.5	1	para
$O_{2}^{2 -}$ (peroxide)	18	1	0	dia

Oxygen series mein jaise-jaise antibonding $π^{*}$ orbitals mein electrons judte hain, bond order 2.5 se 1 tak girta hai, aur O–O bond $O_{2}^{+}$ mein lagbhag $112 p m$ se badhkar peroxide ion mein lagbhag $149 p m$ ho jaata hai.

Example

B₂ paramagnetic kyun hai?

Boron vapour mein $B_{2}$ molecules hote hain, jo paramagnetic hain. Dono orbital orders use karke $B_{2}$ ka configuration likho, decide karo kaun sa order experiment se match karta hai, aur bond order nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

4.9 Hydrogen Bonding

Water 100 °C par boil hota hai, lekin hydrogen sulfide, jo same bent shape ka bhaari molecule hai, −60 °C par boil hota hai. Table 4.6 nitrogen aur fluorine ke hydrides ke liye bhi yahi pattern dikhata hai. Kuch extra cheez $N H_{3}$ , $H_{2} O$ aur $H F$ ke molecules ko saath pakad kar rakhti hai.

Table 4.6Kuch hydrides ke boiling points

Hydride	b.p. / °C	Group mein agla	b.p. / °C
$N H_{3}$	−33	$P H_{3}$	−88
$H_{2} O$	100	$H_{2} S$	−60
$H F$	20	$H C l$	−85

Yeh extra attraction nitrogen, oxygen ya fluorine se jude hydrogen atoms se aata hai. Yeh teenon atoms chhote aur bahut electronegative hain, isliye N–H, O–H ya F–H bond bahut polar hota hai aur uske hydrogen end par asaamanya roop se bada partial positive charge hota hai. Aisa hydrogen doosre molecule ke N, O ya F atom ke lone pair ki taraf kheencha jaata hai. Is link ko hydrogen bond (हाइड्रोजन आबंध) kehte hain, aur ise dotted line se banaya jaata hai. Yeh covalent bond se kaafi kamzor hai, aam taur par uski strength ka dasvaan hissa ya usse bhi kam, lekin inki ginti bahut zyada hoti hai.

Solid aur liquid hydrogen fluoride zigzag chains se bane hote hain, $\dots H - F \dots H - F \dots$ , jinmein har hydrogen do fluorine atoms ke beech baitha hai, ek se chhote covalent bond se aur doosre se lambe hydrogen bond se juda hua (Fig. 4.22).

Definition

Hydrogen bond

F, O ya N se covalent bond se jude hydrogen atom aur kisi doosre F, O ya N atom ke lone pair ke beech ka attractive force, chahe woh atom doosre molecule mein ho ya usi molecule mein.

1/3Molecules ke beech: har H padosi O ke lone pair ki taraf point karta hai.

Fig. 4.22Dotted lines hydrogen bonds hain; solid lines covalent bonds hain.NCERT Fig. 4.22 · p. 131

Dotted lines hydrogen bonds hain, solid lines covalent bonds. Step 1: teen water molecules, ek ka har H agle ke O ki taraf point karta hua, δ+ aur δ− label ke saath. Step 2: HF molecules ki zigzag chain, har H do F atoms ke beech pul banata hua. Step 3: salicylaldehyde, ek benzene ring jiska OH hydrogen usi molecule ke aldehyde oxygen se dotted line se juda hai.

4.9.1 Hydrogen Bond Banne ki Wajah

Sirf F, O aur N hi kyun? H–X bond mein X jitna zyada electronegative hoga, shared pair utna hi hydrogen se door kheencha jaayega; aur hydrogen, jiske paas koi inner electrons nahi hain, lagbhag khula reh jaata hai. Result ek strongly polar unit hai, H par $δ +$ aur X par $δ -$ . Ek unit ka $δ +$ hydrogen doosre unit ke $δ -$ atom ki taraf electrostatically kheencha jaata hai, aur yahi attraction hydrogen bond hai.

Hydrogen bonding kitni hogi, yeh state par depend karta hai: sabse zyada solid mein, jahan har molecule apni jagah par tika rehta hai, liquid mein kam, aur gas mein sabse kam, jahan molecules ek doosre se door hote hain. Iske effects har jagah hain. Ice liquid water se kam dense hai, aur tairti hai, kyunki ice mein har molecule ek khule, cage jaise network mein chaar doosre molecules se hydrogen-bonded hota hai. DNA ke do strands hydrogen bonds se zip ki tarah jude rehte hain.

4.9.2 H-Bonds ke Types

Intermolecular hydrogen bonds ek molecule ko doosre molecule se jodte hain, chahe ek hi substance ke molecules hon (water, ammonia, ethanoic acid) ya do substances ke (water mein ghula ammonia, water ke saath mila ethanol). Yeh boiling points badhate hain aur samjhate hain ki aise liquids itni achchhi tarah kyun mix hote hain.

Intramolecular hydrogen bonds ek hi molecule ke andar rehte hain. Yeh tab bante hain jab O–H ya N–H ka hydrogen usi molecule ke O ya N atom ke bagal mein ho, aur ek ring band ho jaati hai. 2-hydroxybenzaldehyde (salicylaldehyde) mein –OH ka hydrogen bagal wale –CHO group ke oxygen tak pahunchta hai; 2-nitrophenol mein yeh nitro group ke ek oxygen tak pahunchta hai. Jo molecule khud se bond banata hai, woh apne padosiyon ko kam kas kar pakadta hai, isliye 2-nitrophenol lagbhag 216 °C par boil hota hai, jo 4-nitrophenol (lagbhag 279 °C) se kaafi kam hai, jiske –OH groups sirf doosre molecules se hi bond bana sakte hain.

Try karke dekho

Ice kyun tairti hai?

Tumhe chahiye

ek chhota transparent plastic cup
ek marker pen
paani
ek freezer

Pehle guess karo

Tum ek cup aadha bharte ho, paani ka level mark karte ho aur use freeze karte ho. Ice ka top kahan hoga?

mark se neechebilkul mark parmark se upar

Cup ko aadha paani se bharo aur bahar ki taraf paani ka level mark karo.
Cup ko raat bhar freezer mein seedha khada rakho. Plastic use karo, kabhi glass nahi: paani jamte waqt phailta hai aur glass ko tod sakta hai.
Ice ke top ko apne mark se compare karo.
Ek glass paani mein ice cube daalo aur dekho woh kahan tairta hai.

Ice mark se upar khadi hoti hai: paani ka utna hi mass ice ke roop mein zyada jagah leta hai, isliye ice kam dense hai aur lagbhag dasvaan hissa surface se upar rakhkar tairti hai. Ice mein har water molecule chaar padosiyon se ek khule network mein hydrogen-bonded hota hai jiske andar khaali jagah hoti hai. Jab ice pighalti hai, to kuch hydrogen bonds toot jaate hain aur molecules aur paas-paas pack ho jaate hain.

Recap

Summary aur key points

Kössel ne ions ko noble-gas configurations tak electron transfer se samjhaya; Lewis ne covalent bonds ko pairs share karne se samjhaya. Lewis symbols valence electrons dikhate hain (Section 4.1).
Lewis structures ek recipe follow karte hain: electrons gino, skeleton banao, bonds aur lone pairs jodo, phir multiple bonds banao. Formal charge $V - L - \frac{1}{2} S$ sabse achchha structure chunta hai (Section 4.1.4).
Octet rule incomplete octets, odd-electron molecules aur expanded octets ke liye fail hota hai, aur shape ya energy ke baare mein kuch nahi batata (Section 4.1.5).
Ionic compounds tab bante hain jab lattice enthalpy gaseous ions banane ka kharcha wasool kar deti hai; lattice enthalpy ionic charge ke saath badhti hai aur ionic size ke saath ghatti hai (Section 4.2).
Bond length $R = r_{A} + r_{B}$ ; bond angle aur bond enthalpy bonds ko describe karte hain; zyada bond order ka matlab chhote, strong bonds (Section 4.3).
Jab ek Lewis structure fail hota hai, to fixed nuclei wale canonical structures milkar ek single, zyada stable resonance hybrid banate hain (Section 4.3.5).
Polar bonds ke dipole moments $μ = Q \times r$ (debye mein) hote hain; molecule ka dipole bond aur lone-pair dipoles ka vector sum hai. Fajans' rules ionic bonds ko covalent character dete hain (Section 4.3.6).
VSEPR: electron pairs jitna ho sake utna door phailte hain, lp–lp > lp–bp > bp–bp ke saath; lone pairs angles ko dabate hain aur equatorial sites lete hain (Section 4.4).
VB theory: half-filled orbitals overlap karte hain aur electrons pair bante hain; head-on overlap σ bonds deta hai, sideways overlap kamzor π bonds (Section 4.5).
Hybridisation (sp, sp², sp³, sp³d, sp³d²) linear, trigonal, tetrahedral, trigonal bipyramidal aur octahedral geometries aur $C_{2} H_{4}$ aur $C_{2} H_{2}$ ke multiple bonds samjhata hai (Section 4.6).
MO theory: atomic orbitals combine (LCAO) hokar bonding aur antibonding molecular orbitals banate hain; milti-julti energy, same symmetry aur achchha overlap zaroori hai (Section 4.7).
Bond order $= \frac{1}{2} (N_{b} - N_{a})$ ; singly occupied orbitals species ko paramagnetic banate hain, jaise $O_{2}$ aur $B_{2}$ mein (Section 4.7.5).
$H e_{2}$ aur $B e_{2}$ ka bond order 0 hai aur yeh exist nahi karte; $L i_{2}$ , $C_{2}$ aur $N_{2}$ diamagnetic hain (Section 4.8).
Hydrogen bonds F, O ya N par jude H ko F, O ya N ke lone pair se jodte hain, molecules ke beech ya ek molecule ke andar, aur boiling points aur ice par inka bada effect hota hai (Section 4.9).

Practice · 40 questions

Practice questions answers ke saath (NCERT “Exercises”)

Short answerEasyNCERT Exercises 4.1
Samjhao ki do hydrogen atoms molecule kyun banate hain lekin do helium atoms kyun nahi, (a) duplet idea aur (b) atoms ke paas aane par system ki energy, dono ka use karke.
Solution dikhao
(a) Apne single electrons share karke dono H atoms helium ka duplet pa lete hain; helium atom ke paas pehle se duplet hai, isliye share karne se use kuch nahi milta. (b) Paas aate H atoms naye attractions (har nucleus ka doosre ke electron ke liye) mehsoos karte hain jo naye repulsions se zyada hote hain, isliye energy $74 p m$ par minimum tak girti hai: yahi bond hai. Helium ke paas koi half-filled orbital nahi hai, aur $H e_{2}$ ka MO bond order 0 hai, isliye koi minimum nahi banta.
Short answerEasyNCERT Exercises 4.2
Ca, K, Al, S, P aur I ke Lewis symbols likho, aur har ek ki usual group valence batao.
Solution dikhao
Table 4.7
Atom Dots Group valence
Ca 2 2
K 1 1
Al 3 3
S 6 8 − 6 = 2
P 5 8 − 5 = 3
I 7 8 − 7 = 1
Short answerEasyNCERT Exercises 4.3
O aur $O^{2 -}$ , Mg aur $M g^{2 +}$ , N aur $N^{3 -}$ ke Lewis symbols likho. Kin pairs mein atom ne electrons liye hain, aur kin mein khoye hain?
Solution dikhao
O ke 6 dots hain; $O^{2 -}$ ke brackets mein 8 dots aur 2− charge hai. N ke 5 dots hain; $N^{3 -}$ ke brackets mein 8 dots aur 3− charge hai. Dono ne neon ka octet paane ke liye electrons liye. Mg ke 2 dots hain; $M g^{2 +}$ ka koi valence dot nahi hai (uska naya outer shell, n = 2 shell, ek poora octet hai). Magnesium ne electrons khoye.
4
Short answerMediumNCERT Exercises 4.4
Neeche diye gaye ke Lewis structures banao.
1. $P H_{3}$
  Solution dikhao
  8 electrons. P teen P–H bonds banata hai aur ek lone pair rakhta hai.
2. $C S_{2}$
  Solution dikhao
  16 electrons. S=C=S, har sulfur par do lone pairs aur carbon par koi nahi.
3. $S O_{3}^{2 -}$
  Solution dikhao
  $6 + 18 + 2 = 26$ electrons. S teen O atoms se single bonds banata hai aur uske paas ek lone pair hai; har O par teen lone pairs hain. Formal charges: S +1, har O −1, total −2.
4. $H C H O$
  Solution dikhao
  12 electrons. Carbon centre mein, do C–H bonds aur ek C=O double bond ke saath; oxygen par do lone pairs hain.
Short answerEasyNCERT Exercises 4.5
Octet rule batao. Ek example do jahan yeh kaam ka hai, aur iske teen tarah ke failures mein se har ek ka ek example do.
Solution dikhao
Atoms tab tak electrons khote, lete ya share karte hain jab tak unke valence shell mein aath na ho jaayein (H ke liye do). Yeh $M g O$ (transfer) aur $C H_{4}$ (sharing) aur zyada tar organic structures samjhata hai. Failures: incomplete octet, $B F_{3}$ (B par 6 electrons); odd-electron molecule, $N O$ (11 electrons); expanded octet, $P C l_{5}$ (P ke chaaron taraf 10). Yeh $X e F_{4}$ jaise noble-gas compounds, shape aur energy ko bhi ignore karta hai.
Short answerMediumNCERT Exercises 4.6
Lithium fluoride ionic hai, lekin beryllium iodide zyada tar covalent hai. Ionic bonding ko favour karne wale factors se samjhao kyun.
Solution dikhao
Ionic bonding ko favour karte hain: metal ki low ionisation enthalpy, non-metal ki large negative electron gain enthalpy aur badi lattice enthalpy. Li aasani se ionise hota hai, F aasani se electron leta hai, aur chhote ions badi lattice enthalpy dete hain. $B e^{2 +}$ banane mein kaafi zyada energy lagti hai, aur yeh chhota, doubly charged ion bade iodide ion ko bahut zyada polarise karta hai (Fajans' rules), isliye bond zyada tar covalent hai.
7
Short answerMediumNCERT Exercises 4.7
VSEPR theory se har species ka shape predict karo.
1. $S n C l_{4}$
  Solution dikhao
  4 bond pairs, koi lone pair nahi ( $A B_{4}$ ): tetrahedral.
2. $S b C l_{5}$
  Solution dikhao
  5 bond pairs ( $A B_{5}$ ), isliye trigonal bipyramidal.
3. $S e C l_{2}$
  Solution dikhao
  Se ke 6 valence electrons hain; 2 bonds ke baad 4 bachte hain, yaani do lone pairs ( $A B_{2} E_{2}$ ): bent.
4. $A s C l_{3}$
  Solution dikhao
  As ke 5 valence electrons hain; 3 bonds ke baad ek lone pair bachta hai ( $A B_{3} E$ ): trigonal pyramidal.
Short answerMediumNCERT Exercises 4.8
$N H_{4}^{+}$ , $N H_{3}$ aur $N H_{2}^{-}$ , teenon mein nitrogen ke chaaron taraf chaar electron pairs hain. Inke H–N–H bond angles ka order predict karo aur samjhao.
Solution dikhao
Lone pairs: $N H_{4}^{+}$ mein 0, $N H_{3}$ mein 1, $N H_{2}^{-}$ mein 2. Sabmein pairs ka arrangement tetrahedral hai. Lone pairs bond pairs se zyada zor se repel karte hain (lp–lp > lp–bp > bp–bp), isliye har extra lone pair bond angle ko dabata hai: $N H_{4}^{+}$ (109.5°) > $N H_{3}$ (107°) > $N H_{2}^{-}$ (sabse chhota, water ke 104.5° ke kareeb).
Short answerEasyNCERT Exercises 4.9
Bond strength bond order par kaise depend karti hai? $C H_{3} O H$ , $H C H O$ aur $C O$ ke carbon–oxygen bonds se samjhao.
Solution dikhao
C–O bond orders 1, 2 aur 3 hain. Bond order badhne par zyada electron pairs nuclei ko pakadte hain, isliye bond enthalpy badhti hai aur bond length ghatti hai. Bonds lagbhag $143$ se $121$ se $113 p m$ tak chhote hote jaate hain, aur CO ka bond (lagbhag $1076 k J m o l^{- 1}$ ) sabse strong hai.
Short answerEasyNCERT Exercises 4.10
Bond length define karo. Covalent radii N $70 p m$ aur H $37 p m$ use karke N–H bond length ka estimate lagao aur use ammonia mein measured value, $101 p m$ , se compare karo.
Solution dikhao
Bond length kisi bond mein uske energy minimum par do nuclei ke beech ki doori hai. Estimate: $R = 70 + 37 = 107 p m$ , jo measured $101 p m$ se lagbhag 6 % zyada hai. Covalent radii jodna sirf approximate hai; N–H jaise polar bonds aam taur par sum se thode chhote hote hain.
Short answerMediumNCERT Exercises 4.11
Nitrite ion, $N O_{2}^{-}$ , ke dono N–O bonds lagbhag $124 p m$ lambe hain. Ise resonance se samjhao, aur canonical structures ke baare mein yaad rakhne wale main points batao.
Solution dikhao
18 electrons. Octet structure mein ek N=O aur ek N–O⁻ bond hai, aur N par ek lone pair; double bond ko doosre O tak banane se ek doosra, equivalent canonical structure milta hai. Ion inka hybrid hai: har N–O bond ka order 1.5 hai aur har O par aadha negative charge hai, isliye dono bonds barabar hain aur N–O aur N=O ke beech ke hain. Yaad rakho: canonical structures mein nuclei fixed rehte hain; inmein se koi akela exist nahi karta; ion inke beech flip nahi karta; hybrid inmein se kisi se bhi zyada stable hai aur bond properties ka average deta hai.
Short answerMediumNCERT Exercises 4.12
Formula HNCO ke liye do structures likhe gaye hain: $H - N = C = O$ aur $N \equiv C - O - H$ . Kya yeh ek resonance hybrid ke canonical structures ho sakte hain? Samjhao.
Solution dikhao
Nahi. Canonical structures mein har nucleus same jagah hona chahiye aur fark sirf electron positions ka hona chahiye. Yahan ek mein hydrogen nitrogen se juda hai aur doosre mein oxygen se, isliye nuclei ka arrangement alag hai. Yeh do alag compounds hain, isocyanic acid aur cyanic acid, jo isomers (tautomers) ki tarah related hain, resonance forms ki tarah nahi.
13
Short answerMediumNCERT Exercises 4.13
Har species ke resonance structures likho.
1. $S O_{2}$ (octet structures)
  Solution dikhao
  $O = S - O$ aur $O - S = O$ , har ek mein S par ek lone pair; singly bonded O par −1 aur S par +1.
2. $C H_{3} C O O^{-}$
  Solution dikhao
  Do structures: ek oxygen tak C=O aur doosre tak C–O⁻, phir inki jagah badli hui. Hybrid mein dono C–O bonds barabar hain.
3. $N_{3}^{-}$ (azide)
  Solution dikhao
  Teen: $N = N = N$ (formal charges −1, +1, −1), $N \equiv N - N$ (0, +1, −2) aur $N - N \equiv N$ (−2, +1, 0). Pehla, jismein sabse chhote charges hain, sabse zyada contribute karta hai.
14
Short answerMediumNCERT Exercises 4.14
Lewis symbols se dikhao ki har pair ke elements ke react karke ionic compound banane par electrons kaise move karte hain, aur formula do.
1. Li aur O
  Solution dikhao
  Do Li atoms ek-ek electron ek O atom ko dete hain: $2 L i^{+}$ aur $O^{2 -}$ , jisse $L i_{2} O$ banta hai.
2. Mg aur N
  Solution dikhao
  Teen Mg atoms chhe electrons khote hain; do N atoms teen-teen lete hain: $3 M g^{2 +}$ aur $2 N^{3 -}$ , jisse $M g_{3} N_{2}$ banta hai.
3. Al aur S
  Solution dikhao
  Do Al atoms chhe electrons khote hain; teen S atoms do-do lete hain: $2 A l^{3 +}$ aur $3 S^{2 -}$ , jisse $A l_{2} S_{3}$ banta hai.
Short answerMediumNCERT Exercises 4.15
$C S_{2}$ ka dipole moment zero hai, lekin sulfur dichloride, $S C l_{2}$ , ka ek measurable dipole moment hai. Isse dono molecules ke shapes ke baare mein kya pata chalta hai?
Solution dikhao
Dono mein polar bonds hain. $C S_{2}$ mein do C=S bond dipoles cancel hone chahiye, isliye molecule linear hai (S=C=S, 180°). $S C l_{2}$ mein woh cancel nahi hote, isliye molecule bent hai: sulfur par do lone pairs hain ( $A B_{2} E_{2}$ ).
Short answerEasyNCERT Exercises 4.16
Dipole moment measurements ke chaar uses batao, har ek ke example ke saath.
Solution dikhao
1. Molecule polar hai ya nahi, yeh decide karna: $C C l_{4}$ (0) non-polar hai, $C H C l_{3}$ (1.04 D) polar. 2. Shape pata karna: $B F_{3}$ ka zero dikhata hai ki woh planar hai; $N H_{3}$ ka 1.47 D dikhata hai ki woh pyramidal hai. 3. Ionic character ka estimate: measured aur fully ionic dipole moment ka ratio (HCl ke liye lagbhag 18 %). 4. Isomers ko alag pehchanna: cis-1,2-dichloroethene polar hai, trans isomer ka dipole moment zero hai.
Short answerEasyNCERT Exercises 4.17
Electronegativity aur electron gain enthalpy dono atom ki electrons ki bhookh ko describe karte hain. Samjhao ki dono mein kya fark hai: har ek kis cheez ko refer karta hai, kya use measure kiya ja sakta hai, aur uski unit kya hai.
Solution dikhao
Electronegativity bonded state mein kisi atom ki shared electron pair ko apni taraf kheenchne ki ability hai. Yeh ek scale (jaise Pauling ka) par ek relative number hai, iski koi unit nahi hai aur ise seedhe measure nahi kiya ja sakta. Electron gain enthalpy woh enthalpy change hai jab ek isolated gaseous atom ek electron leta hai. Yeh $k J m o l^{- 1}$ mein ek measurable energy hai.
Short answerEasyNCERT Exercises 4.18
Samjhao ki H–Br bond polar kyun hai. Partial charges aur dipole moment ki direction dono conventions mein dikhao.
Solution dikhao
Bromine hydrogen se zyada electronegative hai, isliye shared pair Br ke zyada paas hai: $\overset{δ +}{H} ⁣ - ⁣ \overset{δ -}{B r}$ . Chemists ka crossed arrow H (cross) se Br (head) ki taraf point karta hai, jo electron density ka shift dikhata hai. Physicists ka dipole vector ulti taraf, Br se H ki taraf, point karta hai.
19
Quick answerMediumNCERT Exercises 4.19
Bonds ko badhte ionic character ke order mein lagao.
1. $C l_{2}$ mein Cl–Cl
2. $C s F$ mein Cs–F
3. $H C l$ mein H–Cl
4. $I C l$ mein I–Cl
5. $N a C l$ mein Na–Cl
Answer dekho
Answer
$C l_{2}$ mein Cl–Cl
$I C l$ mein I–Cl
$H C l$ mein H–Cl
$N a C l$ mein Na–Cl
$C s F$ mein Cs–F
Electronegativity differences (Pauling): Cl–Cl 0, I–Cl lagbhag 0.5, H–Cl lagbhag 1.0, Na–Cl lagbhag 2.2, Cs–F lagbhag 3.2. Difference jitna bada, bond utna zyada ionic.
Short answerMediumNCERT Exercises 4.20
Ek student ne ethanal, $C H_{3} C H O$ , ko single C–O bond, oxygen par teen lone pairs aur aldehyde carbon par ek lone pair ke saath banaya. Galti dhoondho aur sahi structure do.
Solution dikhao
Gino: $2 \times 4 + 4 \times 1 + 6 = 18$ electrons. Student ki drawing 6 bonds (12 electrons) plus 4 lone pairs (8) use karti hai, yaani 20: do zyada. Formal charges bhi yahi batate hain: C ka $4 - 2 - 3 = - 1$ aur O ka $6 - 6 - 1 = - 1$ hoga, ek neutral molecule ke liye total −2. Sahi structure: carbon par se lone pair hatao aur C–O bond ko double bond banao, $C H_{3} - C H = O$ , oxygen par do lone pairs ke saath. Ab har atom ka octet (H ke liye duplet) poora hai aur formal charge zero hai.
Short answerMediumNCERT Exercises 4.21
$C C l_{4}$ tetrahedral kyun hai, square planar kyun nahi? Ek VSEPR argument, ek hybridisation argument aur ek experimental evidence do.
Solution dikhao
VSEPR: chaar bond pairs 109.5° (tetrahedron) par sabse door hote hain, 90° (square) par nahi, isliye repulsion sabse kam hota hai. Hybridisation: carbon sp³ hai; iske chaar hybrids tetrahedron ke corners ki taraf point karte hain. Square planar centre ke liye $d s p^{2}$ hybrids chahiye, jo carbon nahi bana sakta. Evidence: $C C l_{4}$ ka dipole moment zero hai, aur $C H_{2} C l_{2}$ sirf ek hi compound ke roop mein milta hai. Square planar carbon $C H_{2} C l_{2}$ ke do isomers deta (Cl atoms bagal mein ya aamne-saamne).
Short answerEasyNCERT Exercises 4.22
Si–F bonds bahut polar hain, phir bhi $S i F_{4}$ ka dipole moment zero hai. Samjhao.
Solution dikhao
$S i F_{4}$ tetrahedral ( $A B_{4}$ ) hai. Iske chaar barabar Si–F bond dipoles regular tetrahedron ke corners ki taraf point karte hain, aur unka vector sum zero hai, isliye molecule non-polar hai.
Short answerMediumNCERT Exercises 4.23
Water ka dipole moment 1.85 D hai, lekin oxygen difluoride, $O F_{2}$ , ka sirf lagbhag 0.30 D hai, jabki fluorine oxygen se zyada electronegative hai. Lone pairs ki madad se samjhao.
Solution dikhao
Dono bent hain aur O par do lone pairs hain. Lone pairs aise dipoles dete hain jo O se door, bonds ki ulti taraf point karte hain. $H_{2} O$ mein oxygen har O–H bond ka negative end hai, isliye bond dipoles ka resultant lone-pair contribution ki hi taraf point karta hai: dono jud jaate hain. $O F_{2}$ mein O–F bond dipoles F ki taraf point karte hain, lone-pair contribution ke ulat, jo unke effect ka zyada tar hissa cancel kar deta hai.
Short answerEasyNCERT Exercises 4.24
Carbon ko example lekar hybridisation samjhao, phir sp, sp² aur sp³ sets ko compare karo: orbitals ki ginti, unke beech ka angle, s character aur har ek ka ek example.
Solution dikhao
Hybridisation ek atom par milti-julti energy wale atomic orbitals ka mixing hai, jisse utne hi equivalent hybrid orbitals bante hain.
Table 4.8
Set Orbitals Angle s character Example
sp 2 180° 50 % $B e F_{2}$
sp² 3 120° 33 % $B F_{3}$
sp³ 4 109.5° 25 % $C H_{4}$
Short answerMediumNCERT Exercises 4.25
Kya reaction $B F_{3} + F^{-} \overset{}{\to} B F_{4}^{-}$ mein boron ka hybridisation badalta hai? Shape ka kya hota hai?
Solution dikhao
$B F_{3}$ mein boron sp² hybridised hai (trigonal planar, 120°), aur uske paas ek khaali 2p orbital hai. $F^{-}$ boron ke khaali orbital mein ek lone pair donate karta hai. Ab boron ke chaar bond pairs hain aur woh sp³ hybridised ho jaata hai: $B F_{4}^{-}$ tetrahedral (109.5°) hai.
Short answerMediumNCERT Exercises 4.26
Jab ammonia ek proton leta hai, $N H_{3} + H^{+} \overset{}{\to} N H_{4}^{+}$ , to kya nitrogen ka hybridisation badalta hai? Shape aur bond angle ka kya hota hai?
Solution dikhao
Nahi: dono mein nitrogen sp³ hai. $N H_{3}$ mein ek hybrid lone pair rakhta hai; $N H_{4}^{+}$ mein woh pair proton se bond banata hai. Shape trigonal pyramidal se tetrahedral ho jaata hai, aur angle 107° se khulkar 109.5° ho jaata hai, kyunki ab bonds ko dabane ke liye koi lone pair nahi hai.
27
Short answerMediumNCERT Exercises 4.27
Har molecule mein σ aur π bonds describe karo, aur batao kaun se orbitals overlap karte hain.
1. methanal, $H C H O$
  Solution dikhao
  C sp² hai: H se do sp²–1s σ bonds aur O se ek sp²–2p σ bond. C ka unhybridised 2p, O ke ek 2p se sideways overlap karke C=O ka π bond banata hai.
2. hydrogen cyanide, $H C N$
  Solution dikhao
  C sp hai: H se ek sp–1s σ bond aur N se ek sp–2p σ bond, ek seedhi line mein. C ke do unhybridised 2p orbitals N ke do 2p orbitals se sideways overlap karke right angles par do π bonds banate hain.
28
NumericalMediumNCERT Exercises 4.28
Har molecule mein σ aur π bonds gino.
1. buta-1,3-diene, $C H_{2} = C H - C H = C H_{2}$ : σ, π
  Answer dekho
  Answer (1) 9, (2) 2
  Chhe C–H aur teen C–C σ bonds; do π bonds.
2. benzene, $C_{6} H_{6}$ : σ, π
  Answer dekho
  Answer (1) 12, (2) 3
  Chhe C–H aur chhe C–C σ bonds; teen π bonds.
29
Quick answerMediumNCERT Exercises 4.29
x ko internuclear axis maano, to inmein se orbitals ke kaun se pairs σ bond bana sakte hain?
$2 s$ aur $2 s$ $2 p_{x}$ aur $2 p_{x}$ $2 p_{z}$ aur $2 p_{z}$ $1 s$ aur $2 p_{y}$
Saare sahi options chuno
Answer dekho
Answer
(A) $2 s$ aur $2 s$
(B) $2 p_{x}$ aur $2 p_{x}$
2s–2s aur $2 p_{x}$ – $2 p_{x}$ , x ke saath head-on overlap karte hain. Do $2 p_{z}$ orbitals axis ke across padte hain aur sideways overlap karte hain (π). 1s aur $2 p_{y}$ zero overlap dete hain.
30
Short answerMediumNCERT Exercises 4.30
Left se right, har carbon atom ka hybridisation batao.
1. $C H_{3} - C \equiv N$
  Solution dikhao
  sp³, sp.
2. $C H_{2} = C = C H_{2}$
  Solution dikhao
  sp², sp, sp² (beech wale carbon ke do double bonds hain).
3. $C H_{3} C O C H_{3}$
  Solution dikhao
  sp³, sp² (C=O wala carbon), sp³.
4. $H C O O H$
  Solution dikhao
  sp² (ek C=O double bond).
Short answerEasyNCERT Exercises 4.31
Bond pairs aur lone pairs kya hote hain? $H_{2} S$ , $P C l_{3}$ aur $X e F_{4}$ mein central atom ke chaaron taraf dono gino.
Solution dikhao
Bond pair do atoms ke beech share hota hai; lone pair ek atom ka hota hai aur bonding mein use nahi hota. $H_{2} S$ : 2 bond pairs, 2 lone pairs. $P C l_{3}$ : 3 bond pairs, 1 lone pair. $X e F_{4}$ : 4 bond pairs, 2 lone pairs.

Atom	Dots	Group valence
Ca	2	2
K	1	1
Al	3	3
S	6	8 − 6 = 2
P	5	8 − 5 = 3
I	7	8 − 7 = 1

Set	Orbitals	Angle	s character	Example
sp	2	180°	50 %	$B e F_{2}$
sp²	3	120°	33 %	$B F_{3}$
sp³	4	109.5°	25 %	$C H_{4}$

Short answerEasyNCERT Exercises 4.32

σ aur π bonds ko paanch points par compare karo: overlap ka type, electron density kahan hai, strength, kya bond akela exist kar sakta hai, aur bond ke around rotation.

Solution dikhao

Table 4.9

Feature	σ bond	π bond
Overlap	head-on, axis ke saath	sideways, parallel p orbitals se
Electron density	axis par, uske around symmetric	axis ke upar aur neeche
Strength	zyada strong (zyada overlap)	kamzor
Akela exist karta hai?	haan, pehla bond	sirf σ bond ke saath
Bond ke around rotation	free	restricted

Short answerMediumNCERT Exercises 4.33
Valence bond theory se describe karo ki hydrogen fluoride ka molecule kaise banta hai.
Solution dikhao
H ke paas ek half-filled 1s orbital hai; F ( $2 s^{2} 2 p^{5}$ ) ke paas ek half-filled $2 p_{z}$ orbital hai. Jab atoms z ke saath paas aate hain, to yeh head-on overlap (s–p overlap) karte hain aur dono electrons opposite spins ke saath pair bante hain. Result ek σ bond hai. Atoms ke paas aane par energy release hoti hai kyunki attractions repulsions se zyada hote hain, aur yeh bond length (lagbhag $92 p m$ ) par minimum tak girti hai; bond polar hai, F par δ− ke saath.
34
Short answerMediumNCERT Exercises 4.34
z ko axis maankar do identical atoms ke liye, inmein se atomic orbitals ke kaun se pairs achchhe se combine karte hain? Jo pair combine nahi karta, uske liye batao ki woh kaun si condition todta hai.
1. 1s aur 1s
  Solution dikhao
  Haan: same energy aur symmetry ( $σ 1 s$ , $σ^{*} 1 s$ ).
2. 1s aur 2s
  Solution dikhao
  Nahi: energies bahut alag hain.
3. $2 p_{z}$ aur $2 p_{x}$
  Solution dikhao
  Nahi: z axis ke around symmetry alag hai.
4. $2 p_{x}$ aur $2 p_{x}$
  Solution dikhao
  Haan: same energy aur symmetry ( $π 2 p_{x}$ , $π^{*} 2 p_{x}$ ).
Short answerMediumNCERT Exercises 4.35
Molecular orbital filling se dikhao ki $N e_{2}$ ke exist karne ki ummeed kyun nahi hai, jabki $N e_{2}^{+}$ exist kar sakta hai.
Solution dikhao
$N e_{2}$ mein 20 electrons hain, jo $σ^{*} 2 p_{z}$ tak har orbital bhar dete hain: $N_{b} = N_{a} = 10$ , isliye bond order 0 hai aur koi stable molecule nahi banta. $N e_{2}^{+}$ mein 19 electrons hain, $σ^{*} 2 p_{z}$ mein sirf ek: bond order $\frac{1}{2} (10 - 9) = \frac{1}{2}$ , isliye ek weakly bound ion exist kar sakta hai (ise gas phase mein dekha gaya hai).
Short answerHardNCERT Exercises 4.36
$C_{2}^{+}$ , $C_{2}$ , $C_{2}^{-}$ aur $C_{2}^{2 -}$ ko badhti stability ke order mein lagao, aur batao ki har ek paramagnetic hai ya diamagnetic.
Solution dikhao
Halke molecules wala order use karo ( $π 2 p$ , $σ 2 p_{z}$ se neeche). $C_{2}$ (12 electrons): $K K (σ 2 s)^{2} (σ^{*} 2 s)^{2} (π 2 p)^{4}$ , bond order 2, diamagnetic. $C_{2}^{+}$ (11): $(π 2 p)^{3}$ , bond order 1.5, paramagnetic. $C_{2}^{-}$ (13): $σ 2 p_{z}$ mein ek electron, bond order 2.5, paramagnetic. $C_{2}^{2 -}$ (14): $(σ 2 p_{z})^{2}$ , bond order 3, diamagnetic ( $N_{2}$ ke saath isoelectronic). Badhti stability: $C_{2}^{+} < C_{2} < C_{2}^{-} < C_{2}^{2 -}$ .
Short answerEasyNCERT Exercises 4.37
Orbital lobes par aksar + aur − mark hote hain. In signs ka kya matlab hai, aur bonding ke liye yeh kyun matter karte hain?
Solution dikhao
Yeh har lobe mein wave function ke signs (phases) hain, electric charges nahi. Same sign wale lobes ka overlap positive hota hai aur nuclei ke beech electron density badhata hai (bonding, $ψ_{A} + ψ_{B}$ ). Opposite signs ka overlap negative hota hai aur node deta hai (antibonding, $ψ_{A} - ψ_{B}$ ). Barabar + aur − contributions cancel hokar zero overlap dete hain, isliye koi bond nahi banta.
Short answerMediumNCERT Exercises 4.38
$A s F_{5}$ mein arsenic ka hybridisation describe karo, aur samjhao ki iske axial bonds equatorial bonds se lambe kyun hain.
Solution dikhao
As ( $[A r] 3 d^{10} 4 s^{2} 4 p^{3}$ ) ek 4s electron ko 4d orbital mein promote karta hai, jisse paanch half-filled orbitals milte hain jo mix hokar trigonal bipyramid ke corners ki taraf point karne wale paanch sp³d hybrids banate hain. Har ek F ke ek half-filled 2p orbital se overlap karta hai. Har axial bond pair ko 90° par teen equatorial bond pairs repel karte hain, jabki har equatorial pair ke 90° par sirf do padosi hain. Zyada repulsion axial bonds ko khinch deta hai.
Short answerEasyNCERT Exercises 4.39
Hydrogen bond kya hai? Strength ke hisaab se ise covalent bonds aur van der Waals forces ke saath rank karo, rough magnitudes ke saath.
Solution dikhao
Hydrogen bond F, O ya N se jude H aur kisi doosre F, O ya N atom ke lone pair ke beech ka attraction hai. Strength: covalent bonds (saikdon $k J m o l^{- 1}$ ) ≫ hydrogen bonds (lagbhag 5–40 $k J m o l^{- 1}$ ) > van der Waals forces (aur bhi kamzor). Isliye hydrogen bond van der Waals forces se strong hai lekin covalent bond se kaafi kamzor.
40
NumericalMediumNCERT Exercises 4.40
Har species ka bond order calculate karo.
1. $F_{2}$
  Answer dekho
  Answer 1
  18 electrons: $N_{b} = 10$ , $N_{a} = 8$ , isliye $\frac{1}{2} (10 - 8) = 1$ .
2. $F_{2}^{+}$
  Answer dekho
  Answer 1.5
  17 electrons: ek antibonding electron kam, isliye $\frac{1}{2} (10 - 7) = 1.5$ .
3. $B_{2}$
  Answer dekho
  Answer 1
  10 electrons: $\frac{1}{2} (6 - 4) = 1$ .
4. $N_{2}^{+}$
  Answer dekho
  Answer 2.5
  13 electrons: $N_{2}$ se ek bonding electron hata diya gaya, isliye $\frac{1}{2} (9 - 4) = 2.5$ .

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercises 4.1	Practice Q1
NCERT Exercises 4.2	Practice Q2
NCERT Exercises 4.3	Practice Q3
NCERT Exercises 4.4	Practice Q4
NCERT Exercises 4.5	Practice Q5
NCERT Exercises 4.6	Practice Q6
NCERT Exercises 4.7	Practice Q7
NCERT Exercises 4.8	Practice Q8
NCERT Exercises 4.9	Practice Q9
NCERT Exercises 4.10	Practice Q10
NCERT Exercises 4.11	Practice Q11
NCERT Exercises 4.12	Practice Q12
NCERT Exercises 4.13	Practice Q13
NCERT Exercises 4.14	Practice Q14
NCERT Exercises 4.15	Practice Q15
NCERT Exercises 4.16	Practice Q16
NCERT Exercises 4.17	Practice Q17
NCERT Exercises 4.18	Practice Q18
NCERT Exercises 4.19	Practice Q19
NCERT Exercises 4.20	Practice Q20
NCERT Exercises 4.21	Practice Q21
NCERT Exercises 4.22	Practice Q22
NCERT Exercises 4.23	Practice Q23
NCERT Exercises 4.24	Practice Q24
NCERT Exercises 4.25	Practice Q25
NCERT Exercises 4.26	Practice Q26
NCERT Exercises 4.27	Practice Q27
NCERT Exercises 4.28	Practice Q28
NCERT Exercises 4.29	Practice Q29
NCERT Exercises 4.30	Practice Q30
NCERT Exercises 4.31	Practice Q31
NCERT Exercises 4.32	Practice Q32
NCERT Exercises 4.33	Practice Q33
NCERT Exercises 4.34	Practice Q34
NCERT Exercises 4.35	Practice Q35
NCERT Exercises 4.36	Practice Q36
NCERT Exercises 4.37	Practice Q37
NCERT Exercises 4.38	Practice Q38
NCERT Exercises 4.39	Practice Q39
NCERT Exercises 4.40	Practice Q40

Quick check: MCQs

4.1.5 Octet Rule ki Limitations4 questions
4.2.1 Lattice Enthalpy3 questions
4.3.4 Bond Order3 questions
4.3.6 Bonds ki Polarity3 questions
4.4 Valence Shell Electron Pair Repulsion (VSEPR) Theory4 questions
4.5.5 Sigma aur pi Bonds ki Strength3 questions
4.6.3 d Orbitals wale Elements ka Hybridisation3 questions
4.7.5 Electronic Configuration aur Molecular Behaviour3 questions
4.8 Kuch Homonuclear Diatomic Molecules mein Bonding4 questions
4.9.2 H-Bonds ke Types3 questions

Glossary · 15

Key terms (English + Hindi)

bond enthalpyआबंध एन्थैल्पीaabandh enthalpy: Kisi khaas tarah ke ek mole bonds todne ke liye zaroori energy, jab saari species gas phase mein hon.
bond orderआबंध कोटिaabandh koti: Do atoms ke beech bonds ki ginti: Lewis structure mein shared pairs, ya MO theory mein $\frac{1}{2} (N_{b} - N_{a})$ .

covalent bondसहसंयोजी आबंधsahsanyoji aabandh: Aisa bond jo tab banta hai jab do atoms electrons ke ek ya zyada pairs share karte hain.Chapter mein pehli baar yahan aaya →

dipole momentद्विध्रुव आघूर्णdvidhruv aaghurn: Charge × positive aur negative charge ke centres ke beech ki doori, $μ = Q \times r$ , aam taur par debye (D) mein.Chapter mein pehli baar yahan aaya →

formal chargeऔपचारिक आवेशaupcharik aavesh: Lewis structure mein free atom ke valence electrons minus uske lone-pair electrons minus uske bonding electrons ka aadha.Chapter mein pehli baar yahan aaya →

hybridisationसंकरणsankaran: Ek atom par milti-julti energy wale atomic orbitals ka mixing, jisse utne hi equivalent hybrid orbitals bante hain.Chapter mein pehli baar yahan aaya →
hydrogen bondहाइड्रोजन आबंधhydrogen aabandh: F, O ya N se jude H aur kisi doosre F, O ya N atom ke lone pair ke beech attraction.Chapter mein pehli baar yahan aaya →

ionic (electrovalent) bondआयनिक आबंधaayanik aabandh: Electrons ke transfer se bane opposite charge wale ions ke beech electrostatic attraction.Chapter mein pehli baar yahan aaya →

lattice enthalpyजालक एन्थैल्पीjaalak enthalpy: Ek mole ionic solid ko poori tarah gaseous ions mein alag karne ke liye zaroori energy.
Lewis symbolलूइस प्रतीकlewis prateek: Element ka symbol, jiske chaaron taraf har valence electron ke liye ek dot ho.Chapter mein pehli baar yahan aaya →

molecular orbitalआण्विक कक्षकaanvik kakshak: Atomic orbitals ko milakar bana orbital jo poore molecule par phaila hota hai; yeh bonding ya antibonding ho sakta hai.Chapter mein pehli baar yahan aaya →

octet ruleअष्टक नियमashtak niyam: Atoms tab tak electrons khote, lete ya share karte hain jab tak unke valence shell mein aath na ho jaayein (hydrogen ke liye do).

pi (π) bondपाई आबंधpai aabandh: Parallel p orbitals ke sideways overlap se bana bond, jiski electron density bond axis ke upar aur neeche hoti hai.Chapter mein pehli baar yahan aaya →

resonance hybridअनुनाद संकरanunaad sankar: Kisi species ka woh ek asli structure jise describe karne ke liye do ya zyada canonical Lewis structures chahiye.Chapter mein pehli baar yahan aaya →

sigma (σ) bondसिग्मा आबंधsigma aabandh: Do nuclei ko jodne wali line ke saath orbitals ke head-on overlap se bana bond.Chapter mein pehli baar yahan aaya →

4.1 Chemical Bonding ka Kössel–Lewis Approach

4.1.1 Octet Rule

4.1.2 Covalent Bond

4.1.3 Simple Molecules ka Lewis Representation (Lewis Structures)

4.1.4 Formal Charge

4.1.5 Octet Rule ki Limitations

4.2 Ionic ya Electrovalent Bond

4.2.1 Lattice Enthalpy

4.3 Bond Parameters

4.3.1 Bond Length

4.3.2 Bond Angle

4.3.3 Bond Enthalpy

4.3.4 Bond Order

4.3.5 Resonance Structures

4.3.6 Bonds ki Polarity

4.4 Valence Shell Electron Pair Repulsion (VSEPR) Theory

4.5 Valence Bond Theory

4.5.1 Orbital Overlap ka Concept

4.5.2 Bonds ki Directional Properties

4.5.3 Atomic Orbitals ka Overlapping

4.5.4 Overlapping ke Types aur Covalent Bonds ka Nature

4.5.5 Sigma aur pi Bonds ki Strength

4.6 Hybridisation

4.6.1 Hybridisation ke Types

4.6.2 sp³, sp² aur sp Hybridisation ke Aur Examples

4.6.3 d Orbitals wale Elements ka Hybridisation

4.7 Molecular Orbital Theory

4.7.1 Molecular Orbitals ka Banna: Linear Combination of Atomic Orbitals (LCAO)

4.7.2 Atomic Orbitals ke Combination ki Conditions

4.7.3 Molecular Orbitals ke Types

4.7.4 Molecular Orbitals ka Energy Level Diagram

4.7.5 Electronic Configuration aur Molecular Behaviour

4.8 Kuch Homonuclear Diatomic Molecules mein Bonding

4.9 Hydrogen Bonding

4.9.1 Hydrogen Bond Banne ki Wajah

4.9.2 H-Bonds ke Types

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