Coordination Compounds: Class 12 Chemistry Chapter 5 Hinglish mein

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

Published 12 Oct 2026 · Updated 12 Oct 2026

100 min
46 practice questions
36 MCQs

Is chapter mein aap kya seekhenge

Werner ki primary aur secondary valences se predict karo ki ek coordination compound paani mein kitne ions deta hai
Ek complex ki coordination entity, central atom, ligands aur unki denticity, coordination number, coordination sphere, polyhedron aur oxidation state pehchano
Mononuclear coordination compounds ke formulas aur IUPAC names likho
Geometrical, optical, linkage, coordination, ionisation aur solvate isomers pehchano aur simple complexes ke stereoisomers gino
Valence bond theory se complexes ki shapes aur magnetic behaviour samjhao aur spin-only magnetic moments calculate karo
Crystal field theory se d orbitals ko split karo, high-spin aur low-spin arrangements mein se chuno aur complexes ke colours samjhao
Metal carbonyls mein carbon monoxide ko metal se jodne wali synergic bonding ko describe karo
Analysis, metallurgy, living things, industry aur medicine mein kaam karte coordination compounds ke examples do

NCERT Chemistry Class 12, Chapter 5, pp. 118–140 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

Silver chloride sabse kam soluble salts mein se ek hai jo tumhe milenge, phir bhi uska dahi jaisa safed precipitate aqueous ammonia ke saath hilane par gayab ho jaata hai. Gold lagbhag har acid ko jhel jaata hai, phir bhi mining companies crushed rock se uske chhote-chhote kan dilute cyanide solution aur hawa se ghol kar nikaal leti hain. Dono cases mein ek metal ion ne apne aas-paas molecules ya ions ka ek fixed set jama kar liya hai aur ek naya, soluble species ban gaya hai.

Aise species hi is unit ka topic hain. Ek coordination compound (उपसहसंयोजन यौगिक) mein ek metal atom ya ion hota hai jo fixed number ke neighbours se juda hota hai, aur har neighbour ek electron pair donate karta hai. Tum dekhoge ki chemists ne pehli baar inhe kaise samjha aur inke naam kaise rakhte hain. Tum yeh bhi jaanoge ki kuch compounds ek hi formula ke saath kai forms mein kyun milte hain, aur kai itne chamakdaar coloured ya magnetic kyun hote hain. Aakhir mein tum inhe kaam karte dekhoge, patton ke hare rang se lekar anticancer drugs tak.

5.1 Coordination Compounds ki Werner's Theory

1800s ke aakhir tak chemists cobalt(III) chloride ko ammonia ke saath mila kar darjanon crystalline solids bana chuke the. Woh inhe dot ke saath likhte the, jaise $C o C l_{3} \cdot 6 N H_{3}$ , $C o C l_{3} \cdot 5 N H_{3}$ aur $C o C l_{3} \cdot 4 N H_{3}$ , aur isi mein ek puzzle chhupa tha. $C o C l_{3}$ aur $N H_{3}$ dono stable compounds hain jinki har valence pehle se use ho chuki hai, to inhe saath kaun pakad kar rakhta hai, aur kya har solid mein chlorine ek hi jagah baithta hai?

Ek simple test jawab deta hai. Thande solution mein silver nitrate daalne par woh free chloride ions ko turant safed $A g C l$ ke roop mein nikaal deta hai, lekin metal se seedhe bonded chlorine atom ko woh kheench nahi sakta. Isliye per mole compound $A g C l$ ke moles ginne se free chloride gina jaata hai. Solution ki electrical conductivity naapne se ek doosra, independent count milta hai: har formula unit kitne total ions release karta hai.

2/3 $C o C l_{3} \cdot 5 N H_{3}$ : ek chlorine chhathi jagah le leta hai, do free chloride ions bachte hain aur 2 mol $A g C l$ banta hai.

Fig. 5.1Cobalt ke chaaron taraf hamesha chhah groups hote hain; sirf sphere ke bahar bacha chloride hi silver ions tak pahunchta hai.NCERT Table 5.1 · p. 119

Ek cobalt ion ek bade dashed circle ke andar hai jis par coordination sphere likha hai, aur uske aas-paas chhah bonded positions hain. Right mein free chloride ions bahar tair rahe hain, add silver nitrate likha ek arrow neeche point karta hai, aur bane silver chloride ke moles diye gaye hain. Step 1 mein chhahon positions par ammonia hai, teen chloride ions free hain aur 3 mol silver chloride banta hai; formula brackets mein Co(NH3)6 aur bahar Cl3 padhta hai. Step 2 mein ek chlorine andar aa gaya hai, do chloride ions free bache hain aur 2 mol silver chloride banta hai. Step 3 mein do chlorines andar hain, ek chloride ion free hai aur sirf 1 mol silver chloride banta hai.

Pattern bilkul saaf hai. Jab bhi ek ammonia molecule kam hota hai, ek chlorine atom cobalt ke paas uski jagah le leta hai, aur woh chlorine free ion ki tarah behave karna band kar deta hai. Ammonia molecules aur bonded chlorines ka total hamesha chhah hota hai. Conductivity bhi yahi kehti hai: pehla compound har formula unit se chaar ions deta hai, doosra teen aur teesra do.

Alfred Werner ne 1893 mein in results ko samjhaya. Unhone propose kiya ki ek metal ion ki do tarah ki combining power hoti hai. Primary valence woh hai jise aaj hum oxidation state kehte hain; yeh negative ions se satisfy hoti hai aur ionisable hoti hai. Secondary valence metal se seedhe jude groups ka number hai, jise aaj coordination number kehte hain. Modern bhasha mein Werner ke ideas aise hain.

Aise compound mein har metal ion ki do alag combining powers hoti hain, ek primary aur ek secondary.
Primary valence sirf anions se poori hoti hai, aur jo anions kahin aur nahi lagte woh solid ke ghulne par free ho jaate hain.
Secondary valence un groups ko ginti hai jo solution mein bhi bonded rehte hain: $N H_{3}$ jaise molecules, $C l^{-}$ jaise anions, ya dono. Har metal ion ka apna characteristic number hota hai, cobalt(III) ke liye chhah.
Secondary bonds fixed directions mein point karte hain, isliye complex ki ek definite shape hoti hai, aur har coordination number kuch khaas shapes ko pasand karta hai.

Aaj $[C o (N H_{3})_{6}] C l_{3}$ jaise formula mein bracket un sab cheezon ko alag karta hai jo secondary valences se pakdi gayi hain: yahi coordination entity hai. Bahar bache ions counter ions hain, aur bonded atoms jo shape banate hain woh coordination polyhedron hai. Isomers gin kar Werner is nateeje par pahunche ki $[C o (N H_{3})_{6}]^{3 +}$ jaisi six-coordinate entities octahedral hoti hain. Four-coordinate entities do tarah ki nikli: tetrahedral, jaise $[Z n (N H_{3})_{4}]^{2 +}$ , aur square planar, jaise $[P t (N H_{3})_{4}]^{2 +}$ .

Example 5.1NCERT Example 5.1 · p. 120

Silver nitrate test se secondary valence

Neeche diye har compound ko paani mein ghola gaya aur excess silver nitrate ke saath treat kiya gaya. Maan lo ki formula ka har neutral molecule metal se bonded hai. Secondary valence nikaalo aur Werner ka formula likho.

Table 5.1

Compound	mol AgCl per mol
violet $C r C l_{3} \cdot 6 H_{2} O$	3
$C r C l_{3} \cdot 5 N H_{3}$	2
$C u C l_{2} \cdot 4 N H_{3}$	2
$I r C l_{3} \cdot 3 N H_{3}$	0
$P t C l_{4} \cdot 4 N H_{3}$	2

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

Compare

Double salt ya complex?

Double salt

Sirf crystal mein ek compound ki tarah rehta hai.
Paani mein poori tarah apne simple ions mein toot jaata hai.
Ammonium alum, $N H_{4} A l (S O_{4})_{2} \cdot 12 H_{2} O$ , $N H_{4}^{+}$ , $A l^{3 +}$ aur $S O_{4}^{2 -}$ ke normal tests deta hai.

Complex

Coordination entity solution mein bhi bani rehti hai.
Sirf counter ions release hote hain.
$K_{3} [F e (C N)_{6}]$ se $K^{+}$ aur $[F e (C N)_{6}]^{3 -}$ milte hain; thiocyanate se koi blood-red colour nahi aata, isliye koi free $F e^{3 +}$ maujood nahi hai.

Quick check

4 mein se 1

[R h C l_{2} (N H_{3})_{4}] C l

ke ek mole ko excess silver nitrate ke saath treat kiya jaata hai.

A g C l

ke kitne moles bante hain?

1230

Answer dekho

Answer (A) 1

Sirf bracket ke bahar wala chloride free hai, isliye 1 mol

A g C l

banta hai. Teeno chlorines ginna (answer 3) rhodium se bonded do chlorines ko ignore karta hai.

5.2 Coordination Compounds se jude kuch Important Terms ki Definitions

Kuch gine-chune words se hum kisi bhi complex ko precisely describe kar sakte hain. Hum kuch laboratory examples ka sahara lenge. Ek hai gehra neela $[C u (N H_{3})_{4}]^{2 +}$ ion jo copper sulphate solution mein ammonia daalne par banta hai. Baaki hain Tollens' reagent ka $[A g (N H_{3})_{2}]^{+}$ ion aur $K_{3} [C o (N O_{2})_{6}]$ , woh peela precipitate jiske aane se potassium ions ka pata chalta hai.

(a) Coordination entity aur (b) central atom

Definition

Coordination entity

Complex ke beech ka metal aur us se seedhe juda har group, jinhe square brackets ke andar likhte hain. Poori unit par positive charge, negative charge ya koi charge nahi ho sakta.

Example $[A g (N H_{3})_{2}]^{+}$ (cation), $[Z n (O H)_{4}]^{2 -}$ (anion), $[F e (C O)_{5}]$ (neutral).

Centre wala metal central atom ya ion hai. Yeh apne neighbours ko ek definite arrangement mein pakad kar rakhta hai aur har ek se electron pair accept karta hai, isliye yeh Lewis acid ki tarah behave karta hai. $[A g (N H_{3})_{2}]^{+}$ mein central ion $A g^{+}$ hai; $[Z n (O H)_{4}]^{2 -}$ mein $Z n^{2 +}$ hai; $[F e (C O)_{5}]$ mein yeh zero oxidation state wala iron atom hai.

(c) Ligands

Definition

Ligand

Ek ion ya molecule jo central atom se ek ya zyada donor atoms ke through bonded hota hai, aur har donor atom ek lone pair deta hai. Ligands Lewis bases ki tarah kaam karte hain.

Example $C l^{-}$ , $O H^{-}$ , $H_{2} O$ , $N H_{3}$ , $C O$ , ethane-1,2-diamine (en).

Ligand kaise bind karta hai yeh uska size tay nahi karta; yeh tay karta hai ki woh kitne donor atoms use karta hai, aur yahi number uski denticity hai. Ammonia ek nitrogen use karta hai, isliye woh unidentate hai. Ethane-1,2-diamine, $H_{2} N C H_{2} C H_{2} N H_{2}$ (short mein en), dono nitrogen atoms se pakadta hai, aur glycinate ion, $H_{2} N C H_{2} C O O^{-}$ , ek nitrogen aur ek oxygen se, isliye dono didentate hain. Polydentate ligand zyada atoms use karta hai: ethylenediaminetetraacetate ion, $E D T A^{4 -}$ , metal ko do nitrogen aur chaar oxygen atoms se pakadta hai, isliye woh hexadentate hai (Fig. 5.18). Ek protein bhi ligand ban sakta hai, jo metal ion ko ek saath kai side-chain atoms se pakadta hai.

2/3en dono nitrogens se bind karta hai aur ek five-membered chelate ring banata hai.

Fig. 5.2Donor atoms tay karte hain ki ligand kaise bind karega: ek atom, ring banate do atoms, ya do alag atoms mein se koi ek.NCERT p. 121

Left mein ek metal atom M hai. Step 1 mein ek ammonia molecule, yaani teen hydrogens wala nitrogen, ek single arrow ke saath apna lone pair M ko donate karta hai, jis par unidentate, 1 donor atom likha hai. Step 2 mein ek ethane-1,2-diamine molecule dono nitrogen atoms se bind karta hai; metal, do nitrogens aur do carbons milkar ek shaded five-membered ring banate hain, jis par didentate likha hai. Step 3 mein akele metal ki jagah do chhote panels aa jaate hain: upar wale mein nitrite ion apne nitrogen se bond karta hai (nitrito-N), aur neeche wale mein ek oxygen se (nitrito-O).

Definition

Chelate

Woh ring jo tab banti hai jab ek ligand same metal ion ko apne do ya zyada donor atoms se pakadta hai; aisa ligand chelating ligand kehlata hai, Greek mein kekde ke panje ke word se.

Example $[N i (e n)_{3}]^{2 +}$ mein teen five-membered chelate rings hain.

Chelate complexes aam taur par same donor atoms wale unidentate ligands se bane milte-julte complexes se kaafi zyada stable hote hain; isi ko chelate effect kehte hain. $[N i (e n)_{3}]^{2 +}$ mein ek metal–nitrogen bond tootne se ligand chhootta nahi, kyunki uska doosra end abhi bhi pakde hue hai, isliye free end ke dobara jud jaane ke chances zyada hain.

Kuch ligands do alag donor atoms offer karte hain aur kisi ek ko bhi use kar sakte hain. Yeh ambidentate ligands hain. Nitrite, $N O_{2}^{-}$ , nitrogen ya oxygen se bond kar sakta hai, aur thiocyanate, $S C N^{-}$ , sulphur ya nitrogen se. Har choice ek alag compound deti hai, jaisa Section 5.4.3 dikhata hai.

(d) Coordination number aur (e) coordination sphere

Definition

Coordination number (CN)

Central atom se $σ$ bonds dwara seedhe jude donor atoms ki ginti. Metal aur ligand ke beech ke koi bhi $π$ bonds count mein nahi judte.

Example $[A g (N H_{3})_{2}]^{+}$ mein CN = 2, $[C u (N H_{3})_{4}]^{2 +}$ mein 4, $[C r (C_{2} O_{4})_{3}]^{3 -}$ mein 6.

Donor atoms gino, ligands nahi. $[C r (C_{2} O_{4})_{3}]^{3 -}$ mein teen oxalate ions chhah oxygen donors dete hain, isliye coordination number 6 hai, aur $[C a (E D T A)]^{2 -}$ mein akela ek $E D T A^{4 -}$ coordination number 6 deta hai.

Square brackets ke andar ki har cheez, yaani central atom aur uske ligands, milkar coordination sphere banate hain. $K_{3} [C o (N O_{2})_{6}]$ ko gholo to bracket wali unit $[C o (N O_{2})_{6}]^{3 -}$ poori bani rehti hai, jabki uske teen $K^{+}$ counter ions alag-alag tair jaate hain.

(f) Coordination polyhedron

Metal ke aas-paas ke donor atoms ko imaginary lines se jodo to coordination polyhedron milta hai. Sabse common hain octahedral (CN 6), tetrahedral aur square planar (dono CN 4); CN 5 ke liye trigonal bipyramidal aur square pyramidal shapes milti hain. Fig. 5.3 har ek ko baari-baari banata hai.

4/4Chhah ligands $\pm x$ , $\pm y$ aur $\pm z$ ke saath, sabse common shape.

Fig. 5.3Har coordination number ki ek ya do typical shapes hoti hain.NCERT Fig. 5.1 · p. 122

Ek metal atom M jiske ligands L perspective mein bane hain, aur dashed lines polyhedron ki outline banati hain. Step 1 mein chaar ligands ek tetrahedron ke corners par hain, coordination number 4, angles 109.5 degrees, example tetraamminezinc(II). Step 2 mein chaar ligands ek plane mein square ke corners par hain, coordination number 4, angles 90 degrees, example tetrachloridoplatinate(II). Step 3 mein ek trigonal bipyramid hai, teen ligands beech mein ghere hue aur ek upar ek neeche, coordination number 5, example pentacarbonyliron(0). Step 4 mein ek octahedron hai, chaar ligands square mein plus ek upar aur ek neeche, coordination number 6, example hexaaquachromium(III).

(g) Central atom ka oxidation number aur (h) homoleptic aur heteroleptic complexes

Har shared electron pair ko us ligand ko wapas de do jisne use donate kiya tha: metal ke paas jo charge bachta hai wahi uska oxidation number hai. Practice mein ise charge balance se nikaalte hain: metal ka oxidation number + ligands ka total charge = entity par charge. Ise metal ke naam ke baad brackets mein Roman numeral se likhte hain: $[C o C l_{4}]^{2 -}$ mein cobalt, cobalt(II) hai, kyunki $x + 4 (- 1) = - 2$ .

Sirf ek kind ke ligand wala complex, jaise $[C u (N H_{3})_{4}]^{2 +}$ ya $[F e (C O)_{5}]$ , homoleptic hai. Do ya zyada kinds wala, jaise $[C r C l (N H_{3})_{5}]^{2 +}$ , heteroleptic hai.

Example

Oxidation number, coordination number aur ligand types

Har entity ke liye metal ka oxidation number aur uska coordination number nikaalo, aur batao ki woh homoleptic hai ya heteroleptic: (a) $K_{3} [C o (N O_{2})_{6}]$ (b) $[F e (C O)_{5}]$ (c) $[C r C l (e n)_{2} (H_{2} O)]^{2 +}$ .

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

Quick check

4 mein se 1

[C r (C_{2} O_{4})_{3}]^{3 -}

mein chromium ka coordination number kya hai?

3496

Answer dekho

Answer (D) 6

Har oxalate didentate hai, isliye teen oxalates 6 donor atoms dete hain. 3 answer karna donor atoms ki jagah ligands ko ginta hai.

5.3 Coordination Compounds ki Nomenclature

Do alag compounds ka formula same ho sakta hai, aur ek compound ko kai tarah se draw kiya ja sakta hai, isliye chemists ko aise names aur formulas chahiye jinmein koi shak na rahe. IUPAC, jo international body chemical names ko standardise karti hai, rules set karti hai. Inhe do chhoti checklists ki tarah seekho, ek formulas likhne ke liye aur ek names bolne ke liye.

5.3.1 Mononuclear Coordination Entities ke Formulas

Is unit ke zyadatar complexes mononuclear hain: centre mein ek metal atom hota hai. Inke formulas ek fixed recipe follow karte hain.

Square bracket kholo aur central atom ka symbol likho.
Ligands ko unke formulas ya abbreviations ke alphabetical order mein jodo, charge chahe jo ho: Cl, $N H_{3}$ se pehle, en, $N O_{2}$ se pehle. en ya py jaisa abbreviation apne pehle letter ke hisaab se aata hai.
Har polyatomic ligand aur har abbreviation ko round brackets mein rakho, aur square brackets ke andar kahin bhi space mat chhodo.
Square bracket band karo. Akeli likhi charged entity apna charge right superscript mein rakhti hai, pehle number: $[C u (N H_{3})_{4}]^{2 +}$ .
Salt ke liye cation ko anion se pehle likho aur charges balance karne ke liye dono ko utna lo jitna chahiye.

5.3.2 Mononuclear Coordination Compounds ke Naam

Naam ek train ki tarah joda jaata hai: aage ligand names, unke peeche metal aur end mein uska oxidation state. Chemists ise additive nomenclature kehte hain. Steps yeh hain:

Har ligand ko uska ligand name do. Anions -ido, -ato ya -ito par khatam hote hain (chlorido, cyanido, oxalato, nitrito); neutral molecules apne naam rakhte hain, sirf aqua ( $H_{2} O$ ), ammine ( $N H_{3}$ ), carbonyl ( $C O$ ) aur nitrosyl ( $N O$ ) ko chhod kar.
Ligand names ko alphabetical order mein lagao. Multiplying prefixes count nahi hote: tetraammine a ke under aata hai.
Har ligand kitne hain yeh di, tri, tetra … se dikhao; jab naam mein pehle se number ho ya naam lamba ho, to bis, tris, tetrakis … use karo aur ligand name brackets mein rakho, jaise bis(ethane-1,2-diamine).
Ligands ke turant baad metal likho. Anionic entity mein uska naam -ate par khatam hota hai (cobaltate, nickelate, zincate) aur kuch metals apne Latin roots use karte hain (table dekho); cation ya neutral molecule mein uska normal naam rehta hai.
End mein metal ka oxidation state brackets mein Roman numerals se likho, zero ke liye (0).
Salt mein cation ko anion se pehle bolo, chahe complex inmein se koi bhi ho. Neutral entity ka koi counter ion nahi hota aur uska naam akele rakha jaata hai.

Table 5.2Common ligands ke names aur unke donor atoms (formula har charge dikhata hai)

Ligand	Name	Donor
$H_{2} O$	aqua	O
$N H_{3}$	ammine	N
$C O$	carbonyl	C
$N O$	nitrosyl	N
$C_{5} H_{5} N$ (py)	pyridine	N
en	ethane-1,2-diamine	2 N
$C l^{-}$ , $B r^{-}$ , $I^{-}$	chlorido, bromido, iodido	halogen
$C N^{-}$	cyanido	C
$O H^{-}$	hydroxido	O
$N O_{2}^{-}$	nitrito-N ya nitrito-O	N ya O
$S C N^{-}$	thiocyanato-S ya thiocyanato-N	S ya N
$S_{2} O_{3}^{2 -}$	thiosulphato	S
$C_{2} O_{4}^{2 -}$ (ox)	oxalato	2 O
$C O_{3}^{2 -}$	carbonato	O
$E D T A^{4 -}$	ethylenediaminetetraacetato	2 N, 4 O

Table 5.3Anionic entities ke andar metal names

Metal	Cation ya neutral	Anion
Fe	iron	ferrate
Cu	copper	cuprate
Ag	silver	argentate
Au	gold	aurate
Pb	lead	plumbate
Sn	tin	stannate
Co	cobalt	cobaltate
Ni	nickel	nickelate
Pt	platinum	platinate
Zn	zinc	zincate

Fig. 5.4 in rules ko sodium cobaltinitrite, $N a_{3} [C o (N O_{2})_{6}]$ , par lagata hai, jo reagent potassium ions ko precipitate karta hai. Ek hi metal do alag naamon se do baar bhi aa sakta hai. Magnus's green salt, $[P t (N H_{3})_{4}] [P t C l_{4}]$ , tetraammineplatinum(II) tetrachloridoplatinate(II) hai: cation mein platinum apna simple naam rakhta hai aur anion mein platinate ban jaata hai.

4/4 $x + 6 (- 1) = - 3$ se cobalt(III) milta hai: sodium hexanitrito-N-cobaltate(III).

Fig. 5.4Formula ka har hissa naam ke ek hisse se match karta hai.NCERT p. 124

Formula Na3 Co NO2 six upar ek shaded frame mein hai. Uske neeche left column formula ke hisse dikhata hai: Na3, (NO2)6, anion mein Co, aur charge sum x plus 6 times minus 1 barabar minus 3. Arrows right mein matching name pieces tak jaate hain. Step 1 sodium dikhata hai, is note ke saath ki cation ka naam pehle aata hai. Step 2 chhah nitrogen-bonded nitrite ions ke liye hexanitrito-N jodta hai. Step 3 cobaltate jodta hai, kyunki metal anion mein hai. Step 4 charge balance se (III) jodta hai, aur poora naam sodium hexanitrito-N-cobaltate(III) neeche aata hai.

Example 5.2NCERT Example 5.2 · p. 124

Names se formulas tak

In ke formulas likho: (a) tetraamminecopper(II) sulphate (b) potassium dicyanidoaurate(I) (c) pentaamminechloridochromium(III) chloride (d) triammineaquadichloridocobalt(III) chloride (e) pentacarbonyliron(0).

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

Example 5.3NCERT Example 5.3 · p. 124

Formulas se names tak

Naam batao: (a) $K_{2} [H g I_{4}]$ (b) $[C r C l_{2} (H_{2} O)_{4}] C l$ (c) $N a_{3} [A g (S_{2} O_{3})_{2}]$ (d) $[C o (N H_{3})_{6}] [F e (C N)_{6}]$ (e) $[C r (H_{2} O)_{5} (N C S)]^{2 +}$ .

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

Quick check

3 mein se 1

K_{4} [F e (C N)_{6}]

ka sahi naam kaun sa hai?

potassium hexacyanidoiron(II)potassium hexacyanidoferrate(II)potassium hexacyanidoferrate(III)tetrapotassium hexacyanoiron

Answer dekho

Answer (B) potassium hexacyanidoferrate(II)

Entity ek anion hai, isliye iron ferrate ban jaata hai;

x + 6 (- 1) = - 4

se (II) milta hai. Simple 'iron' sirf cations aur neutral entities mein use hota hai.

Practice · 2 questions

Practice: formulas aur names: answers ke saath

1
Short answerEasyNCERT Intext Question 5.1
Har compound ka formula do.
1. hexaaquairon(II) sulphate
  Solution dikhao
  Chhah neutral water ligands ke saath Fe(II) ek 2+ entity deta hai, jise ek sulphate balance karta hai: $[F e (H_{2} O)_{6}] S O_{4}$ .
2. sodium hexafluoridoaluminate(III)
  Solution dikhao
  $3 + 6 (- 1) = - 3$ , isliye teen $N a^{+}$ : $N a_{3} [A l F_{6}]$ , mineral cryolite jo aluminium ke extraction mein use hota hai.
3. bis(ethane-1,2-diamine)oxalatocobalt(III) chloride
  Solution dikhao
  Oxalate −2 hai aur en neutral, isliye entity $3 - 2 = + 1$ hai, jise ek chloride balance karta hai. $C_{2} O_{4}$ (C), en (e) se pehle likha jaata hai: $[C o (C_{2} O_{4}) (e n)_{2}] C l$ .
4. potassium amminetrichloridoplatinate(II)
  Solution dikhao
  $2 + 3 (- 1) + 0 = - 1$ , isliye ek $K^{+}$ : $K [P t C l_{3} (N H_{3})]$ .
5. hexaamminenickel(II) hexacyanidocobaltate(III)
  Solution dikhao
  Cation $[N i (N H_{3})_{6}]^{2 +}$ hai aur anion $[C o (C N)_{6}]^{3 -}$ . Teen cations (6+) do anions (6−) ko balance karte hain, isliye formula $[N i (N H_{3})_{6}]_{3}$ $[C o (C N)_{6}]_{2}$ hai, jo bina space ke likha jaata hai.
6. copper(II) hexacyanidoferrate(II)
  Solution dikhao
  Anion $[F e (C N)_{6}]^{4 -}$ ko do $C u^{2 +}$ chahiye: $C u_{2} [F e (C N)_{6}]$ , woh brown precipitate jo copper ions ke test mein use hota hai.
2
Short answerEasyNCERT Intext Question 5.2
IUPAC rules se har compound ka naam rakho.
1. $[A g (N H_{3})_{2}] C l$
  Solution dikhao
  Diamminesilver(I) chloride; yahi cation Tollens' reagent ka active hissa hai.
2. $[C o B r (N H_{3})_{5}] S O_{4}$
  Solution dikhao
  $x - 1 = + 2$ se Co(III) milta hai: pentaamminebromidocobalt(III) sulphate.
3. $L i [A l H_{4}]$
  Solution dikhao
  Hydride, $H^{-}$ , ka naam hydrido hai. $x + 4 (- 1) = - 1$ se Al(III) milta hai, aur anion mein aluminium aluminate ban jaata hai: lithium tetrahydridoaluminate(III), woh reducing agent jise lithium aluminium hydride kehte hain.
4. $[C u (e n)_{2}] S O_{4}$
  Solution dikhao
  en neutral hai, isliye $x = + 2$ : bis(ethane-1,2-diamine)copper(II) sulphate.
5. $K_{2} [P t C l_{6}]$
  Solution dikhao
  $x - 6 = - 2$ se Pt(IV) milta hai: potassium hexachloridoplatinate(IV).
6. $[P t (N H_{3})_{3} (p y)] C l_{2}$
  Solution dikhao
  Ammine, pyridine se pehle: triammine(pyridine)platinum(II) chloride.

5.4 Coordination Compounds mein Isomerism

Do compounds jinka formula same hai lekin unke atoms alag tarah se jude hain, isomers (समावयवी) kehlate hain. Yeh rearrangement alag colour, solubility ya reactivity ke roop mein dikhta hai, aur kabhi-kabhi living cells par bilkul alag effect ke roop mein. Complexes do broad tariko se isomerism dikhate hain.

Stereoisomerism: same atoms same partners se bonded hote hain, lekin space mein alag directions mein point karte hain.
- geometrical isomerism
- optical isomerism
Structural isomerism: bonds hi alag hote hain, kyunki ligand doosra donor atom use karta hai ya groups jagah badal lete hain.
- linkage isomerism
- coordination isomerism
- ionisation isomerism
- solvate isomerism

5.4.1 Geometric Isomerism

Do compounds ka formula $[P t C l_{2} (N H_{3})_{2}]$ hai. Ek, cisplatin, sabse zyada use hone wale anticancer drugs mein se ek hai; doosre ka tumours par koi kaam ka effect nahi. Fark sirf itna hai ki square planar platinum ke aas-paas ligands kahan baithe hain. Cis isomer mein dono chloride ions padosi hain, 90° door; trans isomer mein woh ek doosre ke opposite hain, 180° door. Is tarah ki isomerism ke liye heteroleptic complex chahiye aur yeh coordination numbers 4 aur 6 mein common hai.

Square planar complex $[M A_{2} B_{2}]$ ke liye, jahan A aur B unidentate hain, theek do arrangements hote hain. Chaar alag ligands, $[M A B C D]$ , ke saath yeh poochkar gino ki A ke opposite kaun sa ligand baitha hai: woh B, C ya D ho sakta hai, isliye teen geometrical isomers hain.

2/3Ek A ko ek B se badlo to A ligands ek doosre ke opposite aa jaate hain, 180° door.

Fig. 5.5Cis aur trans mein like ligands ke beech ka angle alag hota hai; tetrahedron mein aisa koi choice nahi hota.NCERT Fig. 5.2 · p. 125

Step 1 ek square planar complex ka top view dikhata hai: centre mein metal Pt, do ligands A paas-paas ke corners par (right aur top) aur do ligands B baaki do corners par; text kehta hai cis isomer, A–M–A angle 90 degrees, drug cisplatin. Step 2 mein upar wala A aur left wala B jagah badal lete hain, to dono A ligands ek doosre ke opposite aa jaate hain; text kehta hai trans isomer, angle 180 degrees. Step 3 square ki jagah ek zinc atom ke aas-paas tetrahedron dikhata hai jis par do A aur do B ligands hain; text kehta hai har angle 109.5 degrees hai, isliye sirf ek form hai.

Tetrahedral complexes kabhi geometrical isomerism nahi dikhate, kyunki tetrahedron ka har corner baaki sab corners ke bagal mein hota hai. Octahedral complexes dikhate hain. $[M A_{4} B_{2}]$ mein, jaise $[C r C l_{2} (N H_{3})_{4}]^{+}$ , do B ligands cis ya trans ho sakte hain. Didentate ligand AA ko do cis positions span karni padti hain, isliye $[M (A A)_{2} B_{2}]$ , jaise $[R h C l_{2} (e n)_{2}]^{+}$ , ke bhi cis aur trans forms hote hain. Har ligand ke teen-teen hon, $[M A_{3} B_{3}]$ , to phir se do forms hote hain. Facial (fac) isomer mein teen B ligands octahedron ka ek triangular face gherte hain. Meridional (mer) isomer mein woh ek aise plane mein hote hain jo metal se hokar guzarta hai.

3/4 $[M A_{3} B_{3}]$ jismein teen B ligands ek face par hain.

Fig. 5.6Wahi chhah positions

[M A_{4} B_{2}]

ke cis aur trans forms, aur

[M A_{3} B_{3}]

ke fac aur mer forms deti hain.NCERT Fig. 5.3, 5.5 · p. 126

Perspective mein bana ek octahedral complex: metal M aur dashed octahedron edges par chhah ligand balls. A wale balls blue hain aur B wale orange, isliye letters aur colours dono se inhe pehchana ja sakta hai. Step 1, cis: do B ligands padosi corners par hain, 90 degrees door, example tetraamminedichloridochromium(III). Step 2, trans: do B ligands upar aur neeche hain, 180 degrees door. Step 3, fac: teen B ligands ek face ke teen corners par hain. Step 4, mer: teen B ligands metal se guzarte ek plane mein hain, unmein se do ek doosre ke opposite, example triamminetrichloridochromium(III).

Example 5.4NCERT Example 5.4 · p. 126

Tetrahedral isomers ke baare mein ek claim ko test karna

Ek student report karta hai ki four-coordinate zinc complex $[Z n C l_{2} (N H_{3})_{2}]$ do geometrical isomers ke roop mein milta hai. Tetrahedron ki geometry se is claim ko check karo, aur batao ki agar is type ke four-coordinate complex ke sach mein do isomers milein to iska kya matlab hoga.

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

Example 5.5NCERT Example 5.5 · p. 127

Octahedral MA4B2 complex ke geometrical isomers

$[C r C l_{2} (H_{2} O)_{4}]^{+}$ ke geometrical isomers draw karo aur batao ki kis ka overall dipole moment nahi hai.

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

5.4.2 Optical Isomerism

Tumhare left aur right haath mirror images hain, phir bhi kitna bhi ghumao, ek doosre par bilkul fit nahi hota. Aisi cheez chiral kehlati hai. Chiral complex do enantiomers ke roop mein milta hai, aisi mirror images jo superimpose nahi ho saktin. Enantiomers ke melting points, solubilities aur colours bilkul same hote hain, lekin woh plane-polarised light ke plane ko barabar angles se opposite directions mein rotate karte hain, jise polarimeter naapta hai. Jo form plane ko right ki taraf ghumata hai woh dextro (d) hai, doosra laevo (l). Dono ka 50 : 50 mixture, yaani racemic mixture, bilkul bhi rotation nahi dikhata.

Six-coordinate complexes mein chirality aam taur par didentate ligands se aati hai. Har $[M (A A)_{3}]$ complex, jaise $[N i (e n)_{3}]^{2 +}$ ya $[C o (C_{2} O_{4})_{3}]^{3 -}$ , chiral hota hai: uske threefold axis ke neeche dekho to uske teen rings ek propeller banate hain jo ya to ek taraf ya doosri taraf mudta hai. Do chelate rings wale $[M (A A)_{2} B_{2}]$ mein sirf cis isomer chiral hai. Trans isomer mein metal aur dono B ligands se hokar ek mirror plane hota hai, isliye uski mirror image wahi molecule hai.

2/3Mirror image anticlockwise mudti hai (Λ).

Fig. 5.7Tris-chelate complex aur uski mirror image opposite directions mein mudte hain, isliye woh enantiomers hain.NCERT Fig. 5.6 · p. 126

Complex ko uske threefold axis ke neeche se dekha gaya hai: chhah donor atoms metal M ke aas-paas hexagon banate hain, paas ke teen atoms solid circles aur door ke teen dashed circles. Teen moti orange bands, har didentate ligand ki ek, propeller ke blades ki tarah ek paas wale atom ko ek door wale atom se jodti hain, aur ek curved arrow twist dikhata hai. Step 1 sirf left-hand form dikhata hai, jis par delta likha hai, jo clockwise mudta hai. Step 2 ek dashed mirror line aur mirror-image form jodta hai, jis par lambda likha hai, jo anticlockwise mudta hai. Step 3 mirror image ko original ke upar khiskata hai: bands match hone ki jagah cross karti hain, aur text kehta hai ki blades kabhi line up nahi hote.

InteractiveIsomer explorer

Perspective mein bana ek interactive octahedron jiske centre mein metal hai aur chhah ligand balls par A, B aur C letters hain, ya didentate ligand AA ke liye ek curved band. Ek menu se formula type chuna jaata hai, jaise MA4B2, MA3B3, MA2B2C2, M(AA)3, M(AA)2B2 ya M(AA)B2C2. Explorer octahedron ke saare 24 rotations try karke har distinct stereoisomer nikaalta hai, unhe cis, trans, fac, mer ya chiral pair jaise naamon wale buttons ki tarah list karta hai, aur ek summary dikhata hai ki kitne geometrical isomers aur kitne stereoisomers hain. Arrow buttons aur dragging model ko ghumate hain, aur Mirror button chune hue isomer ko uski mirror image se badal deta hai, saath mein ek status line batati hai ki mirror image wahi molecule hai ya ek alag enantiomer.

Fig. 5.8Har stereoisomer ki list ke liye ek formula type chuno, phir octahedron ko ghuma kar uski mirror image se test karo.

Example 5.6NCERT Example 5.6 · p. 127

Kaun si entity chiral hai?

Inmein se kaun optically active hai: (a) cis- $[C o B r C l (e n)_{2}]^{+}$ ya (b) trans- $[C o B r C l (e n)_{2}]^{+}$ ?

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

5.4.3 Linkage Isomerism

Ek ambidentate ligand apne do donor atoms mein se kisi se bhi jud sakta hai, aur har choice ek alag compound deti hai. Sulphur se bonded thiocyanate $[C o (N H_{3})_{5} (S C N)]^{2 +}$ deta hai, jabki nitrogen se bonded $[C o (N H_{3})_{5} (N C S)]^{2 +}$ deta hai. Nitrite bhi aisa hi karta hai, aur uske pentaamminecobalt(III) complexes 1890s mein pehchane gaye pehle linkage isomers the. Rakhe rehne par O-bonded (nitrito-O) form dheere-dheere zyada stable N-bonded (nitrito-N) form mein badal jaata hai.

5.4.4 Coordination Isomerism

Jab kisi salt ka cation aur anion dono complexes hon, to ligands do metals ke beech alag-alag tarah se baante ja sakte hain. $[C u (N H_{3})_{4}] [P t C l_{4}]$ mein ammonia molecules copper se aur chloride ions platinum se bonded hain; iske coordination isomer $[P t (N H_{3})_{4}] [C u C l_{4}]$ mein partners badal jaate hain. Dono salts ka overall formula same hai lekin colours aur chemistry alag hain.

5.4.5 Ionisation Isomerism

Kabhi-kabhi bracket ke bahar wala ion bhi metal se bilkul bind kar sakta hai. Use bonded ligands mein se kisi ek se badlo to bilkul same formula wala doosra compound milta hai. $[P t C l_{2} (N H_{3})_{4}] B r_{2}$ aur $[P t B r_{2} (N H_{3})_{4}] C l_{2}$ aise hi ionisation isomers hain. Paani mein yeh alag ions release karte hain, isliye silver nitrate pehle ke saath cream $A g B r$ aur doosre ke saath safed $A g C l$ deta hai.

5.4.6 Solvate Isomerism

Solvate isomers mein same solvent molecules alag tarah se rakhe hote hain: jo solvent molecule ek isomer mein ligand hai, woh doosre mein crystal lattice mein baitha hota hai, metal ko chhue bina. Jab solvent water ho, to ise hydrate isomerism kehte hain. Chromium(III) chloride hexahydrate, $C r C l_{3} \cdot 6 H_{2} O$ , teen forms mein milta hai: violet $[C r (H_{2} O)_{6}] C l_{3}$ , halka hara $[C r C l (H_{2} O)_{5}] C l_{2} \cdot H_{2} O$ aur gehra hara $[C r C l_{2} (H_{2} O)_{4}] C l \cdot 2 H_{2} O$ . Silver nitrate inse 3, 2 aur 1 mol $A g C l$ precipitate karta hai, aur sirf lattice water hi halki drying se aasaani se nikalta hai.

Table 5.4Structural isomerism ke chaar kinds

Kind	Fark	Test
Linkage	ambidentate ligand ka donor atom	infrared spectrum; colour
Coordination	kaun sa metal kaun se ligands pakadta hai	colour; bane ions
Ionisation	ek ligand counter ion se jagah badalta hai	silver nitrate AgBr ya AgCl deta hai
Solvate	water sphere ke andar ya bahar	AgCl ke moles; drying par mass ka loss

Practice · 2 questions

Practice: isomerism: answers ke saath

1
Short answerMediumNCERT Intext Question 5.3
Har complex kis kind ki isomerism dikha sakta hai, uska naam batao aur isomers describe karo.
1. $N a [C o (C_{2} O_{4})_{2} (N H_{3})_{2}]$
  Solution dikhao
  Geometrical: do $N H_{3}$ ligands cis ya trans ho sakte hain. Cis isomer mein koi mirror plane nahi hai, isliye yeh optical pair ke roop mein bhi milta hai. Kul teen stereoisomers: trans, cis-d aur cis-l.
2. $[F e (C_{2} O_{4})_{3}]^{3 -}$
  Solution dikhao
  Sirf optical: har $[M (A A)_{3}]$ complex ki tarah yeh ek propeller hai jo kisi bhi taraf mud sakta hai, jisse enantiomers ka ek pair milta hai. Koi geometrical isomers nahi hain.
3. $[C o (N H_{3})_{5} (S C N)] C l_{2}$
  Solution dikhao
  Linkage: thiocyanate S se ya N se bonded. Ionisation: ek chloride thiocyanate se jagah badal sakta hai, jisse $[C o C l (N H_{3})_{5}]^{2 +}$ milta hai aur chloride aur thiocyanate counter ions bante hain. Inhe milane se aur isomers milte hain.
4. square planar $[P d (N H_{3})_{2} (S C N)_{2}]$
  Solution dikhao
  Geometrical: cis aur trans. Linkage: har thiocyanate S- ya N-bonded ho sakta hai, isliye har geometrical form ke linkage isomers bhi hain.
Short answerMediumNCERT Intext Question 5.4
Simple tests se kaise dikhaoge ki $[C r (N H_{3})_{5} (N O_{3})] S O_{4}$ aur $[C r (N H_{3})_{5} (S O_{4})] N O_{3}$ ionisation isomers hain?
Solution dikhao
Paani mein pehla $[C r (N H_{3})_{5} (N O_{3})]^{2 +}$ aur free $S O_{4}^{2 -}$ deta hai; doosra $[C r (N H_{3})_{5} (S O_{4})]^{+}$ aur free $N O_{3}^{-}$ deta hai.
Barium chloride sirf pehle solution ke saath $B a S O_{4}$ ka safed precipitate deta hai. Brown-ring test sirf doosre mein free nitrate detect karta hai. Inki molar conductivities bhi alag hain, kyunki ek 2 : 2 electrolyte hai aur doosra 1 : 1 electrolyte. Same formula, alag free ions: ionisation isomers.

5.5 Coordination Compounds mein Bonding

Werner ki picture batati hai ki metal se kaun se groups jude hain, lekin kyun nahi batati. Ek achhi bonding theory ko complexes ki shapes bhi samjhani chahiye, aur yeh bhi ki kai complexes magnetic field mein kyun kheenche jaate hain. Use yeh bhi batana chahiye ki itne saare coloured kyun hote hain, aur khaas taur par transition metals itne zyada complexes kyun banate hain. Chemists in sawaalon ka jawab models ki ek seedhi se dete hain. Yeh unit pehli do seedhiyan chadhti hai, valence bond theory (VBT) aur crystal field theory (CFT). Do aur advanced models, ligand field theory aur molecular orbital theory, tum aage padhoge.

5.5.1 Valence Bond Theory

Valence bond theory maanti hai ki har ligand ek lone pair offer karta hai, isliye metal ko har ligand ki taraf aim kiya ek khaali orbital dena padta hai. Woh apne kuch khaali orbitals ko mila kar utne hi identical hybrid orbitals banata hai jo polyhedron ke corners ki taraf point karte hain. Yeh set ya to andar ke $(n - 1) d$ orbitals ko $n s$ aur $n p$ ke saath milakar banta hai, ya $n s$ aur $n p$ ko bahar ke $n d$ orbitals ke saath milakar. Phir har lone pair ek hybrid ko bharta hai aur coordinate $σ$ bond banata hai, aur hybridisation ka kind shape tay karta hai.

Table 5.5Hybrid orbital sets aur unki di hui shapes (CN = coordination number)

Hybrid set	Shape (CN)	Example
$s p^{3}$	tetrahedral (4)	$[Z n (N H_{3})_{4}]^{2 +}$
$d s p^{2}$	square planar (4)	$[N i (C N)_{4}]^{2 -}$
$s p^{3} d$	trigonal bipyramidal (5)	$[C u C l_{5}]^{3 -}$
$s p^{3} d^{2}$	octahedral, outer orbital (6)	$[F e (H_{2} O)_{6}]^{2 +}$
$d^{2} s p^{3}$	octahedral, inner orbital (6)	$[F e (C N)_{6}]^{4 -}$

Iron(II) dikhata hai ki ek hi ion do octahedral sets mein se kaise chun sakta hai (Fig. 5.9). Free $F e^{2 +}$ ion mein chhah 3d electrons hain, jinmein se chaar unpaired hain. Cyanide ions 3d electrons ko teen orbitals mein ikattha kar dete hain, jisse do 3d orbitals khaali ho jaate hain. Yeh 4s aur teen 4p orbitals ke saath $d^{2} s p^{3}$ hybridisation mein judte hain, aur $[F e (C N)_{6}]^{4 -}$ mein koi unpaired electron nahi bachta: yeh diamagnetic hai. Kyunki yeh andar ke 3d orbitals use karta hai, ise inner orbital, low spin ya spin-paired complex kehte hain.

Water molecules electrons ko pair hone par majboor nahi karte. Tab ion bahar ke do 4d orbitals tak pahunchta hai, $s p^{3} d^{2}$ hybrids bante hain, aur $[F e (H_{2} O)_{6}]^{2 +}$ apne chaaron unpaired electrons rakhta hai. Yeh outer orbital, high spin ya spin-free complex hai, aur paramagnetic hai.

3/3Koi pairing nahi; uski jagah do 4d orbitals use hote hain: $s p^{3} d^{2}$ , outer orbital, chaar unpaired.

Fig. 5.9Strong-field cyanide 3d electrons ko pair karta hai aur inner orbital complex deta hai; weak-field water aisa nahi karta.NCERT p. 129

3d, 4s, 4p aur 4d label wale orbital boxes ki rows, jinmein electrons up aur down arrows se bane hain; ligands ke diye electron pairs shaded boxes mein bane hain. Step 1 free iron(II) ion dikhata hai: paanch 3d boxes mein ek pair aur chaar single electrons, 4 unpaired, baaki sab khaali. Step 2 cyanide complex jodta hai: chhah 3d electrons teen boxes mein paired hain, aur chhah shaded ligand pairs aakhri do 3d boxes, 4s aur teen 4p boxes bharte hain, jinpar d2sp3 ka bracket hai, 0 unpaired. Step 3 water complex jodta hai: 3d electrons free ion jaise hi rehte hain, aur chhah ligand pairs 4s, teen 4p boxes aur do 4d boxes bharte hain, jinpar sp3d2 ka bracket hai, 4 unpaired.

Four-coordinate complexes bhi yahi logic follow karte hain (Fig. 5.10). Nickel(II) $3 d^{8}$ hai, jismein do unpaired electrons hain. Chloride ions ke saath yeh 4s aur teen 4p orbitals use karta hai; $s p^{3}$ hybrids tetrahedron ke corners ki taraf point karte hain, aur $[N i C l_{4}]^{2 -}$ do unpaired electrons ke saath paramagnetic rehta hai. Cyanide 3d electrons ko pair kar deta hai, jisse ek 3d orbital khaali hota hai, isliye $d s p^{2}$ hybrids bante hain; yeh ek plane mein hote hain, aur $[N i (C N)_{4}]^{2 -}$ square planar aur diamagnetic hai. $[N i (C O)_{4}]$ mein nickel ka oxidation state zero hai ( $3 d^{8} 4 s^{2}$ ). Carbon monoxide do 4s electrons ko 3d mein dhakel deta hai, jisse $3 d^{10}$ banta hai; khaali 4s aur 4p orbitals $s p^{3}$ hybrids banate hain, isliye molecule tetrahedral aur diamagnetic hai.

3/3Pairing se ek 3d orbital khaali hota hai: $d s p^{2}$ , square planar $[N i (C N)_{4}]^{2 -}$ , diamagnetic.

Fig. 5.10Wahi nickel(II) ion ek paramagnetic tetrahedral chloride complex aur ek diamagnetic square planar cyanide complex banata hai.NCERT p. 130

3d, 4s aur 4p ke boxes electron arrows ke saath; shaded boxes mein ligands ke electron pairs hain. Step 1 free nickel(II) ion dikhata hai jiske 3d mein teen pairs aur do single electrons hain, 2 unpaired. Step 2 tetrahedral chloride complex jodta hai: 3d wahi rehta hai, chaar ligand pairs 4s aur teen 4p boxes mein, jinpar sp3 ka bracket hai, 2 unpaired. Step 3 square planar cyanide complex jodta hai: aath 3d electrons chaar boxes mein paired hain, aur chaar ligand pairs aakhri 3d box, 4s aur do 4p boxes bharte hain, jinpar dsp2 ka bracket hai, 0 unpaired, aur ek 4p box khaali bachta hai.

5.5.2 Coordination Compounds ki Magnetic Properties

Paramagnetic sample magnetic field mein kheencha jaata hai, diamagnetic sample halka sa bahar dhakela jaata hai. Sample ko strong magnet ke paas aur uske bina tol kar (magnetic balance) uska magnetic moment milta hai, aur moment se unpaired electrons ka number $n$ pata chalta hai. First-row complexes ke liye achhe approximation mein sirf electron spins hi contribute karte hain:

spin-only magnetic moment

μ = \sqrt{n (n + 2)} B M

(5.1)

jahan

$μ$: Bohr magnetons (BM) mein magnetic moment
$n$: unpaired electrons ka number

Table 5.6Spin-only magnetic moments

Unpaired electrons, n	μ (BM)
1	1.73
2	2.83
3	3.87
4	4.90
5	5.92

Ab neeche table ke pairs compare karo. Wahi ion cyanide ki tulna mein water ke saath zyada unpaired electrons rakhta hai, isliye kuch ligands d electrons ko pair hone par majboor karte hain aur kuch nahi. Orbitals ginne se pata chalta hai ki valence bond theory ke liye yeh kyun matter karta hai. $d^{2} s p^{3}$ hybrids ko do khaali 3d orbitals chahiye. Teen ya usse kam d electrons wale ion, jaise $C r^{3 +}$ , ke paas yeh pehle se hote hain, isliye $[C r (N H_{3})_{6}]^{3 +}$ apne teeno unpaired electrons rakhta hai. Chaar, paanch ya chhah d electrons wala ion do 3d orbitals sirf kuch electrons ko pair karke hi khaali kar sakta hai.

Table 5.7Same ion, alag ligands: outer orbital (water) aur inner orbital (cyanide) complexes

Ion	Water ke saath	Cyanide ke saath
$C r^{2 +}$ ( $d^{4}$ )	$[C r (H_{2} O)_{6}]^{2 +}$ : 4 unpaired	$[C r (C N)_{6}]^{4 -}$ : 2 unpaired
$M n^{2 +}$ ( $d^{5}$ )	$[M n (H_{2} O)_{6}]^{2 +}$ : 5 unpaired	$[M n (C N)_{6}]^{4 -}$ : 1 unpaired
$F e^{2 +}$ ( $d^{6}$ )	$[F e (H_{2} O)_{6}]^{2 +}$ : 4 unpaired	$[F e (C N)_{6}]^{4 -}$ : 0 unpaired

Valence bond theory cyanide complexes ko inner orbital ( $d^{2} s p^{3}$ ) aur aqua complexes ko outer orbital ( $s p^{3} d^{2}$ ) batati hai, lekin use batana padta hai ki kaun se ligands pairing karwate hain. $N i^{2 +}$ jaisa $d^{8}$ ion kabhi do 3d orbitals khaali nahi kar sakta, isliye uske octahedral complexes, jaise $[N i (N H_{3})_{6}]^{2 +}$ , hamesha do unpaired electrons wale outer orbital complexes hote hain.

Example 5.7NCERT Example 5.7 · p. 131

Magnetism se geometry

Magnetic measurements four-coordinate ion $[C o C l_{4}]^{2 -}$ mein teen unpaired electrons dikhate hain. Kya yeh tetrahedral hai ya square planar? Iska spin-only moment nikaalo.

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

5.5.3 Valence Bond Theory ki Limitations

Valence bond theory kai complexes ki shapes aur magnetism samjhati hai, lekin iski saaf kamzoriyan hain:

Yeh kuch nahi batati ki complexes coloured kyun hote hain.
Yeh pehle se weak-field aur strong-field ligands mein fark nahi kar sakti; use batana padta hai ki kaun se ligands pairing karwate hain.
Yeh predict nahi kar sakti ki four-coordinate complex tetrahedral hoga ya square planar.
Yeh sirf unpaired electrons ginti hai aur measured magnetic moments ka exact size explain nahi kar sakti.
Yeh koi numbers nahi deti ki complex kitna stable hai, na thermodynamically na kinetically.
Yeh assumptions par tiki hai, jaise kaun se orbitals hybridise hote hain, jo predict karne ki jagah known facts se fit karne ke liye chune jaate hain.

Quick check

3 mein se 1

[N i (C N)_{4}]^{2 -}

diamagnetic hai. Valence bond theory ise kaun si hybridisation aur shape deti hai?

s p^{3}

, tetrahedral

d s p^{2}

, square planar

s p^{3} d^{2}

, octahedral

d^{2} s p^{3}

, octahedral

Answer dekho

Answer (B) $d s p^{2}$ , square planar

Aath 3d electrons ko pair karne se ek 3d orbital khaali hota hai, jo 4s aur do 4p orbitals ke saath milta hai:

d s p^{2}

, square planar. Tetrahedral

s p^{3}

complex do unpaired electrons rakhta.

5.5.4 Crystal Field Theory

Valence bond theory ko batana padta tha ki kaun se ligands electrons pair karte hain. Crystal field theory ise predict karti hai, yeh poochkar ki jab ligands metal ke aas-paas ikatthe hote hain to paanch d orbitals ki energies kaise badalti hain. Har d orbital ko negative charge ka ek badal socho jiski apni shape aur direction hai. Akele ion mein paanchon badalon ki energy barabar hoti hai: woh degenerate hain. Ab ligands ko paas lao, jinhe sirf negative point charges (anions) ya andar ki taraf negative end wale dipoles ( $H_{2} O$ , $N H_{3}$ ) maana jaata hai. Repulsion har orbital ko upar uthata hai, lekin barabar nahi: jo badal seedha ligand ki taraf aim kiya hai woh ligands ke beech se nikalne wale badal se zyada upar dhakela jaata hai. Agar charge ek sphere par barabar phaila ho to paanchon saath-saath upar uthte. Yeh model covalent bonding ko poori tarah ignore karta hai, ek kamzori jis par hum Section 5.5.6 mein wapas aayenge.

1/2 $d_{x^{2} - y^{2}}$ seedha ligands ki taraf point karta hai. $d_{z^{2}}$ $z$ ke saath yahi karta hai.

Fig. 5.11Octahedral field mein ligands ki taraf aim kiye orbitals upar uthte hain; unke beech aim kiye orbitals neeche jaate hain.NCERT p. 131

Ek metal nucleus wahan hai jahan dashed x aur y axes cross karte hain, aur har axis par same distance par ek ligand L hai. Step 1 d x squared minus y squared orbital dikhata hai: chaar blue lobes axes ke saath hain aur seedha chaar ligands ki taraf point karte hain; text kehta hai strong repulsion, energy rises, aur yeh ki yeh orbital e g set ka hai. Step 2 lobes ko 45 degrees ghumata hai aur unhe green rang deta hai taaki d x y orbital dikhe, jiske lobes ligands ke beech point karte hain; text kehta hai weaker repulsion, energy falls, aur yeh ki yeh t 2 g set ka hai.

(a) Octahedral field mein splitting

Chhah ligands ko $\pm x$ , $\pm y$ aur $\pm z$ ke saath rakho. $d_{x^{2} - y^{2}}$ aur $d_{z^{2}}$ orbitals unki taraf point karte hain, sabse zyada repulsion mehsoos karte hain aur upar uthte hain; yeh $e_{g}$ set banate hain. $d_{x y}$ , $d_{x z}$ aur $d_{y z}$ orbitals axes ke beech point karte hain aur neeche jaate hain; yeh $t_{2 g}$ set banate hain. Yahi crystal field splitting (क्रिस्टल क्षेत्र विपाटन) hai, aur energy gap ko $Δ_{o}$ (o octahedral ke liye) likhte hain. Average energy nahi badalti, isliye do $e_{g}$ orbitals total mein utna hi upar uthte hain jitna teen $t_{2 g}$ orbitals neeche jaate hain:

\begin{aligned} E (e_{g}) & = + \frac{3}{5} Δ_{o} \\ E (t_{2 g}) & = - \frac{2}{5} Δ_{o} \end{aligned}

(5.2)

jahan

$Δ_{o}$: octahedral crystal field splitting energy
$E$: spherical-field average ke relative energy

3/4Chhah ligands: $e_{g}$ $\frac{3}{5} Δ_{o}$ upar, $t_{2 g}$ $\frac{2}{5} Δ_{o}$ neeche.

Fig. 5.12Wahi paanch orbitals octahedron ke aas-paas ek tarah split hote hain aur tetrahedron ke aas-paas ulti tarah, kam split ke saath.NCERT Fig. 5.8, 5.9 · p. 132

Left mein ek energy axis upar point karta hai. Paanch chhote bars paanch d orbitals ke liye hain: teen blue (dxy, dxz, dyz) aur do orange (dx2−y2, dz2). Step 1: paanchon same neeche level par hain, jis par free ion likha hai. Step 2: paanchon saath-saath ek dashed line tak upar jaate hain jis par average likha hai, yaani spherical field. Step 3: octahedral field mein do orange bars Delta o ke teen-paanchwe hisse jitna upar jaakar e g banate hain aur teen blue bars Delta o ke do-paanchwe hisse jitna neeche jaakar t 2 g banate hain, aur ek double arrow Delta o mark karta hai. Step 4: tetrahedral field mein pattern ulat jaata hai aur chhota ho jaata hai: teen blue bars Delta t ke do-paanchwe hisse jitna upar jaakar t 2 banate hain, do orange bars Delta t ke teen-paanchwe hisse jitna neeche jaakar e banate hain, aur ek note kehta hai ki Delta t, Delta o ka chaar-nauwan hissa hai.

$Δ_{o}$ ka size metal ion aur ligands par depend karta hai. Metal par zyada charge ligands ko paas kheenchta hai aur splitting badhata hai, aur group mein neeche 4d aur 5d metals ki taraf jaane se bhi yahi hota hai. Kisi given metal ion ke liye ligands ko unki splitting ke hisaab se rank kiya ja sakta hai, jo unke complexes ki absorb ki hui light se naapi jaati hai. Yeh ranking spectrochemical series hai:

spectrochemical series (weak field → strong field)

\begin{aligned} I^{-} < B r^{-} < S C N^{-} \\ < C l^{-} < S^{2 -} < F^{-} \\ < O H^{-} < C_{2} O_{4}^{2 -} \\ < H_{2} O < N C S^{-} \\ < E D T A^{4 -} < N H_{3} \\ < e n < C N^{-} < C O \end{aligned}

Definition

Spectrochemical series

Ligands ko badhti crystal field splitting ke order mein lagaya gaya, jo unke complexes ke absorption spectra se experimentally mila hai.

ExampleWater ammonia se kam splitting deta hai, aur ammonia cyanide se kam.

Ab d electrons bharo. Octahedral complex mein pehle teen electrons Hund's rule ke hisaab se teen $t_{2 g}$ orbitals mein akele-akele jaate hain. Chauthe electron ke saamne ek choice hai. Woh $e_{g}$ orbital mein ja sakta hai, jiski cost $Δ_{o}$ hai, ya $t_{2 g}$ orbital mein pair ho sakta hai, jiski cost pairing energy $P$ hai, yaani ek orbital mein do electrons ka extra repulsion.

Compare

Chhota gap ya bada gap?

$Δ_{o} < P$ : weak field

$e_{g}$ tak chadhna pairing se sasta padta hai.
Chautha electron upar jaata hai: $t_{2 g}^{3} e_{g}^{1}$ .
Weak-field ligands high-spin complexes dete hain.

$Δ_{o} > P$ : strong field

Pairing chadhne se sasti padti hai.
Chautha electron pair hota hai: $t_{2 g}^{4} e_{g}^{0}$ .
Strong-field ligands low-spin complexes dete hain.

2/3Weak field: chautha electron $e_{g}$ mein jaata hai. Strong field: woh $t_{2 g}$ mein pair hota hai.

Fig. 5.13

d^{4}

se aage

Δ_{o}

ka size

P

ke compare mein high spin aur low spin ke beech faisla karta hai.NCERT p. 132

Do energy diagrams aamne-saamne. Left mein weak field: do e g bars teen t 2 g bars se thoda sa hi upar hain, Delta o less than P mark kiya hai. Right mein strong field: e g bars kaafi upar hain, Delta o greater than P mark kiya hai. Step 1 (d3): dono diagrams mein har t 2 g bar mein ek up arrow hai, 3 unpaired. Step 2 (d4): left mein chautha electron upar e g mein jaata hai, 4 unpaired, high spin, configuration t2g3 eg1; right mein woh t 2 g mein pair hota hai, 2 unpaired, low spin, t2g4. Step 3 (d6): left mein t2g4 eg2 hai jismein 4 unpaired hain; right mein chhahon electrons t 2 g mein paired hain, 0 unpaired.

Yahi muqabla $d^{5}$ , $d^{6}$ aur $d^{7}$ ions ka faisla karta hai. $d^{8}$ , $d^{9}$ aur $d^{10}$ ke liye levels bharne ka sirf ek tareeka hai, isliye ligand strength unpaired electrons ka number nahi badal sakti. Hum har arrangement ko uski crystal field stabilisation energy (CFSE) se score kar sakte hain, $t_{2 g}$ mein $x$ electrons aur $e_{g}$ mein $y$ ginkar:

E_{C F} = (\frac{3}{5} y - \frac{2}{5} x) Δ_{o}

(5.3)

jahan

$x, y$: $t_{2 g}$ aur $e_{g}$ mein electrons ke numbers
$E_{C F}$: free ion mein pehle se maujood pairs ke alawa har electron pair ke liye $P$ jodo

$d^{4}$ se $d^{7}$ tak kisi bhi ion ke liye yeh numbers chalao aur faisla hamesha same aata hai: strong field low-spin arrangement ko favour karta hai aur weak field high-spin ko.

Example

High spin ya low spin?

$F e^{2 +}$ ke liye pairing energy lagbhag $210 k J m o l^{- 1}$ hai. Splitting $[F e (H_{2} O)_{6}]^{2 +}$ mein lagbhag $124 k J m o l^{- 1}$ aur $[F e (C N)_{6}]^{4 -}$ mein lagbhag $393 k J m o l^{- 1}$ hai. Har ion ka configuration, unpaired electrons ka number aur spin-only moment predict karo.

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

(b) Tetrahedral field mein splitting

Tetrahedral complex mein chaar ligands axes ke beech, metal ke aas-paas ek cube ke alternate corners par baithe hote hain. Ab $d_{x y}$ , $d_{x z}$ aur $d_{y z}$ ligands ke zyada paas point karte hain bajaye $d_{x^{2} - y^{2}}$ aur $d_{z^{2}}$ ke. Isliye pattern ulta ho jaata hai: neeche ek pair $e$ aur upar ek trio $t_{2}$ (Fig. 5.12, aakhri step). Sirf chaar ligands ke saath, jinmein se koi bhi kisi orbital ki taraf seedha aim nahi kiya, splitting kaafi chhoti hoti hai. Metal, ligands aur bond length same rakho, to theory deti hai

Δ_{t} = \frac{4}{9} Δ_{o}

(5.4)

Pairing energies aam taur par itne chhote gap se badi hoti hain, isliye low-spin tetrahedral complexes kam milte hain. Labels se subscript g bhi hat jaata hai. Woh letter, German gerade (even) se, batata hai ki metal se hokar invert karne par orbital kaise behave karta hai, ek symmetry jo octahedron mein hai aur tetrahedron mein nahi; isliye tetrahedral sets ko simply $e$ aur $t_{2}$ likhte hain.

InteractiveCrystal field lab

Octahedral ya tetrahedral complex ke liye ek energy diagram jismein d-orbital levels ke do sets bars ki tarah bane hain. Electrons up aur down arrows ki tarah dikhte hain. Buttons d electrons ka number 1 se 10 tak badalte hain aur geometry switch karte hain; ek slider splitting Delta ko pairing energy P ke multiple ke roop mein 0.2 se 3 tak set karta hai. Lab levels ko sabse kam energy wale tareeke se bharta hai aur configuration (jaise t2g4 eg2), unpaired electrons ka number, complex high spin hai ya low spin, Bohr magnetons mein spin-only magnetic moment aur crystal field stabilisation energy dikhata hai, aur utne d electrons wale real ions ki list deta hai.

Fig. 5.14d electrons ka number aur geometry chuno, phir splitting ko pairing energy ke relative slide karo aur electrons ko rearrange hote dekho.

5.5.5 Coordination Compounds mein Colour

Sunlight ya safed lamp ki light mein har visible wavelength hoti hai. Ise ek complex ke solution se guzaaro to wavelengths ka ek band absorb ho jaata hai aur baaki nikal jaata hai. Hamari aankhein bache hue hisse ko hataye gaye band ka complementary colour samajhti hain, yaani colour wheel par uske opposite wala colour (Fig. 5.15). Isliye peeli light absorb karne wala complex violet dikhta hai, aur neeli light absorb karne wala yellow-orange dikhta hai.

2/3 $[C r (H_{2} O)_{6}]^{3 +}$ 575 nm ke paas peeli light absorb karta hai aur violet dikhta hai.

Fig. 5.15Complex us colour ke opposite wala colour dikhta hai jo woh absorb karta hai.NCERT Table 5.3 · p. 133

Ek ring chhah coloured sectors mein bati hai, har ek par uska naam likha hai: upar red, phir clockwise orange, yellow, green, blue aur violet. Step 1 note karta hai ki opposite colours complementary hote hain. Step 2 hexaaquachromium(III) dikhata hai: yellow sector absorbed ke roop mein outline hai, 575 nanometres ke paas, ek arrow wheel ke paar violet sector tak jaata hai, jo dashed line se outline hai, aur text kehta hai ki tumhe violet dikhta hai. Step 3 hexaamminechromium(III) dikhata hai: blue, 465 nanometres ke paas, absorb hota hai aur arrow orange ki taraf point karta hai; text kehta hai ki tumhe yellow-orange dikhta hai.

Crystal field ki bhasha mein colour tab aata hai jab ek electron neeche wale d level se upar wale level par jump karta hai, yaani d–d transition. Absorb hone wale photon mein bilkul gap jitni energy honi chahiye, isliye absorb hui wavelength seedha splitting ko naapti hai. Sabse simple case paani mein titanium(III) jaisa $d^{1}$ ion hai. Sirf ek jump possible hai, $t_{2 g}^{1} e_{g}^{0} \to t_{2 g}^{0} e_{g}^{1}$ , isliye ek hi absorption band hai, green–blue mein 500 nm ke paas, aur $[T i (H_{2} O)_{6}]^{3 +}$ solutions purple dikhte hain. General taur par,

Δ = \frac{N_{A} h c}{λ}

(5.5)

jahan

$λ$: absorb hui wavelength
$N_{A} h c$: $1.196 \times 10^{5} k J m o l^{- 1} n m$ , isliye $k J m o l^{- 1}$ mein $Δ$ = $119 600 / λ$ , jahan $λ$ nm mein hai

Example

Colour se splitting

$[C r (H_{2} O)_{6}]^{3 +}$ sabse zyada 575 nm ke paas absorb karta hai aur $[C r (N H_{3})_{6}]^{3 +}$ 465 nm ke paas. In $d^{3}$ ions ke liye sabse kam energy wala absorption $Δ_{o}$ ke barabar hai. Har ek ke liye $Δ_{o}$ $k J m o l^{- 1}$ mein nikaalo aur result ko spectrochemical series se check karo.

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

Kyunki splitting ligands se aati hai, ligands badalne se colour badalta hai, jaisa Fig. 5.15 ke do chromium(III) complexes ne dikhaya. Kuch ions ka koi d–d colour nahi hota, chahe aas-paas kuch bhi ho: bina d electrons ( $S c^{3 +}$ ) ya poore bhare d set ( $Z n^{2 +}$ ) ke saath koi d–d jump possible nahi, isliye unke complexes colourless hote hain. Ligands hatane se bhi colour ja sakta hai. Neele copper(II) sulphate crystals garam karne par lagbhag safed ho jaate hain: water chale jaane par har $C u^{2 +}$ sulphate ions se ghira hota hai, jinka kaafi weaker field absorption ko visible range se bahar infrared mein le jaata hai. Neeche ki do activities tumhe dono effects dikhati hain.

Try karke dekho

Water se colour

Tumhe chahiye

copper(II) sulphate pentahydrate (blue vitriol)
sookhi hard-glass test tube aur holder
spirit lamp ya burner
paani ke saath dropper

Pehle guess karo

Paani wapas daalne par safed powder ka colour kya ho jaayega?

safed hi rahegaphir se neelaharakaala

Copper(II) sulphate pentahydrate ke kuch neele crystals ko sookhi test tube mein halka garam karo jab tak woh safed na ho jaayein. Tube ko thanda hone do, phir do-teen boondein paani daalo aur dekho.

Garam karne se water nikal jaata hai. Tab har $C u^{2 +}$ sirf sulphate ions se ghira hota hai, jinka weaker field absorption ko infrared mein le jaata hai, isliye solid lagbhag safed dikhta hai. Paani daalne se copper ions ke aas-paas aqua ligands wapas aa jaate hain. Splitting badhti hai, absorption visible spectrum ke red end par wapas aata hai, aur neela colour achhi-khaasi heat release ke saath laut aata hai.

Try karke dekho

Ek gehra neela complex

Tumhe chahiye

dilute copper(II) sulphate solution
dropper bottle mein dilute aqueous ammonia
test tube

Pehle guess karo

Ammonia water se stronger-field ligand hai. Jab

C u^{2 +}

ke aas-paas ammonia water ki jagah leta hai, to colour kaise badalna chahiye?

fade hokar colourless ho jaata haiaur gehra, zyada violet-blue ho jaata haipeela ho jaata hainahi badalta

Ek test tube mein lagbhag 2 mL dilute copper(II) sulphate solution lo aur uska colour note karo.
Dilute aqueous ammonia ek-ek boond karke daalo, har boond ke baad hilao.
Tab tak daalte raho jab tak aur koi change na ho, aur final colour ko shuru wale colour se compare karo.

5.5.6 Crystal Field Theory ki Limitations

Yeh ek fact hai jise point-charge picture explain nahi kar sakti. Fluoride ek chhota ion hai jis par poora negative charge hai, jabki carbon monoxide uncharged aur lagbhag non-polar hai, isliye sirf electrostatics ke hisaab se fluoride kahin zyada strong ligand hona chahiye. Spectrochemical series ulta kehti hai: CO d orbitals ko kisi bhi halide se kaafi zyada split karta hai. Missing ingredient covalency hai. Asli metal–ligand bonds electrons share karte hain, aur khaas taur par CO metal ke d electrons ko apne khaali $π^{*}$ orbitals mein accept karta hai (Section 5.6), jisse gap chauda ho jaata hai. Crystal field theory magnetism aur colour ke liye ek quick, reliable guide bani rehti hai; ligand field theory aur molecular orbital theory, jo tum shayad aage padhoge, covalent hissa jodti hain.

Practice · 6 questions

Practice: bonding, magnetism aur colour: answers ke saath

Short answerMediumNCERT Intext Question 5.5
Nickel(II) laal, square planar $[N i (d m g)_{2}]$ banata hai, jise magnet attract nahi karta, aur tetrahedral $[N i B r_{4}]^{2 -}$ , jise karta hai. Valence bond theory se is fark ko samjhao.
Solution dikhao
Dono mein nickel $N i^{2 +}$ hai ( $3 d^{8}$ , do unpaired electrons). $[N i B r_{4}]^{2 -}$ mein weak bromide ligands 4s aur 4p use karte hain ( $s p^{3}$ , tetrahedral), aur do unpaired 3d electrons bache rehte hain: paramagnetic. $[N i (d m g)_{2}]$ mein aath 3d electrons chaar orbitals mein pair hote hain, aur $d s p^{2}$ hybridisation ke liye ek 3d orbital khaali hota hai: square planar, koi unpaired electron nahi, isliye diamagnetic.
Short answerMediumNCERT Intext Question 5.6
Magnet $[M n C l_{4}]^{2 -}$ ko zor se attract karta hai lekin $[N i (C O)_{4}]$ ko bilkul nahi. Shape reason nahi ho sakti, kyunki dono tetrahedral hain. To reason kya hai?
Solution dikhao
$[M n C l_{4}]^{2 -}$ mein Mn +2 hai: $3 d^{5}$ , aur chloride pairing nahi karwata, isliye paanch unpaired electrons bachte hain ( $s p^{3}$ ). $[N i (C O)_{4}]$ mein Ni 0 hai: $3 d^{8} 4 s^{2}$ ; CO 4s electrons ko 3d mein dhakel kar $3 d^{10}$ deta hai, sab paired, aur khaali 4s aur 4p $s p^{3}$ hybrids banate hain.
Short answerMediumNCERT Intext Question 5.7
$[C r (H_{2} O)_{6}]^{2 +}$ ka magnetic moment 4.9 BM ke kareeb hai, lekin $[C r (C N)_{6}]^{4 -}$ ka sirf lagbhag 3 BM. Is fark ko samjhao.
Solution dikhao
Water weak ligand hai: koi pairing nahi, chaar unpaired electrons wala $s p^{3} d^{2}$ outer orbital complex (4.90 BM). Cyanide strong hai: chaar 3d electrons teen orbitals mein jitna ho sake pair hote hain, do unpaired electrons aur $d^{2} s p^{3}$ hybridisation ke liye do khaali 3d orbitals bachte hain (2.83 BM).
Short answerMediumNCERT Intext Question 5.8
Valence bond theory $[C o (N O_{2})_{6}]^{3 -}$ ko inner orbital complex aur $[N i (N O_{2})_{6}]^{4 -}$ ko outer orbital complex maanti hai, jabki dono metals chhah N-bonded nitrite ligands pakadte hain. Kyun?
Solution dikhao
$C o^{3 +}$ $d^{6}$ hai. Nitrite electrons ko teen 3d orbitals mein pair kar deta hai, aur $d^{2} s p^{3}$ ke liye do khaali ho jaate hain: inner orbital, diamagnetic. $N i^{2 +}$ $d^{8}$ hai; poori tarah pair hone par bhi uske electrons ko chaar 3d orbitals chahiye, isliye sirf ek khaali hai. Isliye octahedral bonding ko 4d orbitals use karne padte hain: $s p^{3} d^{2}$ , outer orbital, do unpaired electrons.
Quick answerEasyNCERT Intext Question 5.9
Gold(III) chloride hydrochloric acid mein square planar ion $[A u C l_{4}]^{-}$ ke roop mein ghulta hai. Iske kitne electrons unpaired hain?
Answer dekho
Answer 0
Au +3 hai ( $5 d^{8}$ ). Square planar $d s p^{2}$ bonding ko ek khaali 5d orbital chahiye, isliye aath electrons chaar orbitals mein pair hote hain: koi unpaired electron nahi.
Short answerMediumNCERT Intext Question 5.10
$[C o (H_{2} O)_{6}]^{2 +}$ aur $[C o (N O_{2})_{6}]^{4 -}$ dono mein cobalt(II) hai, phir bhi ek mein teen unpaired electrons hain aur doosre mein sirf ek. Kaun sa kaun hai? Crystal field theory se samjhao.
Solution dikhao
Water weak-field ligand hai, isliye $Δ_{o} < P$ : $[C o (H_{2} O)_{6}]^{2 +}$ mein saat electrons $t_{2 g}^{5} e_{g}^{2}$ ki tarah phailte hain, teen unpaired. N-bonded nitrite strong-field ligand hai, isliye $Δ_{o} > P$ : saatwa electron $e_{g}$ mein jaane se pehle chhah electrons $t_{2 g}$ bharte hain, jisse $[C o (N O_{2})_{6}]^{4 -}$ mein ek unpaired electron ke saath $t_{2 g}^{6} e_{g}^{1}$ milta hai.

5.6 Metal Carbonyls mein Bonding

Carbon monoxide zyadatar transition metals se seedha judkar homoleptic carbonyls banata hai, yaani aise compounds jinmein CO hi akela ligand hai. Zyadatar volatile liquids ya kam melting point wale solids hote hain, aur unki shapes is baat se tay hoti hain ki har metal kitne CO groups pakadta hai (Fig. 5.16 aur table).

Table 5.8Kuch homoleptic metal carbonyls

Carbonyl	Name	Structure
$[N i (C O)_{4}]$	tetracarbonylnickel(0)	tetrahedral
$[F e (C O)_{5}]$	pentacarbonyliron(0)	trigonal bipyramidal
$[C r (C O)_{6}]$	hexacarbonylchromium(0)	octahedral
$[M n_{2} (C O)_{10}]$	decacarbonyldimanganese(0)	Mn–Mn bond se jude do $M n (C O)_{5}$ square pyramids; koi CO bridges nahi
$[C o_{2} (C O)_{8}]$	octacarbonyldicobalt(0)	Co–Co bond, jismein do CO groups do cobalt atoms ke beech bridge banate hain

Fig. 5.16Har chhoti ball ek CO ligand hai, jo carbon ke through metal se bonded hai.NCERT Fig. 5.13 · p. 136

Chaar panels. Upar left: ek nickel atom jiske chaar chhote CO balls tetrahedron ke corners par hain, label Ni(CO)4, tetrahedral. Upar right: ek iron atom paanch CO balls ke saath, teen beech mein ghere hue aur ek upar ek neeche, label Fe(CO)5, trigonal bipyramidal. Neeche left: ek chromium atom jiske chhah CO balls octahedron ke corners par hain, label Cr(CO)6, octahedral. Neeche right: bond se jude do manganese atoms, har ek par paanch CO balls square pyramid ki tarah lage hain, label Mn2(CO)10, two square pyramids.

Carbon monoxide ek kamzor Lewis base hai jiska dipole lagbhag nahi hai, phir bhi yeh low oxidation states wale metals se bahut mazbooti se bind karta hai. Reason yeh hai ki metal–carbon bond ek saath dono directions mein kaam karta hai.

Forward (σ) donation: carbon ka lone pair, oxygen ka nahi, metal ke ek khaali orbital mein jaata hai aur ligand se metal ki taraf $σ$ bond banata hai.
Back (π) donation: zero oxidation state par d electrons se bhara metal us charge ka kuch hissa side mein CO ke khaali antibonding $π^{*}$ orbitals mein dhakel deta hai, jinki shape uske d orbitals se overlap karne ke liye bilkul sahi hai.

Dono effects ek doosre ki madad karte hain, isiliye bonding ko synergic kehte hain. σ donation se mile electrons metal ko electrons mein ameer banate hain aur kuch wapas dene layak banate hain; electrons wapas dene se metal phir se behtar acceptor ban jaata hai. Aakhir mein metal–carbon bond kisi ek effect se jitna banta usse zyada strong ho jaata hai. Kyunki back-donated electrons CO ke antibonding orbital mein jaate hain, C–O bond thoda weaker ho jaata hai: C–O stretching vibration free CO mein $2143 c m^{- 1}$ se girkar $[C r (C O)_{6}]$ mein lagbhag $2000 c m^{- 1}$ ho jaati hai.

2/3Metal ka ek bhara d orbital CO ke khaali $π^{*}$ orbital mein electrons bhejta hai.

Fig. 5.17σ donation aur π back-donation ek doosre ko mazboot karte hain: M–C bond strong hota hai jabki C–O bond thoda weak ho jaata hai.NCERT Fig. 5.14 · p. 136

Left mein ek metal M triple line se bane C≡O group ke carbon se bonded hai. Step 1: carbon par ek blue lone-pair lobe metal ki taraf point karta hai, aur upar ek arrow, jis par sigma donation likha hai, C se M tak jaata hai. Step 2: M ke aas-paas bhare metal d orbital ke chaar lobes dikhte hain, aur carbon aur oxygen ke upar-neeche CO ke khaali pi-star orbital ke lobes dikhte hain; neeche ek curved arrow, jis par pi back-donation likha hai, metal lobe se CO lobe tak jaata hai. Step 3 right mein do labels jodta hai: M–C stronger, aur C–O weaker, lower stretching frequency ke saath.

Quick check

3 mein se 1

Metal carbonyl ki synergic bonding mein π bond tab banta hai jab

CO apna carbon lone pair metal ko donate karta haioxygen metal ko ek lone pair donate karta haimetal CO ke σ orbital mein electrons donate karta haimetal ka bhara d orbital CO ke khaali

π^{*}

orbital mein electrons donate karta hai

Answer dekho

Answer (D) metal ka bhara d orbital CO ke khaali $π^{*}$ orbital mein electrons donate karta hai

π component metal d orbitals se CO ke antibonding

π^{*}

orbitals mein back-donation hai. Carbon lone pair σ bond banata hai, π bond nahi.

5.7 Coordination Compounds ki Importance aur Applications

Patte ke hare rang se lekar chemotherapy drip ki drug tak, complexes woh kaam karte hain jo simple salts nahi kar sakte. Table inmein se kuch ko ikattha karti hai, is hisaab se group karke ki woh kahan kaam karte hain.

Table 5.9Kaam karte coordination compounds

Area	Substance	Use
Living things	chlorophyll (Mg), haemoglobin (Fe), vitamin $B_{12}$ (Co)	photosynthesis, oxygen transport, healthy red blood cells (pernicious anaemia se bachata hai)
Enzymes	carbonic anhydrase (Zn), cytochromes (Fe)	tissues se $C O_{2}$ jaldi hatate hain; respiration mein electrons le jaate hain
Medicine	cisplatin; EDTA, D-penicillamine, desferrioxamine B	tumours ki growth dheemi karta hai; extra lead, copper ya iron nikaalte hain
Metals nikaalna	$[A u (C N)_{2}]^{-}$ , $[A g (C N)_{2}]^{-}$	crushed ore se gold aur silver ghol kar nikaalte hain; phir zinc metal ko free karta hai
Metals purify karna	$[N i (C O)_{4}]$	impure nickel se volatile carbonyl banta hai aur garam karne par decompose hokar pure nickel deta hai
Industrial catalysis	Wilkinson's catalyst, ek rhodium complex	alkenes ka hydrogenation
Electroplating	$[A g (C N)_{2}]^{-}$ , $[A u (C N)_{2}]^{-}$ baths	smooth, ek jaisi coatings kyunki ek time par sirf thode free metal ions hote hain
Ions detect karna	dimethylglyoxime; thiocyanate; Nessler's reagent, $K_{2} [H g I_{4}]$	rosy-red solid $N i^{2 +}$ dikhata hai; blood-red colour $F e^{3 +}$ dikhata hai; brown colour ammonia dikhata hai
Paani ki hardness	$N a_{2} H_{2} E D T A$ titration	$C a^{2 +}$ aur $M g^{2 +}$ ko bind karta hai; unki alag stabilities se har ek ko estimate kiya ja sakta hai
Photography	$[A g (S_{2} O_{3})_{2}]^{3 -}$	film fix karte waqt hypo unexposed silver bromide ko ghol deta hai

Gold extraction dikhata hai ki complex formation asli kaam kaise karta hai. Crushed ore ko dilute sodium cyanide ke saath hilaya jaata hai aur usmein hawa bheji jaati hai. Oxygen gold ko oxidise karta hai, aur cyanide har $A u^{+}$ ko turant bahut stable linear ion $[A u (C N)_{2}]^{-}$ ke roop mein pakad leta hai:

\begin{aligned} 4 A u + 8 C N^{-} + O_{2} \\ + 2 H_{2} O \\ \overset{}{\to} 4 [A u (C N)_{2}]^{-} \\ + 4 O H^{-} \end{aligned}

Phir saaf solution mein zinc dust daalne par woh gold ko displace kar deta hai, kyunki zinc aur bhi zyada favoured $[Z n (C N)_{4}]^{2 -}$ banata hai:

\begin{aligned} 2 [A u (C N)_{2}]^{-} + Z n \\ \overset{}{\to} [Z n (C N)_{4}]^{2 -} \\ + 2 A u \end{aligned}

Nickel ko bhi isi tarah Mond process se purify kiya jaata hai. Impure nickel par lagbhag 330 K par carbon monoxide guzaarne se gaseous $[N i (C O)_{4}]$ banta hai aur impurities peeche reh jaati hain. Gas ko lagbhag 450 K tak garam karne par pure nickel jam jaata hai aur CO dobara use ke liye free ho jaata hai.

Hardness titration mein EDTA har $C a^{2 +}$ ya $M g^{2 +}$ ion ko chhah donor atoms se lapet leta hai (Fig. 5.18). Paanch chelate rings complexes ko bahut stable banati hain, isliye end point sharp hota hai. Yahi pakad calcium disodium EDTA ko lead poisoning ka ilaaj banati hai: yeh lead ions ko urine ke saath body se bahar le jaata hai. D-penicillamine extra copper nikaalta hai, desferrioxamine B extra iron nikaalta hai, aur cisplatin, Section 5.4.1 wala cis isomer, divide hoti cancer cells mein DNA se bind karta hai aur unhe use copy karne se rok deta hai.

Fig. 5.18Ek EDTA ion octahedral complex ke saare chhah donor atoms deta hai, aur paanch chelate rings banata hai.NCERT p. 137

Centre mein ek calcium ion Ca hai, jo apne aas-paas lage chhah donor atoms se bonded hai: left mein do nitrogen atoms aur upar, right aur neeche chaar oxygen atoms. Moti hari bands EDTA chain dikhati hain: ek band bilkul left mein do nitrogens ko jodti hai, aur har nitrogen se do bands mud kar do oxygens tak jaati hain, ek upar aur ek neeche. Labels kehte hain six donor atoms, two N and four O, aur five chelate rings, saath mein calcium–EDTA complex ion ka formula.

Quick check

3 mein se 1

Ore se leach karne par gold solution mein kis form mein rehta hai?

[A g (S_{2} O_{3})_{2}]^{3 -}

[N i (C O)_{4}]

[Z n (C N)_{4}]^{2 -}

[A u (C N)_{2}]^{-}

Answer dekho

Answer (D) $[A u (C N)_{2}]^{-}$

Gold dicyanidoaurate(I),

[A u (C N)_{2}]^{-}

, ke roop mein ghulta hai.

[Z n (C N)_{4}]^{2 -}

baad mein banta hai, jab zinc gold ko displace karta hai.

Recap

Summary aur key points

Coordination compound mein ek metal atom ya ion ligands ka ek definite set pakadta hai, aur har ligand ek electron pair donate karta hai (Section 5.2).
Werner ki primary valence (ionisable, yaani oxidation state) aur secondary valence (coordination number, direction mein fixed) silver nitrate aur conductivity results ko samjhati hain (Section 5.1).
Ligands uni-, di- ya polydentate ho sakte hain; do ya zyada atoms se bind karne wala ligand chelate banata hai, aur ambidentate ligand do donor atoms mein se koi bhi use kar sakta hai ((c) Ligands).
Coordination number donor atoms ko ginta hai; oxidation number charge balance se nikalta hai (coordination number, oxidation number).
IUPAC formulas mein pehle metal, phir ligands alphabetically aate hain; names mein pehle cation, ligands alphabetically, metal (anions mein -ate ke saath) aur uska oxidation state aata hai (Section 5.3).
Geometrical isomers (cis/trans, fac/mer) square planar aur octahedral complexes mein bante hain, tetrahedral mein kabhi nahi (Section 5.4.1).
$[M (A A)_{3}]$ aur cis- $[M (A A)_{2} B_{2}]$ jaise chiral complexes enantiomers ke roop mein milte hain jo polarised light ko opposite directions mein rotate karte hain (Section 5.4.2).
Linkage, coordination, ionisation aur solvate isomers mein fark yeh hai ki kaun se atoms bonded hain (Section 5.4.3 se Section 5.4.6 tak).
Valence bond theory hybridisation ko shape se jodti hai aur inner orbital (low spin) aur outer orbital (high spin) complexes samjhati hai; spin-only moment $\sqrt{n (n + 2)}$ BM hai (Section 5.5.1, Section 5.5.2).
Crystal field theory octahedral field mein d orbitals ko $Δ_{o}$ se $t_{2 g}$ aur $e_{g}$ mein, aur tetrahedral field mein $Δ_{t} = \frac{4}{9} Δ_{o}$ se $e$ aur $t_{2}$ mein split karti hai; $Δ_{o}$ versus $P$ high ya low spin tay karta hai (Section 5.5.4).
Complexes coloured hote hain kyunki d–d transitions visible light absorb karte hain; dikhne wala colour absorb hue colour ka complement hai (Section 5.5.5).
Metal carbonyls synergic σ donation aur π back-donation se jude rehte hain; complexes analysis, metallurgy, biology, catalysis aur medicine mein important hain (Section 5.6, Section 5.7).

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 Intext Question 5.1	Practice: formulas aur names · Q1
NCERT Intext Question 5.2	Practice: formulas aur names · Q2
NCERT Intext Question 5.3	Practice: isomerism · Q1
NCERT Intext Question 5.4	Practice: isomerism · Q2
NCERT Intext Question 5.5	Practice: bonding, magnetism aur colour · Q1
NCERT Intext Question 5.6	Practice: bonding, magnetism aur colour · Q2
NCERT Intext Question 5.7	Practice: bonding, magnetism aur colour · Q3
NCERT Intext Question 5.8	Practice: bonding, magnetism aur colour · Q4
NCERT Intext Question 5.9	Practice: bonding, magnetism aur colour · Q5
NCERT Intext Question 5.10	Practice: bonding, magnetism aur colour · Q6
NCERT Exercises 5.1	Exercises · Q1
NCERT Exercises 5.2	Exercises · Q2
NCERT Exercises 5.3	Exercises · Q3
NCERT Exercises 5.4	Exercises · Q4
NCERT Exercises 5.5	Exercises · Q5
NCERT Exercises 5.6	Exercises · Q6
NCERT Exercises 5.7	Exercises · Q7
NCERT Exercises 5.8	Exercises · Q8
NCERT Exercises 5.9	Exercises · Q9
NCERT Exercises 5.10	Exercises · Q10
NCERT Exercises 5.11	Exercises · Q11
NCERT Exercises 5.12	Exercises · Q12
NCERT Exercises 5.13	Exercises · Q13
NCERT Exercises 5.14	Exercises · Q14
NCERT Exercises 5.15	Exercises · Q15
NCERT Exercises 5.16	Exercises · Q16
NCERT Exercises 5.17	Exercises · Q17
NCERT Exercises 5.18	Exercises · Q18
NCERT Exercises 5.19	Exercises · Q19
NCERT Exercises 5.20	Exercises · Q20
NCERT Exercises 5.21	Exercises · Q21
NCERT Exercises 5.22	Exercises · Q22
NCERT Exercises 5.23	Exercises · Q23
NCERT Exercises 5.24	Exercises · Q24
NCERT Exercises 5.25	Exercises · Q25
NCERT Exercises 5.26	Exercises · Q26
NCERT Exercises 5.27	Exercises · Q27
NCERT Exercises 5.28	Exercises · Q28
NCERT Exercises 5.29	Exercises · Q29
NCERT Exercises 5.30	Exercises · Q30
NCERT Exercises 5.31	Exercises · Q31

Quick check: MCQs

5.1 Coordination Compounds ki Werner's Theory4 questions
(g) Central atom ka oxidation number aur (h) homoleptic aur heteroleptic complexes4 questions
5.3.2 Mononuclear Coordination Compounds ke Naam3 questions
5.4.2 Optical Isomerism4 questions
5.4.6 Solvate Isomerism4 questions
5.5.3 Valence Bond Theory ki Limitations3 questions
(b) Tetrahedral field mein splitting4 questions
5.5.6 Crystal Field Theory ki Limitations4 questions
5.6 Metal Carbonyls mein Bonding3 questions
5.7 Coordination Compounds ki Importance aur Applications3 questions

Glossary · 15

Key terms (English + Hindi)

ambidentate ligand: Do alag possible donor atoms wala ligand jo kisi ek se bhi bind karta hai, jaise $N O_{2}^{-}$ (N ya O) ya $S C N^{-}$ (S ya N).Chapter mein pehli baar yahan aaya →

central atom ya ionकेंद्रीय परमाणुcentral atom or ionkendriya paramanu: Coordination entity ke centre wala metal atom ya ion; yeh ligands se electron pairs accept karta hai, isliye Lewis acid ki tarah kaam karta hai.Chapter mein pehli baar yahan aaya →
chelateकीलेटkeelet: Woh ring jo tab banti hai jab ek ligand same metal ion se do ya zyada donor atoms ke through bind karta hai; chelate complexes aam taur par milte-julte non-chelated complexes se zyada stable hote hain.
coordination compoundउपसहसंयोजन यौगिकupsahsanyojan yaugik: Aisa compound jismein ek metal atom ya ion fixed number ke padosi ions ya molecules pakadta hai, aur har ek electron pair donate karta hai, jaise $[C u (N H_{3})_{4}] S O_{4}$ .Chapter mein pehli baar yahan aaya →
coordination entity: Central atom aur us se bonded ligands, square brackets mein likhe; yeh cation, anion ya neutral molecule ho sakta hai.Chapter mein pehli baar yahan aaya →
coordination numberउपसहसंयोजन संख्याupsahsanyojan sankhya: Central atom se $σ$ bonds dwara jude ligand donor atoms ki ginti.Chapter mein pehli baar yahan aaya →
coordination polyhedron: Central atom ke aas-paas donor atoms jo shape banate hain, jaise octahedral, tetrahedral ya square planar.Chapter mein pehli baar yahan aaya →
coordination sphere: Central atom aur uske ligands, square brackets mein band; brackets ke bahar likhe ions counter ions hain.Chapter mein pehli baar yahan aaya →
crystal field splittingक्रिस्टल क्षेत्र विपाटनcrystal kshetra vipatan: Jab ligands metal ion ko gherte hain to d orbitals ke groups ke beech khulne wala energy gap; octahedral field mein $Δ_{o}$ aur tetrahedral mein $Δ_{t}$ .Chapter mein pehli baar yahan aaya →

enantiomer: Aise molecules ya ions ke pair mein se ek jo mirror images hain lekin superimpose nahi ho sakte (yeh chiral hain); dono plane-polarised light ko opposite directions mein rotate karte hain.Chapter mein pehli baar yahan aaya →

homoleptic complex: Sirf ek kind ke ligand wala complex, jaise $[F e (C O)_{5}]$ . Heteroleptic complex mein do ya zyada kinds hote hain.Chapter mein pehli baar yahan aaya →

inner orbital complex: Aisa complex jiske hybrid orbitals andar ke $(n - 1) d$ orbitals use karte hain; $d^{4}$ – $d^{6}$ ions mein yeh d electrons ko pair hone par majboor karta hai (low spin). Outer orbital complex $n d$ orbitals use karta hai aur apne electrons unpaired rakhta hai (high spin).Chapter mein pehli baar yahan aaya →

ligandलिगन्डligand: Aisa ion ya molecule jo central atom ko kam se kam ek electron pair donate karta hai. Yeh jitne donor atoms use karta hai, wahi iski denticity hai.

spectrochemical series: Ligands ki list unki splitting ke order mein, jo unke complexes ki absorb ki hui light se milti hai: halide ions weak end par, $C N^{-}$ aur $C O$ strong end par.Chapter mein pehli baar yahan aaya →
synergic bonding: Metal carbonyls mein do-taraf bonding: CO metal ko lone pair donate karta hai ( $σ$ ) aur metal d electrons wapas CO ke khaali $π^{*}$ orbitals mein donate karta hai ( $π$ ); dono ek doosre ko mazboot karte hain.Chapter mein pehli baar yahan aaya →

Coordination Compounds: Class 12 Chemistry Chapter 5 Hinglish mein

Is chapter mein aap kya seekhenge

5.1 Coordination Compounds ki Werner's Theory

5.2 Coordination Compounds se jude kuch Important Terms ki Definitions

(a) Coordination entity aur (b) central atom

(c) Ligands

(d) Coordination number aur (e) coordination sphere

(f) Coordination polyhedron

(g) Central atom ka oxidation number aur (h) homoleptic aur heteroleptic complexes

5.3 Coordination Compounds ki Nomenclature

5.3.1 Mononuclear Coordination Entities ke Formulas

5.3.2 Mononuclear Coordination Compounds ke Naam

Practice: formulas aur names: answers ke saath

5.4 Coordination Compounds mein Isomerism

5.4.1 Geometric Isomerism

5.4.2 Optical Isomerism

5.4.3 Linkage Isomerism

5.4.4 Coordination Isomerism

5.4.5 Ionisation Isomerism

5.4.6 Solvate Isomerism

Practice: isomerism: answers ke saath

5.5 Coordination Compounds mein Bonding

5.5.1 Valence Bond Theory

5.5.2 Coordination Compounds ki Magnetic Properties

5.5.3 Valence Bond Theory ki Limitations

5.5.4 Crystal Field Theory

(a) Octahedral field mein splitting

(b) Tetrahedral field mein splitting

5.5.5 Coordination Compounds mein Colour

5.5.6 Crystal Field Theory ki Limitations

Practice: bonding, magnetism aur colour: answers ke saath

5.6 Metal Carbonyls mein Bonding

5.7 Coordination Compounds ki Importance aur Applications

Summary aur key points

Practice questions answers ke saath (NCERT “Exercises”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye