CBO-4 ResoSir Qt0q96a

Chemical Bonding-IV
 Marked questions are recommended for Revision.

 fpfUgr iz'u nksgjkus ;ksX; iz'u gSA
PART - I : SUBJECTIVE QUESTIONS
Hkkx - I : fo"k;kRed iz'u ¼SUBJECTIVE QUESTIONS½
Section (A) : MOT
[k.M (A) : vkf.od d{kd fl)kUr (MOT)
A-1. Find out the bond order of :
fuEu dk cU/k Øe Kkr dhft,A
(a) H2 (b) H2+ (c) He2 (d) Li2 (e) Be2 (f) B2
Ans. (a) 1 (b) 1/2 (c) 0 (d) 1 (e) 0 (f) 1
Sol. (a) Hydrogen molecule (H2) : H2 : (1s)2
N  Na 2  0
Its bond order, therefore, is = b  1
2 2
(b) Cation of hydrogen molecule (H2+) : H2+ : (1s)1
Its bond order, therefore, is = 1/2 (1 – 0) = 1/2
(c) Helium molecule (He2) : He2 : (1s)2 (*1s)2
Its bond order, therefore, is ½(2 – 2) = 0
(d) Lithium molecule (Li2) : (1s)2 (*1s)2 (2s)2
Its bond order, therefore, is 1/2(4 – 2) = 1.
(e) Beryllium (Be2) : (1s)2 (*1s)2 (2s)2 (*2s)2
(f) Boron (B2) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p1x = 2p1y) (pz)0
gy % (a) gkbMªkstu v.kq (H2) : H2 : (1s)2
N N 20
bldk ca/kØe = b a  1
2 2
(b) gkbMªkstu v.kq dk /kuk;u (H2+) : H2+ : (1s)1
bldk ca/kØe = 1/2 (1 – 0) = 1/2
(c) ghfy;e v.kq (He2) : He2 : (1s)2 (*1s)2
bldk ca/kØe ½(2 – 2) = 0
(d) fyfFk;e v.kq (Li2) : (1s)2 (*1s)2 (2s)2
bldk ca/kØe 1/2(4 – 2) = 1.
(e) csjhfy;e (Be2) : (1s)2 (*1s)2 (2s)2 (*2s)2
bldk ca/kØe 1/2(4 – 4) = 0.
(f) cksjkWu (B2) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p1x = 2p1y) (pz)0
bldk ca/kØe 1/2(6 – 4) = 1.
A-2. Identify the molecules or atoms or ions from the following molecular orbital energy level formulations.
The species should be selected from (B2, C2, O22+, O2, F2, N2)
fuEu vkf.od d{kd ÅtkZ Lrj lw=k ls v.kq vFkok ijek.kq vFkok vk;u dh igpku dhft, A bu iztkfr;ksa (B2, C2,
O22+, O2, F2, N2) esa ls p;fur dhft,A
(a) KK  (2s)2  * (2s)2 (2p x )1 (2p y )1
(b) KK  (2s)2  * (2s)2 (2p x )2 (2p y )2
(c) KK  (2s)2  * (2s)2 (2p z )2 (2p x )2 (2p y )2
(d) KK  (2s)2  * (2s) (2p z )2 (2p x )2 (2p y )2 (2p x )1  * (2p y )1
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Chemical Bonding-IV
(e)
(f) KK σ (2s)2 σ *(2s)2 π( 2p y )2 (2p x )2 σ( 2p z )2
Ans. (a) B2 (b) C2 (c) O22+ (d) O2 , (e) F2 (f) N2
A-3. What is the bond order of underlined species in NO [BF4]?
;kSfxd NO [BF4] esa js[kkafdr iztkfr dk ca/k Øe D;k gS \
Ans. 3
Sol. Complex exists as NO+ and [BF4]– . NO+ is isoelectronic with N2 ; so 1s2, *1s2, 2s2, *2s2, 2p2x =
10 – 4
2p2y, 2pz2 , then its bond order is =3
2
gy % ladqy ;kSfxd NO+ vkSj [BF4]– ds :i esa vfLrRo esa jgrk gSA NO+ rFkk N2 lebysDVªkWfud gS] vr% 1s2, *1s2,
10 – 4
2s2, *2s2, 2p2x = 2p2y, 2pz2 , rc bldk ca/k Øe = 3.
2
Section (B) : Applilcation of MOT

[k.M (B) : MOT ds vuqiz;ksx
B-1. How would you explain that B2 molecule is not diamagnetic?
vki dSls le>kvksxsa fd B2 v.kq izfrpqEcdh; ugh gksrk gSA
Sol. Boron (B2) : B2 is a good example of the energy level shift caused by the mixing of s and p orbitals. In
the absence of mixing, the g (2p) orbital is expected to be lower in energy than the u(2p) orbitals and
the resulting molecule would be diamagnetic. However, mixing of the g(2s) orbital with the g(2p)
orbital lowers the energy of the g(2s) orbital and increases the energy of the g(2p) orbital to a higher
level than the  orbitals, giving the order of energies shown below. As a result, the last two electrons
are unpaired in the degenerate (having the same energy)  orbitals, and the molecule is paramagnetic.
(1s)2 (*1s)2 (2s)2 (*2s)2 (2p1x = 2p1y) (pz)0.
gy- cksjkWu (B2) : B2 ÅtkZ Lrj foLFkkiu dk ,d loksZÙke mnkgj.k gSa ftlesa ;g foLFkkkiu s o p-d{kdksa ds feykus
(mixing) ds ifj.kkeLo:i izkIr gksrk gSaA s o p-d{kdksa ds ugha feyus dh ifjfLFkfr esa] ;g ekuk tkrk gSa fd g(2p)
d{kdksa dh ÅtkZ u(2p) d{kdksa dh rqyuk esa vis{kkd`r de gksrh gSa] ftlds ifj.kkeLo:i v.kq izfrpqEcdh; gksxkA
blh izdkj ls] g(2p) d{kd ,oe~ g(2s) d{kd ds feyus ij g(2s) d{kd dh ÅtkZ vis{kkd`r de gks tkrh gSa rFkk
g(2p) d{kd dh ÅtkZ  d{kd dh vis{kk mPprj Lrj rd c<+ tkrh gSa] ftlds ifj.kkeLo:i vfUre nks bysDVªkWu
leHkza'k ¼leku ÅtkZ okys½  d{kdksa esa v;qfXer fLFkfr esa gksrs gSa rFkk ;g v.kq vuqpqEcdh; gksrk gSaA
(1s)2 (*1s)2 (2s)2 (*2s)2 (2p1x = 2p1y) (pz)0.
B-2. Explain why NO+ is more stable towards dissociation into its atoms than NO ?
NO+ blds ijek.kqvksa esa fo;ksftr gksus ds izfr] NO dh rqyuk esa vf/kd LFkk;h gS] D;ksa \
Sol. NO+ and NO are derivative of N2; so NO+ bond order = 3 and NO bond order = 2.5; B.O. bond
strength. 
gy % NO+ vkSj NO, N2 ds O;qRiUu gSaA vr% NO+ ca/k Øe = 3 rFkk NO ca/k Øe = 2.5 ; ca/k Øe (B.O) ca/k lkeF;ZA
B-3. Which of the following are gerade molecular orbitals?
fuEu esa ls dkSuls ftjsM vkf.od d{kd (gerade molecular orbitals) gS \
(i) *2s (ii) 2pz (iii) 2py (iv) *2px
Ans. (ii) & (iv)
B-4. Arrange following compounds in the order of increasing order of O–O bond length.
(i) O2 (ii) O2[BF4] (iii) KO2
fuEu ;kSfxdksa dks O–O cU/k yEckbZ ds c<+rs gq, Øe esa O;ofLFkr dhft,A
(i) O2 (ii) O2[BF4] (iii) KO2
Ans. O–O bond length order is ii < i < iii
Ans. O–O ca/k yEckbZ (bond length) dk Øe ii < i < iii gSA
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Chemical Bonding-IV
Section (C) : Metallic bonding
[k.M (C) : /kkfRod cU/k
C-1. Zinc has lowest melting point in 3d-series elements. Why ?
3d-Js.kh ds rRoksa esa ftad dk xyukad U;wu gksrk gS] D;ksa \
Sol. Weakest metallic bonding amongst the 3d-series elements  no unpaired electrons available for
metallic bonding in case of zinc.
3d-Js.kh rRoksa esa nqcZyre /kkfRod cU/k  ftad dh ifjfLFkfr esa /kkfRod cU/ku ds fy, dksbZ v;qfXer bysDVªkWu
miyC/k ugha gksrk gSA
C-2. Among Be and Li, which should have higher melting point and why ?
Be o Li esa ls fdldk xyukad mPp gksrk gS o D;ksa ?
Sol. Be should have higher melting point as it contain 2 electrons for metallic bonding where as Li contain
only one. Further more, size of Be is smaller than that of Li.
Be dk xyukad mPp gksuk pkfg, D;ksafd ;g /kkfRod cU/ku ds lkFk 2 bysDVªkWu j[krk gS tcfd Li dsoy 1e– j[krk
gSA vkxs Be dk vkdkj Li ls de gksrk gSA
PART - II : ONLY ONE OPTION CORRECT TYPE

Hkkx - II : dsoy ,d lgh fodYi çdkj (ONLY ONE OPTION CORRECT TYPE)
Section (A) : MOT
[k.M (A) : vkf.od d{kd fl)kUr (MOT)
A-1. During the formation of a molecular orbital from atomic orbitals of the same atom, probability of electron
density is :
(A) none zero in the nodal plane (B) maximum in the nodal plane
(C*) zero in the nodal plane (D) zero on the surface of the lobe
leku ijek.kq ds ijek.oh; d{kd ls ,d vkf.od d{kd ds fuekZ.k ds nkSjku] bysDVªkWu ?kuRo dh izkf;drk fuEu gSa %
(A) uksMy ry esas v'kwU; (B) uksMy ry esa vf/kdre
(C*) uksMy ry esa 'kwU; (D) ikyh dh lrg ij 'kwU;
Sol. The electron density is zero in the nodal plane during the formation of a molecular orbital from atomic
orbitals of the same atom.
gy % leku ijek.kq ds ijek.oh; d{kd ls ,d vkf.od d{kd ds fuekZ.k esa uksMy ry esa bysDVªkWu ?kuRo 'kwU; gksrk gSA
A-2. If Z-axis is the molecular axis, then -molecular orbitals are formed by the overlap of
;fn Z-v{k vkf.od v{k gS] rc -vkf.od d{kd fdlds vfrO;kiu ls cusxs µ
(A) s + pz (B) px + py (C) pz + pz (D*) px + px
A-3. Bond order is a concept in the molecular orbital theory. It depends on the number of electrons in the
bonding and antibonding orbitals. Which of the following statements is true about it ? The bond order
(A) Can have a negative quantity
(B) Has always an integral value
(C*) Can assume any positive or integral or fractional value including zero
(D) Is a non zero quantity
ca/k Øe] vkf.od d{kd fl}kUr esa ,d vo/kkj.kk gSaA ;g ca/kh rFkk izfrca/kh d{kdksa esa bysDVªkWuksa dh la[;k ij fuHkZj
djrk gSaA ca/k Øe ds lUnHkZ esa dkSulk dFku lgh gS \
(A) ca/k Øe ,d _.kkRed ek=kk gks ldrh gSA
(B) ca/k Øe lnSo ,d iw.kk±d eku gksrk gSa
(C*) ca/k Øe /kukRed ;k iw.kk±d ;k fHkUukRed eku ;k 'kwU; gks ldrk gSA
(D) ca/k Øe ,d v'kwU; ek=kk gSA
A-4. Which of the following pairs have identical values of bond order ?
(A*) N2+ and O2+ (B) F2 and Ne2 (C) O2 and B2 (D) C2 and N2
fuEu esa ls fdl ;qXe ds fy, cU/k&Øe dk eku leku gSa \
(A*) N2+ rFkk O2+ (B) F2 rFkk Ne2 (C) O2 rFkk B2 (D) C2 rFkk N2
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Chemical Bonding-IV
Sol. (A) Bond order of N2+ = 2.5 ; The bond order of O2+ 1/2(10 – 5) = 2.5.
(B) Bond order of F2 = 1 ; The bond order of Ne2 = 0.
(C) Bond order of O2 = 2 ; The bond order of B2 = 1.
(D) Bond order of C2 = 2 ; The bond order of N2 = 3.
Sol. (A) N2+ dk ca/k Øe = 2.5 ; O2+ dk ca/k Øe 1/2(10 – 5) = 2.5.
(B) F2 dk ca/k Øe = 1 ; Ne2 dk ca/k Øe = 0.
(C) O2 dk ca/k Øe = 2 ; B2 dk ca/k Øe = 1.
(D) C2 dk ca/k Øe = 2 ; N2 dk ca/k Øe = 3.
A-5. Which of the following molecules/ions exhibit sp mixing?
(A) B2 (B) C22– (C) O2+ (D*) Both (A) and (B)
fuEu esa ls dkSuls v.kq@vk;u sp feJ.k n'kkZrs gS\
(A) B2 (B) C22– (C) O2+ (D*) (A) o (B) nksuksa
A-6. The common features of the species N22– , O2 and NO– are :
(A) bond order three and isoelectronic. (B*) bond order two and isoelectronic.
(C) bond order three but not isoelectronic. (D) bond order two but not isoelectronic.
N22– , O2 rFkk NO– Lih'kht ds fy, leku vfHkyk{kf.kd xq.k gSa %
(A) ca/k Øe rhu rFkk lebysDVªkWfud gSaA (B*) ca/k Øe nks rFkk lebysDVªkWfud gSaA
(C) ca/k Øe rhu ysfdu lebysDVªkWfud ugha gSaA (D) ca/k Øe nks ysfdu lebysDVªkWfud ughaA
Sol. N2 : 1s *1s 2s *2s 2px  2py 2pz 2px1 *2py1.
2– 2 2 2 2 2 2 2
10 – 6 10 – 6
B.O. N22– = = 2. ; B.O. O2 = = 2.
2 2
10 – 6
NO– isoelectronic with O2 so B.O. = = 2.
2
All have same number of electrons (i.e. 16) so isoelectronic.
gy- N22– : 1s2 *1s2 2s2 *2s2 2px2  2py2 2pz2 2px1 *2py1.
10 – 6 10 – 6
N22– dk ca/k Øe = = 2. ; O2 dk ca/k Øe = = 2.
2 2
10 – 6
NO– , O2 dk lebysDVªkWfud gS] vr% ca/k Øe = = 2.
2
lHkh leku la[;k esa bysDVªkWu j[krs gS] (vFkkZr~ 16) vr% lebysDVªkWfud gSaA
A-7. Which of the following molecular orbitals has two nodal planes.
fuEu esa ls dkSuls vkf.od d{kd esa nks uksMy ry gS \
(A) 2s (B) 2py (C*) *2py (D) *2px
Section (B) : Applilcation of MOT
[k.M (B) : MOT ds vuqiz;ksx
B-1. Among the following species, which has the minimum bond length ?
fuEu Lih'kht esa ls fdldh cU/k yEckbZ U;wure gS \
(A) B2 (B*) C2 (C) F2 (D) O2–
–
Sol. B2 bond order = 1 ; C2 bond order = 2 ; F2 bond order = 1 ; O2 bond order = 1.5
bond order 1/bond length.
B2 ca/k Øe = 1 ; C2 ca/k Øe = 2 ; F2 ca/k Øe = 1 ; O2– ca/k Øe = 1.5
ca/k Øe 1/ca/k yEckbZ
B-2. Which of the following species is paramagnetic ?
fuEu esa ls dkSulh iztkfr vuqpqEcdh; gSa \
(A*) NO– (B) O22– (C) CN– (D) CO
–
Sol. (A) NO is derivative of O2 and isoelectronic with O2.
So (1s)2 (*1s)2 (2s)2 (*2s)2 2pz)2 (2p2x = 2p2y ) (*2px1 = *2p1y) and 2 unpaired electrons.
(B) O22– : (1s)2 (*1s)2 (2s)2 (*2s)2 2pz)2 (2p2x = 2p2y) (*2px2 = *2p2y) and no unpaired
electrons.
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Chemical Bonding-IV
(C) CN– is derivative of and isoelectronic with N2 : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y) 2pz)2
and no unpaired electron.
(D) CO is derivative of and isoelectronic with N2 : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y) 2pz)2
and no unpaired electron.
gy % (A) NO–, O2 dk O;qRiUu gS rFkk O2 ds lkFk lebysDVªksfud (1s)2 (*1s)2 (2s)2 (*2s)2 2pz)2 (2p2x =
2p2y) (*2px1 = *2p1y) vkSj 2 v;qfXer bysDVªkWu gSA
(B) O22– : (1s)2 (*1s)2 (2s)2 (*2s)2 2pz)2 (2p2x = 2p2y) (*2px2 = *2p2y) dksbZ v;qfXer bysDVªksu ugha gSA
(C) CN– dk O;qRiUu gS rFkk N2 ds lkFk lebysDVªksfud : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y) 2pz)2
vkSj dksbZ v;qfXer bysDVªkWu ugha gSA
(D) CO, N2 dk O;qRiUu ,oa lebysDVªkWfud gS (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y) 2pz)2 vkSj dksbZ
v;qfXer bysDVªkWu ugha gSA
B-3. The following molecules / species have been arranged in the order of their increasing bond orders,
Identify the correct order.
fuEu v.kq@Lih'kht dks muds c<+rs gq, cU/k&Øe ds vk/kkj ij O;ofLFkr fd;k x;k gS] lgh Øe dks igpkfu;sA
(I) O2 (II) O2– (III) O22– (IV) O2+
(A*) I I I < I I < I < IV (B) IV < I I I < I I < I (C) I I I < I I < IV < I (D) I I < I I I < I < IV
Sol. O2 O2– O22– O2+
Bond Order 2 1.5 1 2.5
–
gy. cU/k Øe O2 O2 O2 2–
O2+
2 1.5 1 2.5
B-4. Which one is paramagnetic from the following
(A) O2– (B) NO (C*) Both (A) and (B) (D) CN–
fuEu essa ls vuqpqEcdh; gS
(A) O2– (B) NO (C*) (A) rFkk (B) nksuksa (D) CN–
B-5. Which of the following orders is correct in respect of bond dissociation energy ?
(A) N2+ > N2– (B) O2+ > O3 (C) NO+ > NO (D*) All of these
ca/k fo;kstu ÅtkZ ds lanHkZ esa fuEu esa ls lgh Øe dkSulk gS \
(A) N2+ > N2– (B) O2+ > O3 (C) NO+ > NO (D*) mijksDr lHkh
Sol. From MOT & bond order values.
MOT o ca/k Øe ekuks ls
B-6. S1  The HOMO in F2– is *2px = *2py molecular orbitals.

S2  Bond order of O2– is more then O2+.
S3  NO+ is more stable than N2+.
S4  C2 is more stable than C2+.
State, in order, whether S1, S2, S3, S4 are true or false
S1  F2– esa HOMO, *2px = *2py vkf.od d{kd gksrs gSA
S2  O2– esa ca/k Øe O2+ ls vf/kd gksrk gSA
S3  NO+, N2+ dh rqyuk esa vf/kd LFkk;h gSA
S4  C2, C2+ dh rqyuk esa vf/kd LFkk;h gSA
S1, S2, S3, S4 esa lR; ;k vlR; dk Øe gSA
(A) FFFT (B) FTTT (C) FTFT (D*) FFTT
Section (C) : Metallic bonding
[k.M (C) : /kkfRod cU/k
C-1. Iron is harder than sodium because :
(A) iron atoms are smaller. (B) iron atoms are more closely packed.
(C) metallic bonds are stronger in sodium. (D*) metallic bonds are stronger in iron.
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Chemical Bonding-IV
vk;ju] lksfM;e ls dBksj gksrk gSa D;ksfd %
(A) vk;ju ijek.kq NksVs gksrs gSaA (B) vk;ju ijek.kq cgqr ikl ladqfyr gksrs gSaA
(C) lksfM;e esa /kkfRod cU/k izcy gksrk gSaA (D*) vk;ju esa /kkfRod cU/k izcy gksrk gSaA
Sol. The strength of metallic bonds depends upon the number of mobile electron(s) per atom. Sodium has
only one mobile electron per atom where as iron has 8 mobile electrons per atom.
fdlh Hkh /kkfRod ca/k dh izcyrk bl ckr ij fuHkZj djrh gS fd mlesa izfr ijek.kq xfr'khy bysDVªkWu fdrus gSA Na
esa izfr ijek.kq ek=k ,d rFkk vk;ju esa izfr ijek.kq vkB (8) xfr'khy bysDVªkWu gSaA
C-2. The enhanced force of cohesion in metals is due to :
(A) The covalent linkages between atoms
(B) The electrovalent linkages between atoms
(C) The lack of exchange of valency electrons
(D*) The delocalization of valence electron between metallic kernels.
fuEu eas ls fdl dkj.k ls /kkrqvksa esa llatd cy izcy@vf/kd gksrk gS %
(A) ijek.kqvksa ds e/; lgla;kstd ca/k ds dkj.k
(B) ijek.kqvksa ds e/; fo|qr lgla;kstd fyadst ds dkj.k
(C) la;kstdrk bysDVªkWuksa ds fofue; dh gkfu ds dkj.k
(D*) /kkfRod dusZy ds e/; la;ksth bysDVªkWu ds foLFkkuhdj.k ds dkj.k
Sol. Electron sea model of metallic bonding.
/kkfRod ca/k dk bysDVªkWu leqnz ekWMyA
C-3. In the following metals which one has lowest probable interatomic forces.
(A) Copper (B) Silver (C) Zinc (D*) Mercury
fuEu esa ls U;wure laHkkfor vUrj&ijekf.od cyksa okyh /kkrq gS
(A) dkWij (B) flYoj (C) ftad (D*) edZjh
Sol. Cu, Ag, Zn are solids at room temperature where as mercury is liquid.
Cu,Ag, Zn dejs ds rki ij Bksl gS tcfd edZjh nzo gSA
PART - III : MATCH THE COLUMN

Hkkx - III : dkWye dks lqesfyr dhft, (MATCH THE COLUMN)
1. Match the following :
Column – I Column – II
(A) O2 and NO– (p) Same magnetic property and bond order as that in N2+
(B) O2+ and NO (q) Same bond order but not same magnetic property as that in O2
(C) CO and CN– (r) Same magnetic property and bond order as that N22 –
(D) C2 and CN+ (s) Same magnetic property and bond order as that in NO+
fuEu dks lqesfyr dhft,A
dkWye – I dkWye – II
(A) O2 rFkk NO– (p) N2+ ds leku pqEcdh; xq.k rFkk cU/k Øe
(B) O2+ rFkk NO (q) O2 ds leku cU/kØe ysfdu leku pqEcdh; xq.k ugha
(C) CO rFkk CN– (r) N22– ds leku pqEcdh; xq.k rFkk cU/k Øe
(D) C2 rFkk CN+ (s) NO+ ds leku pqEcdh; xq.k rFkk cU/k Øe
Ans. (A – r) ; (B – p) ; (C – s) ; (D – q)
ADVCBO - 6
Chemical Bonding-IV
 Marked questions are recommended for Revision.

 fpfUgr iz'u nksgjkus ;ksX; iz'u gSA
PART - I : ONLY ONE OPTION CORRECT TYPE
Hkkx - I : dsoy ,d lgh fodYi çdkj (ONLY ONE OPTION CORRECT TYPE)
1. Number of antibonding electrons in N2 is :

N2 esa izfrca/kh (antibonding) bysDVªksuksa dh la[;k gSa %
(A*) 4 (B) 10 (C) 12 (D) 14
Sol. (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y) 2pz)2 ; number of anti bonding electrons in N2 is 4.
* represents antibonding molecular orbitals.
gy % (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y) 2pz)2 ; N2 esa izfrca/kh bysDVªkWu dh la[;k 4 gSA
* izfrca/kh vkf.od d{kd dks n'kkZrk gSA
2. Following is the molecular orbital configuration of a diatomic molecule

,d f}ijekf.od v.kq dk vkf.od d{kd vfHkfoU;kl fuEu gS
 2p2y
 1s2 * 1s2  2s2 * 2s2  2p2x 
  2pz
2
Its bond order is : (bldk ca/k Øe gS%)

(A*) 3 (B) 2.5 (C) 2 (D) 1
10  4 6
Sol. B.O. ca/k Øe = = =3
2 2
3. The bond order of He2 molecule ion is :

He2 v.kq vk;u dk ca/k Øe gSA
(A) 1 (B) 2 (C*) 1/2 (D) 1/4
2 1 1
Sol. He = (1s2)(1*s1) B.O. = =
2 2 2
2 1 1
He = (1s2)(1*s1) ca/k Øe = =
2 2 2
4. Which species can exist among the following :
fuEu esa ls dkSulh Lih'kht dk vfLrRo gksxk \
(A*) B2 (B) Be2 (C) Ne2 (D) He2
Sol. B2 : (1s)2( 1s)2( 2s)2( 2s)2(2Px1 = 2Py1)
B.O. ca/k Øe = 5  4 = 1 > 0
2 2
Rest all have zero B.O.
ckdh lHkh ds ca/k Øe 0 gSA
5. Among the following which one will have the largest O – O bond length ?
fuEu esa ls fdlesa vf/kdre O – O cU/k yEckbZ gS %
(A) KO2 (B) O2 (C) O2+ [AsF6] – (D*) K2O2
–
Sol. (A) O2 B.O = 1½ (B) O2 B.O = 2 (C) O2+ = 2½ (D) O22 – = 1
1
Bond order  
bond length
gy- (A) O2– B.O = 1½ (B) O2 B.O = 2 (C) O2+ = 2½ (D) O22 – = 1
1
ca/k Øe  
ca/k yEckb Z
ADVCBO - 7
Chemical Bonding-IV
6. The correct order in which the O–O bond length increases in the following is :
fuEu esa ls dkSuls fodYi esa O–O ca/k yEckbZ ds c<+us dk lgh Øe gS\
(A) H2O2 < O2 < O3 (B) O2 < H2O2 < O3 (C*) O2 < O3 < H2O2 (D) O3 < H2O2 < O2
1
Sol. B.O. 
Bond length
1
ca/k Øe 
ca/k yEckb Z
7. Which of the following is a wrong order with respect to the property mentioned against each ?
(A*) O22– > O2 > O2+ [Paramagnetic moment] (B) (NO)¯ > (NO) > (NO)+ [bond length]
(C) H2 > H2+ > He2+ [bond energy] (D) NO2+ > NO2 > NO2¯ [bond angle]
Lih'kht ds lkeus of.kZr fd, x, xq.kksa ds lanHkZ esa] fuEu esa ls dkSu xyr gS \
(A*) O22– > O2 > O2+ [vuqpqEcdh; vk?kw.kZ] (B) (NO)¯ > (NO) > (NO)+ [cU/k yEckbZ]
(C) H2 > H2+ > He2+ [cU/k ÅtkZ] (D) NO2+ > NO2 > NO2¯ [cU/k dks.k]
Sol. NO– > NO > NO+ (bond length)
Bond order 2.0 2.5 3
H2 > H2+ > He2+ (bond energy)
Bond order 1 0.5 0.5
(In He2+ more electron in antibonding MO's)
NO2+ > NO2 > NO2– (bond angle)
Bond angle 180º 133º 115º
O22– < O2+ < O2 (paramagnetic moment)
No. of unpaired e– 0 1 2
gy% NO– > NO > NO+ (ca/k yEckbZ)
ca/k Øe 2.0 2.5 3
H2 > H2+ > He2+ (ca/k ÅtkZ)
ca/k Øe 1 0.5 0.5
(He2 esa] vkcU/kh MO's esa vf/kd bysDVªkWu gksrs gSA)
+
NO2+ > NO2 > NO2– (ca/k dks.k)

ca/k dks.k 180º 133º 115º
O22– < O2+ < O2 (vuqpEq cdh; vk?kw.kZ)
v;qfXer bysDVªkWuksa dh la[;k 0 1 2
8. Which of the following option with respect to increasing bond dissociation energies is correct ?
cU/k fo;kstu ÅtkZ ds c<+rs gq;s Øe ds lUnHkZ esa dkSulk fodYi lgh gS \
(A) NO < C2 < O2– < He2+ (B) C2 < NO < He2+ < O2–
–
+
(C) He2 < O2 < NO < C2 (D*) He2+ < O2– < C2 < NO
Sol. He2+ bond order = 2  1  1 ; O2– bond order = 10  7  1.5
2 2 2
C2 bond order = 8  4 10  5
 2 ; NO bond order =  2 .5
2 2
Bond order  bond dissociation energy.
gy. He2+ cU/k Øe = 2  1  1 ; O2– cU/k Øe = 10  7  1.5

2 2 2
C2 cU/k Øe 8  4 10  5
=  2 ; NO cU/k Øe =  2 .5
2 2
cU/k Øe  cU/k fo;kstu ÅtkZ
9. Pick out the incorrect statement.
(A) N2 has greater dissociation energy than N2+ (B) O2 has lower dissociation energy than O2+
(C*) Bond length in N2+ is less than N2 (D) Bond length in NO+ is less than in NO.
ADVCBO - 8
Chemical Bonding-IV
vlR; oDrO; NkafV;sA
(A) N2 dh fo;kstu ÅtkZ N2+ ls vf/kd gSaA (B) O2 dh fo;kstu ÅtkZ O2+ ls de gksrh gSA
(C*) N2+ dh cU/k yEckbZ N2 ls de gSaA (D) NO dh cU/k yEckbZ NO ls de gksrh gSA
+
Sol. N2 : (1s) (*1s) (2s) (*2s) (2p x = 2p2y) 2pz)2

2 2 2 2 2
The bond order of N2 is 1/2(10 – 4) = 3.

N2+ : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y) 2pz)1
The bond order of N2+ is 1/2(9 – 4) = 2.5.
O2 : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y) (*2px1 = *2p1y)
The bond order of O2 1/2(10 – 6) = 2.
O2– : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px2 = *2p1y)
The bond order of O2 1/2(10 – 7) = 1.5.
NO+ derivative of O2 and isoelectronic with O22+ ; so (1s)2 (*1s)2 (2s)2 (*2s)2 2pz)2 (2p2x = 2p2y)
The bond order of NO+ 1/2(10 – 4) = 3.
NO derivative of O2 and isoelectronic with O2+; (1s)2 (*1s)2 (2s)2 (*2s)2 2pz)2 (2p2x=2p2y),
*2px)1
The bond order of NO is 1/2(10 – 5) = 2.5.
Bond order 1/bond length  bond dissociation energy.
gy % N2 : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y) 2pz)2
N2 ¼ukbVªkstu½ dk ca/k Øe gS 1/2(10 – 4) = 3.
N2+ : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y) 2pz)1
N2+ dk ca/kØe gSA 1/2(9 – 4) = 2.5.
O2 : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y) (*2px1 = *2p1y)
vkWDlhtu O2 dk ca/k Øe gSA 1/2(10 – 6) = 2.
O2– : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px2 = *2p1y)
O2 dk ca/k Øe gSA 1/2(10 – 7) = 1.5.
NO+ tks O2 dk O;qRiUu gS vkSj O22+; ls lebysDVªksfud gS] (1s)2 (*1s)2 (2s)2 (*2s)2 2pz)2 (2p2x=2p2y)
NO+ dk ca/k Øe gS 1/2(10 – 4) = 3.
NO tks O2 dk O;qRiUu gS rFkk O2+ ; (1s)2 (*1s)2 (2s)2 (*2s)2 2pz)2 (2p2x=2p2y), *2px)1
NO dk ca/k Øe gS is 1/2(10 – 5) = 2.5.
ca/k Øe 1/ca/k yEckbZ  ca/k fo;kstu ÅtkZA
10. The species which are diamagnetic :
çtkfr tks fd çfrpqEcdh; gS %
(A) O2– (B) NO2 (C) ClO2 (D*) N2O4
Sol. N2O4
O2–, NO2 and ClO2 are odd electron species.

O2–, NO2 rFkk ClO2 ds ikl v;qfXer e– gSA
11. Which of the following is observed in metallic bonds ?

(A*) Mobile valence electrons (B) Localised electrons
(C) Highly directed bond (D) None of these
/kkfRod cU/kksa esa fuEu esa ls fdldks izsf{kr fd;k tkrk gS \
(A*) xfr'khy la;kstdrk bySDVªkWu (B) LFkkuh;d`r bySDVªkWu
(C) mPp funsZf'kr ¼fn'kkRed½ cU/k (D) dksbZ ugha
Sol. Metallic bond results from the electrical attractions among positively charged metal ions and mobile,
delocalised electrons belonging to the crystal as a whole.
gy % /kuvkosf'kr /kkfRod vk;uksa rFkk xfr'khy foLFkkuhd`r bysDVªkWu ds e/; oS|qr vkd"kZ.k ds QyLo:i /kkfRod ca/k
curk gSA
ADVCBO - 9
Chemical Bonding-IV
PART - II : SINGLE OR DOUBLE INTEGER TYPE
Hkkx - II : ,dy ;k f}&iw.kk±d eku izdkj (SINGLE OR DOUBLE INTEGER TYPE)
1. Find the no. of species having fractional bond order ?
fuEu esa ls fdruh Lih'kht esa fHkUukRed ca/k dks.k mifLFkr gS\
(a) N2+ (b) N2– (c) O2 (d) O2+
(e) F2 (f) B2 (g) C2+ (h) CN–
(i) NO+
Ans. 4 (a, b, d, g)
Sol. N2+, N2– , O2+, C2+ have fractional bond order.
Sol. N2+, N2– , O2+, C2+ fHkUukRed ca/k dks.k j[krs gSA
2. Find out the no. of correct statements :

(a) Bond length N2+ > Bond length N2 (b) Bond length NO+ < Bond length of NO
(c) Bond length CN– < Bond length of CN (d) Bond length O2– < Bond length of O2–2
(e) Bond length O2 > Bond length of O2+ (f) Bond length B2 > Bond length of B2–
lgh dFkuksa dh la[;k Kkr dhft, %
(a) N2+ dh ca/k yEckbZ > N2 dh ca/k yEckbZ (b) NO+ dh ca/k yEckbZ < NO dh ca/k yEckbZ
(c) CN– dh ca/k yEckbZ < CN dh ca/k yEckbZ (d) O2– dh ca/k yEckbZ < O2–2 dh ca/k yEckbZ
(e) O2 dh ca/k yEckbZ > O2+ dh ca/k yEckbZ (f) B2 dh ca/k yEckbZ > B2– dh ca/k yEckbZ
Ans. 6 (a, b, c, d, e, f)
3. In how many conversions, the bond length increases ?

fuEu esa ls fdrus vUrZifjorZuksa esa ca/k yEckbZ c<+rh gSa \
(i) NO NO+ (ii) N2+ N2– (iii) O2  O2+ (iv) H2  H2+
(v) NH3  NH4+ (vi) NH3NH2– (vii) BF3BF4–
Ans. 4 (ii, iv, vi, vii)
4. Which of the following have bond order less than two ?

fuEu esa ls fdruh Lih'kht ds ca/k Øe dk eku nks ls de gS \
(a) NO3– (b) CO32 – (c) F2 (d) Cl2
(e) Br2 (f) O22– (g) O2– (h) N2–
2+
(i) O2 (j) Li2+ (k) He2+
Ans. 9 (a, b, c, d, e, f, g, j, k)
PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE

Hkkx - III : ,d ;k ,d ls vf/kd lgh fodYi çdkj
1. Which of the following have bond order three ?
fuEu esa ls fdldk cU/k Øe rhu gSa \
(A*) O22+ (B*) NO+ (C*) CN– (D) CN+
Sol. O22+ bond order = 3, NO+ bond order = 3, CN– bond order = 3, CN+ bond order = 2
O22+ ca/k Øe = 3, NO+ ca/k Øe = 3, CN– ca/k Øe = 3, CN+ ca/k Øe = 2
2. The species which are paramagnetic is/are :

çtkfr tks fd vuqpqEcdh; gS %
(A*) NO (B*) NO2 (C*) ClO2 (D) N2O4
.
Sol. (A) : N  O (B) (C) (D)
ADVCBO - 10
Chemical Bonding-IV
3. Which of the statement(s) are correct ?
(A) There is a single bond in FO+
(B*) The F and O are further apart in FO– than in FO+.
(C) There is a double bond in FO–.
(D*) It would take more energy to break F–O bond in FO+ than in FO–.
fuEu esa ls dkSulk @ dkSuls dFku lgh gS@gSa \
(A) FO+ esa ,d ,dy ca/k gksrk gSA
(B*) FO– esa F rFkk O ijek.kq FO+ dh rqyuk esa ijLij vf/kd nqjh ij gksrs gSA
(C) FO– esa ,d f}ca/k gksrk gSA
(D*) FO+ esa F–O ca/k dks fo;ksftr djus ds fy, FO– dh rqyuk esa vf/kd ÅtkZ dh vko';drk gksrh gSA
Sol. In FO+ total no. of electrons = 16, so bond order will be 2.
In FO– total number of electrons = 18, so bond order will be 1.
gy. FO+ esa dqy bysDVªkWuksa dh la[;k = 16, vr% cU/k Øe 2 gksxkA
FO– esa dqy bysDVªkWuksa dh la[;k 18 gS] vr% cU/k Øe 1 gksxkA
4. Among the following, the species with one unpaired electron are :
fuEu esa ls fdl Lih'kht esa ,d v;qfXer bysDVªkWu gS %
(A*) O2+ (B*) NO (C*) O2– (D) B2
Sol. (A) O2 : (1s) (*1s) (2s) (*2s) (2pz) (2p2x = 2p2y ) (*2px1 = *2py)
+ 2 2 2 2 2
(B) NO is derivative of O2: NO(O2+) (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y) (*2px1 = *2py)
(C) O2– : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px2 = *2p1y)
(D) B2 : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p1x = 2p1y) (pz)0
gy- (A) O2+ : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px1 = *2py)
(B) NO, O2 dk O;qRiUu gS : NO(O2+) (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y) (*2px1 = *2py)
(C) O2– : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px2 = *2p1y)
(D) B2 : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p1x = 2p1y) (pz)0
5. Identify correct statements
(A*) Down the group strength of metallic bond increases nearly all in transition elements.
(B) Down the group strength of metallic bond increases in alkali metals.
(C*) Down the group strength of metallic bond decreases in alkali metals.
(D) Down the group strength of metallic bond decreases in transition metals.
lgh dFkuksa dh igpku dhft,A
(A*) laØe.k rRoksa esa] oxZ esa] uhps tkus ij] /kkfRod ca/k lkeF;Z esa yxHkx o`f) gksrh gSA
(B) {kkjh; /kkrqkvsa esa] oxZ esa] uhps tkus ij] /kkfRod ca/k lkeF;Z esa o`f) gksrh gSA
(C*) {kkjh; /kkrqkvsa esa] oxZ esa] uhps tkus ij] /kkfRod ca/k lkeF;Z esa deh gksrh gSA
(D) laØe.k rRoksa esa] oxZ esa] uhps tkus ij] /kkfRod ca/k lkeF;Z esa o`f) gksrh gSA
Sol. Fact.
6. Which of the following statements are correct for band theory of metallic bond.
(A) Valence band is empty or half filled in metal.
(B*) Conduction band is empty in metal
(C*) Energy gap between conduction and valence band is very large in non-conductors.
(D) Overlapping of conduction & valence band occurs in semi-conductors
/kkfRod ca/k ds cS.M fl)kUr ds fy, fuEu esa ls dkSuls dFku lgh gSA
(A) /kkrqvksa esa la;kstd cS.M fjDr vFkok v)Ziwfjr gksrk gSA
(B*) /kkrqvksa esa pkyu cS.M fjDr gksrk gSA
(C*) dqpkydksa esa] pkyu cS.M rFkk la;kstd cS.M ds e/; ÅtkZ vUrjky cgqr vf/kd gksrk gSA
(D) v)Zpkydksa esa] pkyu cS.M rFkk la;ksdrk cS.M ds e/; vfrO;kfir ik;k tkrk gSA
Sol. Facts.
ADVCBO - 11
Chemical Bonding-IV
7. The force that binds a metal atom to a number of electrons with in its sphere of influence is known as a
metallic bond. Now, which of these is /are true for this found.
(A*) Metallic bond is non-directional in nature.
(B*) Metallic bonds are weaker than covalent bond.
(C) Energy required to vapourise a mole of metal (say, copper) to the vapour state is larger than the
energy required to vapourise a mole of a covalent substance (say, graphite)
(D*) The valency electrons in a metallic bond are mobile.
,d /kkrq ijek.kq rFkk blds xksyh; izHkkoh {ks=k esa ik;s tkus okyk cy] /kkfRod ca/k dgykrk gSA vc bl cy ds fy,
fuEu esa ls dkSulk@dkSuls fodYi lgh gS@gSa %
(A*) /kkfRod ca/k dh izd`fr v&fn'kkRed gksrh gSA
(B*) /kkfRod ca/k lgla;kstd ca/k ls nqcZy gksrs gSA
(C) ,d eksy /kkrq (tSls dkWij) dks ok"i izkoLFkk esa ok"ihd`r djus ds fy, vko';d ÅtkZ] ,d eksy lgla;kstd
inkFkZ (tSls xzsQkbV) dks ok"i izkoLFkk esa ok"ihd`r djus ds fy, vko';d ÅtkZ dh rqyuk esa vf/kd gksrh gSA
(D*) /kkfRod ca/k esa la;kstdrk bysDVªkWu xfreku gksrs gSA
Sol. Facts rF;A
PART - IV : COMPREHENSION
Hkkx - IV : vuqPNsn (COMPREHENSION)
Read the following passage carefully and answer the questions.
fuEu vuqPNsn dks /;kuiwoZd if<+;s rFkk iz'uks a ds mÙkj nhft,A
Comprehension # 1
The distribution of electrons among various molecular orbitals is called the electronic configuration of
the molecule which provides us the following very important informations about the molecule.
(A) Stability of molecule : The molecule is stable if number of bonding molecular orbital electrons
(Nb) is greater than the number of antibonding molecular orbital electrons (Na) and vice- versa.
1
(B) Bond order : Bond order = (Nb – Na)
2
A positive bond order means a stable molecule while a negative or zero bond order means an unstable
molecule.
(C) Nature of the bond : Bond order 1, 2,or 3 corresponds to single, double or triple bonds
respectively.
(D) Bond length : Bond length decreases as bond order increases.
(E) Magnetic nature : Molecular orbitals in a molecule are doubly occupied, the substance is
diamagnetic and if one or more molecular orbitals are singly occupied, it is paramagnetic.
vuqPNsn # 1
fofHkUu vkf.od d{kdksa esa bysDVªksuksa ds forj.k dks v.kq dk vkf.od bysDVªkWfud foU;kl dgk tkrk gSa tks gesa v.kqvksa
ds ckjs esa cgqr egRoiw.kZ tkudkjh nsrk gSaA
(A) v.kq dk LFkkf;Ro % v.kq LFkk;h gksrk gSa ;fn cfU/kr vkf.od d{kd bysDVªkWuksa (Nb) dh la[;k çfrcU/kh vkf.od
d{kd bysDVªkWuksa (Na) dh la[;k ls vf/kd gksrh gSa rFkk bldk foijhr Hkh lR; gksrk gSaA
1
(B) cU/k&Øe % cU/k&Øe % = (Nb – Na)
2
/kukRed cU/k Øe dk vFkZ gS] v.kq LFkk;h v.kq gksrk gSa tcfd _.kkRed vFkok 'kwU; cU/k Øe dk vFkZ gS] v.kq vLFkk;h
v.kq gksrk gSA
(C) cU/k dh izd`fr % cU/k Øe 1 , 2, vFkok 3 Øe'k% ,dy] f}cU/k vFkok f=kcU/k ls lEcfU/kr gSa
(D) cU/k Øe c<+us ds lkFk cU/k yEckbZ de gksrh tkrh gSaA
(E) pqEcdh;&izd`fr % ,d v.kq esa ;fn vkf.od d{kd ;qfXer bysDVªkWu ls Hkjs gks] rc inkFkZ izfrpqEcdh; gksrk gSa rFkk
,d vFkok ,d ls vf/kd vkf.od d{kd ,dy bysDVªkWu ¼v;qfXer½ ls Hkjs gks] rc ;g vuqpEq cdh; gksrk gSA
ADVCBO - 12
Chemical Bonding-IV
1. Which of the following statements is incorrect ?
(A) Among O2+, O2 and O2– the stability decreases as O2+ > O2 > O2–
(B) He2 molecule does not exit as the effect of bonding and anti-bonding molecular orbitals cancel each
other
(C) C2, O22– and Li2 are diamagnetic
(D*) In F2 molecule, the energy of 2 Pz is more than  2px and  2 Py
fuEu esa ls dkSulk dFku xyr gSa \
(A) O2+, O2 rFkk O2– esa ls LFkkf;Ro dk vojksgh Øe O2+ > O2 > O2– gksrk gSA
(B) He2 v.kq ugha ik;k tkrk gSa D;ksafd cfU/kr o çfrcU/kh d{kdksa dk çHkko ,d&nwljs dks fujLr dj nsrs gSaA
(C) C2 ,O22– rFkk Li2 izfrpqEcdh; gSa
(D*) F2 v.kq esa 2 Pz dh ÅtkZ]  2px rFkk  2 Py dh vis{kk vf/kd gksrh gSaA
Sol. (A)
Bond order  stability (i.e., bond strength)

(B) Helium molecule (He2) : He2 : (1s)2 (*1s)2
Bond order of He2 is ½(2 – 2) = 0
The molecular orbital description of He2 predicts two electrons in a bonding orbital and two electrons in
an antibonding orbital, with a bond order of zero - in other words, no bond. The noble gas He has not
significant tendency to form diatomic molecules and, like the other noble gases, exists in the form of
free atoms.
(C) Carbon molecule (C2) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y) or KK (2s)2 (*2s)2 (2p2x
= 2p2y )
Lithium molecule (Li2) : (1s)2 (*1s)2 (2s)2
Peroxide (O22–) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y) (*2px2 = *2p2y)
As all electrons are paired so C2, Li2 and O22– are diamagnetic.
(D) Fluorine molecule (F2) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y) (*2px2 = *2p2y)
Sol. (A)
vkca/k Øe LFkkf;Ro (vFkkZr~ vkca/k lkeF;Z)

(B) ghfy;e v.kq (He2) : He2 : (1s)2 (*1s)2
He2 dk ca/k Øe ½(2 – 2) = 0 gSaA
He2 v.kq ds vkf.od d{kd o.kZu esa] nks bysDVªkWu ca/kh d{kd esa rFkk nks bysDVªkWu çfrca/kh d{kd esa mifLFkr gSA
bldk caèk Øe 'kwU; gksrk gSA nwljs 'kCnksa esa dksbZ ca/k ugha curk gSaA vU; ukscy xSlksa ds leku vfØ;@ukscy xSl
He Hkh f}ijek.kqd v.kq esa vfLrRo ugha j[krk gSa rFkk ,dy•ijek.kq ds :i esa vfLrRo j[krk gSaA
(C) dkcZu v.kq (C2) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y) or KK (2s)2 (*2s)2 (2p2x = 2p2y )
fyfFk;e v.kq (Li2) : (1s)2 (*1s)2 (2s)2
ijvkWDlkbM (O22–) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y) (*2px2 = *2p2y)
pwafd lHkh bysDVªkWu ;qfXer gS] vr% C2, Li2 rFkk O22– çfrpqEcdh; gSaA
(D) ¶yksjhu v.kq (F2) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y) (*2px2 = *2p2y)
ADVCBO - 13
Chemical Bonding-IV
2. The bromine (Br2) is coloured because:
(A) the difference in energy (E) between HOMO and LUMO is large and the electronic excitation take
place by absorption of light which falls in ultra violet region.
(B) the difference in energy (E) between HOMO and LUMO is small and the electronic excitation take
place by absorption of light which falls in infrared region.
(C) the bromine molecule is paramagnetic and the difference in energy (E) is such that the electronic
excitation take place in visible light.
(D*) the difference in energy (E) between HOMO and LUMO is such that the electronic excitation take
place by absorption of light which falls in visible region and bromine molecule is diamagnetic.
czksehu (Br2) jaxhu gksrk gSa D;ksafd&
(A) HOMO rFkk LUMO ds chp ÅtkZ esa vUrj (E) vf/kd gksrk gSa rFkk izdk'k ds vo'kks"k.k }kjk bysDVªkWfud
mÙkstu izkIr gksrk gSa tks ijkcSaxuh {ks=k esa gksrk gSaA
(B) HOMO rFkk LUMO ds chp ÅtkZ esa vUrj (E) de gksrk gSa rFkk izdk'k ds vo'kks"k.k }kjk bysDVªkWfud mÙkstu
izkIr gksrk gSa tks vojDr {ks=k esa gksrk gSaA
(C) czksehu v.kq vuqpqEcdh; gksrk gSa rFkk ÅtkZ esa vUrj (E) bl izdkj gksrk gSa fd bysDVªkWfud mÙkstu n`'; {ks=k esa
izkIr gksrk gSA
(D*) HOMO rFkk LUMO ds chp ÅtkZ esa vUrj (E) bl izdkj gksrk gSa fd izdk'k dk vo'kks"k.k }kjk bysDVªkWfud
mÙkstu bl izdkj ls fd;k tkrk gSa fd tks n`'; {ks =k esa izkIr gksrk gSa rFkk czksehu v.kq izfrpqEcdh; gksrk gSA
Sol. (D) It is correct statement. (D) ;g lgh dFku lgh gSaA
3. N2 has greater bond dissociation energy than N2+, where as O2 has a lower bond dissociation energy
than O2+ because:
(A) Bond order is reduced when O2 is ionized to O2+ and bond order is increased when N2 is ionized to N2+
(B*) Bond order is increased when O2 is ionized to O2+ and bond order is decreased when N2 is ionized to
N2+
(C) Bond order is deceased when O2 is ionized to O2+ and bond order is decreased when N2– is ionized to
N2+
(D) None of these.
N2 dh ca/k fo;kstu ÅtkZ N2+ ls vf/kd gksrh gSa tcfd O2 dh ca/k fo;kstu ÅtkZ O2+ ls de gksrh gSa D;ksafd&
(A) tc O2 dks O2+ esa vk;uhd`r fd;k tkrk gSa] rc cU/k Øe de gks tkrk gSa rFkk tc N2 dks N2+ esa vk;uhd`r
fd;k tkrk gSa] rc cU/k Øe esa o`f) gksrh gSa
(B*) tc O2 dks O2+ esa vk;uhd`r fd;k tkrk gSa] rc cU/k Øe c<+ tkrk gSa rFkk tc N2 dks N2+ esa vk;uhd`r fd;k
tkrk gSa] rc cU/k Øe esa deh gksrh gSa
(C) tc O2 dks O2+ esa vk;uhd`r fd;k tkrk gS] rc cU/k Øe de gks tkrk gS rFkk tc N2– dks N2+ esa vk;uhd`r
gksrk gSa] rc cU/k Øe esa deh gksrh gSa
(D) mijksDr esa ls dksbZ ugha
Sol. (B) Oxygen molecule (O2) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y) (*2px1 = *2p1y)
Bond order = 1/2(10 – 6) = 2.0,
O2+ : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px1 = *2p0y)
Bond order = 1/2(10 – 5) = 2.5.
Nitrogen molecule (N2) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y ) 2pz)2
The bond order of N2 is 1/2(10 – 4) = 3.
N2+ : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y ) 2pz)1
Bond order = 1/2(9 – 4) = 2.5.
gy- (B) vkWDlhtu v.kq (O2) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y) (*2px1 = *2p1y)
ca/k Øe = 1/2(10 – 6) = 2.0,
O2+ : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px1 = *2p0y)
ca/k Øe = 1/2(10 – 5) = 2.5.
ukbVªkstu v.kq (N2) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y ) 2pz)2
N2 dk ca/k Øe 1/2(10 – 4) = 3 gSaA
N2+ : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y ) 2pz)1
ca/k Øe = 1/2(9 – 4) = 2.5.
ADVCBO - 14
Chemical Bonding-IV
Comprehension # 2
In a molten metal, the metallic bond is still present, although the order structure has been broken down.
The metallic bond isn’t fully broken until the metal boils. That means boiling point is actually a better
guide to the strength of the metallic bond then melting point is. On melting the bond is loosened, not
broken.
vuqPNsn # 2
,d xfyr /kkrq esa] /kkfRod ca/k mifLFkr gksrs gS ;)fi] Øfed fu;fer lajpuk foyqIr gks tkrh gSA tc rd /kkrq
mcyuk izkjEHk ugha djrh gS rc rd /kkfRod ca/k iw.kZr% foyqIr ¼VwVrs½ ugha gksrs gSA vFkkZr~ bldk vFkZ ;g gqvk dh
okLro esa DoFkukad fcUnq] xyukad fcUnq dh rqyuk esa /kkfRod ca/k lkeF;Z ds lanHkZ ls lgh] tkudkjh nsrk gSA xyus
ij /kkfRod ca/k nqcZy gksrs gSA foyqIr ugha gksrs gSA
4. Order of boiling point of K, Ca, Sc is
K, Ca, Sc DoFkukad dk Øe gS&
(A) K > Ca > Sc (B) Ca > K > Sc (C*) Sc > Ca > K (D) K > Sc > Ca
5. Order of boiling point & melting point of Zn, Cd, Hg, respectively is :
Zn, Cd, Hg ds Øe'k% DoFkukad rFkk xyukad dk lgh Øe gS&
(A*) Zn > Cd > Hg & Zn > Cd > Hg (B) Hg > Cd > Zn & Zn > Cd > Hg
(C) Hg > Cd > Zn & Hg > Cd > Zn (D) Zn > Cd > Hg & Hg > Cd > Zn
Comprehension # 3
Two models are considered to explain metallic bonding :
(A) Band model (B) Electron-sea model
(A) Band Model :
The interaction of two atomic orbitals, say the
3s-orbitals of two sodium atoms, produces two
molecular orbitals, one bonding orbital and one
antibonding orbital. If N atomic orbitals interact,
N molecular orbitals are formed. Atoms interact
more strongly with nearby atoms than with
those farther away. The energy that separates
bonding and antibonding molecular orbitals
decreases as the interaction (overlap) between
the atomic orbitals decreases. When we
consider all the possible interactions among one Figure-1. The band of orbitals resulting from
mole of Na atoms, there is formation of series of interaction of the 3s-orbitals in a crystal of sodium
very closely spaced molecular orbitals (3s and
3*s). This consists of a nearly continuous band of orbitals belonging to the crystal as a whole. One
mole of Na atoms contributes one mole (6.02 × 1023) of valence electrons thus, 6.02 × 1023 orbitals in
the band are half-filled.
The empty 3 p atomic orbitals of Na atoms also
interact to form a wide band of 3 × 6.07 × 1023
orbitals.
The 3s and 3p atomic orbitals are quite close in
energy, so that these bands of molecular orbitals
overlap. The two overlapping bands contain 4 × 6.02
×1023 orbitals. Because each orbital can hold two
electrons, the resulting combination of bands is only
one-eighth full.
According to band theory, the highest-energy
electrons of metallic crystals occupy either a partially
filled band or a filled band that overlaps an empty
band. A band within which (or into which) electrons
must move to allow electrical conduction is called a
conduction band. The electrical conductivity of a
metal decreases as temperature increases. The Figure-2. Overlapping of a half–filled “3s” band
increase in temperature causes thermal agitation of with an empty “3p” band of NaN crystal
ADVCBO - 15
Chemical Bonding-IV
the metal ions. This impedes the flow of electrons
when an electric field is applied.
Crystalline non-metals, such as diamond and phosphorus, are insulators, they do not conduct
electricity. It is due to the fact that their highest-energy electrons occupy filled bands of molecular
orbitals that are separated from the lowest empty bond (conduction band) by an energy difference
called the band gap. In an insulator, this band gap is an energy difference that is too large for electrons
to jump to get to the conduction band.
Elements that are semiconductors have filled bands that are only slightly below, but do not overlap with
empty bands. They do not conduct electricity at low temperatures, but a small increase in temperature
is sufficient to excite some of the highest-energy electrons into the empty conduction band.
(B) Electron-Sea Model :
Metals have ability to conduct electricity, ability to conduct heat, ease of deformation [that is, the ability
to be flattened into sheets (malleability) and to be drawn into wires (ductility)] and lustrous appearance.
One over simplified model that can account for some of these properties is the electron-sea model. The
metal is pictured as a network of positive ions immersed in a “sea of electrons”. In lithium the ions
would be Li+ and one electron per atom would be contributed to the sea. These free electrons account
for the characteristic metallic properties. If the ends of a bar of metal are connected to a source of
electric current, electrons from the external source enter the bar at one end. Free electrons pass
through the metal and leave the other end at the same rate.
In thermal conductivity no electrons leave or enter the metal but those in the region being heated gain
kinetic energy and transfer this to other electrons.
According to the electron-sea model, the case of deformation of metals can be thought of in this way : If
one layer of metal ions is forced across another, perhaps by hammering, the internal structure remains
unchanged as the sea of electrons rapidly adjusts to the new situation.
vuqPNsn # 3
/kkfRod ca/k dks le>us ds fy, nks çfr:i (models) dk voyksdu djrs gSaA
(A) cS.M çfr:i (Band model) (B) bysDVªkWu leqnz çfr:i (Electron-sea model)
(A) cS.M çfr:i (Band Model) :
nks ijekf.od d{kdksa dh vUr% fØ;k crkrh gS fd nks
lksfM;e ijek.kqvksa ds 3s-d{kdksa ls nks vkf.od d{kd
çkIr gksrs gSA ,d ca/kh vkf.od d{kd o ,d foifjr
ca/kh vkf.od d{kdA ;fn N ds ijekf.od d{kd
vfHkfØ;k djrs gS rc N vkf.od curs gSaA ijek.kq nwj
okys ijek.kqvksa dh vis{kk ikl okys ijek.kqvksa ls vfr
çcy :i ls vfHkfØ;k djrs gSaA ca/kh o foifjr ca/kh
vkf.od d{kdksa dks i`Fkd djus okyh ÅtkZ ijekf.od
d{kdksa ds e/; vUr% fØ;k ¼vfrO;kiu½ ?kVus ls ?kVrh fp=k-1 % lksfM;e ds fØLVy esa 3s-d{kdksa dh vUr% fØ;k ls
gSaA tc ge ,d eksy Na ijek.kqvksa ds e/; lHkh çkIr d{kdksa ds cS.M
lEHko vUr%fØ;kvksa ij fopkj djrs gS] rc vkf.od
d{kdksa (3s rFkk 3*s) dh ,d Js.kh dk fuekZ.k gksrk gSaA ;g fØLVyksa ls lEcaf/kr d{kdksa ds lrr~ cS.M dk fuekZ.k
djrh gSaA Na ijek.kqvksa ds ,d eksy ls ,d eksy (6.02 × 1023) la;ksth bysDVªkWu çkIr gksrs gSaA vr% cS.M esa 6.02 ×
1023 d{kd v)Ziwfjr gSaA
ADVCBO - 16
Chemical Bonding-IV
Na ijek.kqvksa ds fjDr 3 p ijekf.od d{kd vUr%
fØ;k dj 3 × 6.07 × 1023 d{kdksa ds cS.M cukrs
gSaA 3s o 3p d{kdksa dh ÅtkZ yxHkx leku gksrh
gSaA blfy, bu d{kdksa ds vkf.od d{kd cS.M
vfrO;kiu djrs gSA nks vfrO;kfir cS.M 4 ×
6.02 ×1023 d{kd ;qDr gSaA D;ksfd çR;sd d{kd
nks bysDVªkWu j[krk gSaA cS.Mksa dk ifj.kkeh
la;kstu dk ek=k 1/8 Hkkx Hkjk gqvk gksrk gSaA
cS.M fl)kar ds vuqlkj] /kkfRod fØLVy ds mPp
ÅtkZ okys bysDVªkWu ;k rks vkaf'kd Hkjs cS.M esa
vFkok iw.kZ iwfjr cS.M esa tks fd fjDr cS.M ds
lkFk vfrO;kfir gksrk gS esa mifLFkr gksrs gS cS.M
ftlesa dh fo|qr pkyu ds fy, bysDVªkWu dk
xfreku gksuk vko';d gksrk gSa] pkyu cS.M
fp=k-2 % NaN fØLVy ds ,d v)Ziwfjr “3s” cS.M dk fjDr “3p”
(conduction band) dgykrk gSaA rki esa o`f) ds
cS.M ds lkFk vfrO;kiu
lkFk /kkrq dh fo|qr pkydrk esa deh vkrh gSA
rkieku esa o`f) /kkrq vk;uksa ds rkih; xfr'khyrk (thermal agitation) dk dkj.k gksrk gS tc fo|qr {ks=k yxk;k
tkrk gS rc ;g bysDVªkWu dh xfr dks vo:) djrk gSaA
fØLVyh; v/kkrq,sa tSls fd ghjk rFkk QkWLQksjl dqpkyd gS ;g fo|wr /kkjk dk pkyu ugha djrs gSaA ;g bl rF; ds
dkj.k gksrk gS fd buds mPp ÅtkZ okys bysDVªkWu vkf.od d{kdksa ds iw.kZiwfjr cS.Mksa esa mifLFkr gksrs gS tks fd
fuEure ÅtkZ okys fjDr cS.M ¼pkyu cS.M½ (conduction band) ls ,d ÅtkZ vUrj }kjk i`Fkd jgrs gSaA ;g ÅtkZ
vUrj cS.M vUrjky ¼ÅtkZ vUrjky½ dgykrk gSaA dqpkyd esa ;g cS.M vUrjky ,d vR;f/kd cM+k ÅtkZ vUrjky gS
tks fd bysDVªkWu ds pkyu cS.M esa tkus ds fy, cgqr vf/kd gksrk gSaA
rRo tks v)Zpkyd gS muesa iw.kZiwfjr cS.M gksrs gS tks fd fjDr cS.M ls dsoy FkksMk lk uhps jgrs gS ijUrq buds lkFk
vfrO;kfir ugha gksrs gSaA fuEu rki ij ;g fo|qr dk pkyu ugha djrs gS ijUrq rkieku esa FkksMh lh o`f) dqN mPp
ÅtkZ bysDVªkWuksa dks fjDr pkyu cS.M esa mÙksftr djus ds fy, i;kZIr gksrh gSaA
(B) bysDVªkWuµleqnz çfr:i (Electron-Sea Model) :
/kkrqvksa esa fo|qr pkydrk] Å"ek pkydrk] iquZfuekZ.k dh çd`fr ¼'khV cukus dh çd`fr ¼vk?kkro/kZuh;rk½½ rkj cukus
dh ço`fr ¼rU;rk½ o ped n'kkZus dh ço`fr ik;h tkrh gSaA bu xq.kksa dh ,d ljy çfr:i }kjk O;k[;k dh tk ldrh
gSA ;g çfr:i bysDVªkWu leqnz çfr:i gSA /kkrq dks /kukosf'kr vk;u ds :i esa çnf'kZr fd;k tk ldrk gS] tks fd
bysDVªkWu ds leqnz esa lekosf'kr (immersed) gksrs gSA fyfFk;e esa /kkrq vk;u Li+ rFkk çfr ijek.kq ,d bysDVªkWu leqnz
ds fy, ;ksxnku nsxkA ;s eqDr bysDVªkWu /kkrqvksa ds yk{kf.kd /kkfRod xq.kksa ds fy, mÙkjnk;h gSaA ;fn /kkrq dh ,d
NM+ ds ,d fljs dks oS|qr /kkjk ds L=kksr ls tksMk tkrk gS rc bysDVªkWu ckg~; L=kksr ls /kkrq ds ,d fljs esa ços'k
djrs gS rFkk eqDr bysDVªkWu /kkrq esa çokfgr gksrs gSA rFkk nwljs fljs ls leku nj ls eqDr gksr s gSaA
rkih; pkydrk esa dksbZ bysDVªkWu /kkrq esa ços'k vFkok fu"dkf"kr ugha gksrk gSaA ijUrq og Hkkx tks fd xeZ fd;k tkrk
gS xfrt ÅtkZ xzg.k djrk gSaA rFkk blsa vU; bysDVªkWu dks LFkkukUrfjr dj fn;k tkrk gSaA
bysDVªkWu leqnz çfr:i ds vuqlkj /kkrqvksa ds fod`rhdj.k dks fuEu çdkj ls le>k;k tk ldrk gSaA ;fn /kkrq vk;uksa
dh ,d ijr dks nwljh ijr ds Åij cyiwoZd foLFkkfir fd;k tk;s vFkkZr~ gFkksMs ls pksV igq¡pk;h tk;s
(hammering) rc vkUrfjd lajpuk vifjofrZr cuh jgrh gS D;ksafd bysDVªkWu dk leqnz rhozrk ls u;h ifjfLFkfr esa
lek;ksftr gks tkrk gSA
6. Considering band model, select the incorrect statement :
(A) Li metal should have partilly filled valence band and empty conduction band.
(B) Mg metal should have fully filled valence band and overlapping conduction band.
(C) Electrical conductivity of a metal decreases as temperature increases.
(D*) The energy spread of each atomic energy level of an element behaving like a semiconductor is
infinitisimally small.
ADVCBO - 17
Chemical Bonding-IV
cS.M izfr:i ij fopkj djrs gq,] vlR; dFkuksa dk p;u dhft,A
(A) Li /kkrq vkaf'kd Hkjs la;kstdrk cS.M rFkk fjDr pkyu cS.M j[krh gSA
(B) Mg /kkrq] iw.kZ Hkjs la;ksdrk cS.M rFkk vfrO;kfir pkyu cS.M j[krh gSA
(C) ,d /kkrq dh oS|qr pkydrk] rki o`f) ds lkFk de gksrh gSA
(D*) v)Zpkyd ds lkFk O;ogkj djus okys ,d rRo izR;sd ijek.kq ijek.fod ÅtkZ Lrj dk ÅtkZ vUrjky vuUr
:i ls NksVk gksrk gSA
7. All metal written below have usually low melting points except :
(A) Caesium (B) Gallium (C*) Gold (D) Mercury
uhps fy[kh x;h lHkh /kkrq;sa ,d /kkrq ds vfrfjDr lkekU;r% U;wu DoFkukad j[krh gS] og /kkrq gS %
(A) flft;e (B) xsfy;e (C*) xksYM (D) edZjh
8. Which of the following physical properties can be explained by electron sea model :
(A) Electrical conduction (B) Thermal conduction
(C) Malleability (D*) All of these
fuEu esa ls dkSuls HkkSfrd xq.k/keZ dks bySDVªkWu leqnz izfr:i }kjk le>k;k tk ldrk gSA
(A) oS|qr pkydrk (B) m"ek pkydrk (C) v?kkro/kZuh;rk (D*) ;s lHkh
Comprehension # 4
Answer Q.9, Q.10 and Q.11 by appropriately matching the information given in the three
columns of the following table.
Observe the three columns in which column-1 represents molecule, column-2 represents bond orders
while column-3 represents molecule properties. Properties of molecule explained by molecular orbital
theory. Like megnatic nature, orbital mixing etc.
Column 1 Column 2 Column 3
(I) B2 (i) Bond Order = 2 (P) Diamagnetic in Nature
(II) O 2 (ii) Bond Order = 2.5 (Q) SP Mixing Occure
(III) F2 (iii) Bond Order = 1 (R) Paramagnetic in Nature
(S) Highest Occupied Molecular orbital (HOMO)
(IV) C2 (iv) Bond Order = 3
is Bonding Molecular orbital (BMO)
vuqPNsn # 4
uhps nh x;h Vscy ds rhu dk¡yeksa esa miyC/k lwpuk dk mi;qDr <ax ls lqesy dj iz'uksa Q.9, Q.10 vkSj Q.11 ds
mÙkj nhft;sA
rhu dkWye dk izs{k.k dhft, ftuesa dkWye -1 v.kq dks iznf'kZr djrk gS] dkWye-2 ca/k Øe dks iznf'kZr djrk gS tcfd
dkWye-3 vkf.od xq.k/keksZ dks iznf'kZr djrk gSa v.kqvksa ds xq.kksa dks vkf.od d{kd fl)kUr }kjk le>k;k tkrk gSA
tSls pqEcdh; izd`fr] d{kd lfEeJ.k bR;kfnA
dkWye 1 dkWye 2 dkWye 3
(I) B2 (i) ca/k Øe = 2 (P) izfrpqEcdh; izd`fr

(II) O 2 (ii) ca/k Øe = 2.5 (Q) SP lfEeJ.k izkIr gksrk gSA
(III) F2 (iii) ca/k Øe = 1 (R) vuqpqEcdh; izd`fr
(S) mPpre Hkjs gq, vkf.od d{kd (HOMO) caf/kr
(IV) C2 (iv) ca/k Øe = 3
vkf.od d{kd (BMO) gksrs gSA
9. Which is incorrect combination?

fuEu esa ls dkSulk xyr la;kstu gS\
(A) (I) (iii) (Q) (B*) (II) (ii) (P) (C) (III) (iii) (P) (D) (IV) (i) (P)
10. Which is correct combination for Diamagnetic species ?

izfrpqEcdh; Lih'kht ds fy, dkSulk la;kstu lgh gS \
(A) (III) (i) (P) (B) (IV) (ii) (R) (C*) (IV) (i) (S) (D) (III) (iii) (Q)
ADVCBO - 18
Chemical Bonding-IV
11. Which is correct combination?
fuEu esa ls dkSulk lgh la;kstu gS\
(A) (III) (ii) (S) (B) (III) (iii) (Q) (C) (II) (ii) (P) (D*) (I) (iii) (R)
2 2 2 2 1 1
Sol. B2 (10 e) 1s <  * 1s < 2 s <  * 2s < ( 2p x  2p y ) < *2pz.
Nb  Na 64
Nb = 6 e, Na = 4 , B.O. = = =1
2 2
And S.P. mixing occurs, Paramagnetic in nature, HOMO is BMO.
(rFkk S.P. lfEeJ.k izkIr gksrk gS] vuqpEq cdh; izd`fr , HOMO, BMO gSA)
2 2 2 2 2 2 2 1
 O 2 (15 e) 1s <  * 1s < 2 s <  * 2s < 2p z < ( 2p x  2p y ) < (  * 2p x = *2py) < *2pz
Nb  Na 10  5
Nb = 10 e, Na = 5 , B.O. = = = 2.5
2 2
And no S.P. mixing occur, Paramagnetic in nature, HOMO is ABMO.
(rFkk S.P. lfEeJ.k izkIr ugha gksrk gS] vuqpqEcdh; izd`fr, HOMO, ABMO gSA)
2 2 2 2 2 2 2 2 2
 F2 (18 e) 1s <  * 1s < 2 s <  * 2s < 2p z < ( 2p x  2p y ) < (  * 2p x =  * 2p y ) < *2pz
Nb  Na 10  8
Nb = 10 e, Na = 8 , B.O. = = =1
2 2
And no S.P. mixing occur, Diamagnetic in nature, HOMO is ABMO 
(rFkk S.P. lfEeJ.k izkIr gksrk gS] izfrpqEcdh; izd`fr , HOMO, ABMO gSA)
2 2 2 2 2 2
 C2 (12 e) 1s <  * 1s < 2 s <  * 2s < ( 2p x  2p y ) < *2pz.
Nb  Na 84
Nb = 8 e, Na = 4 , B.O. = = =2
2 2
And S.P. mixing occur, Diamagnetic in nature, HOMO is BMO.
(rFkk S.P. lfEeJ.k izkIr gksrk gS] izfrpqEcdh; izd`fr , HOMO, BMO gSA) 
* Marked Questions may have more than one correct option.

* fpfUgr iz'u ,d ls vf/kd lgh fodYi okys iz'u gS -
PART - I : JEE (ADVANCED) / IIT-JEE PROBLEMS (PREVIOUS YEARS)

Hkkx - I : JEE (ADVANCED) / IIT-JEE ¼fiNys o"kksZ½ ds iz'u
1. Write the Molecular orbital electron distribution of O2. Specify its bond order and magnetic property.
[JEE–2000(M), 3/135]
O2 ds vkf.od d{kdks esa bysDVªkWu forj.k fyf[k, rFkk bldk cU/k Øe rFkk pqEcdh; xq.k/keZ crkb;sA
[JEE–2000(M), 3/135]
Sol. Molecular orbital electronic configuration of O2 is as follows (Z is taken as molecular axis).
1s2  *1s 2  2s 2  *2s 2  2 p 2  2 p 2   2p 2  *2 p 1   *2p 1 
z x y x y
10 – 6
Bond order = = 2.
2
As it contains two unpaired electrons in bonding  molecular orbitals O2 is paramagnetic.
So, Magnetic moment = n (n  2) = 2 (2 + 2) = 2.83 B.M.
ADVCBO - 19
Chemical Bonding-IV
gy- O2 vkf.od d{kdks dk bysDVªksfud foU;kl fuEu gS (Z dks vkf.od v{k ysrs gS)
1s2  *1s 2  2s 2  *2s 2  2 p 2  2 p 2   2p 2  *2 p 1   *2p 1 
z x y x y
10 – 6
ca/k Øe = = 2.
2
D;ksafd cU/kh  vkf.od d{kdksa esa nks v;qfXer bysDVªkWu gS vr% O2 vuqpqEcdh; gksxk
blfy;s, pqEcdh; vk?kw.kZ = n (n  2) = 2 (2  2) = 2.83 B.M.
2. Which of the following molecular species has unpaired electron(s) ? [JEE–2002(S), 3/150]
fuEu esa ls fdl vkf.od iztkfr esa v;qfXer bysDVªkWu gS \ [JEE–2002(S), 3/150]
(A) N2 (B) F2 (C*) O2– (D) O22–
Sol. –
O2 is derivative of O2 and has 17 electrons. So its molecular orbital electronic configuration is
 1s2  *1s 2  2s 2  *2s 2  2 p 2  2 p 2  2p 2  *2 p 2  *2p 1
z x y x y
As it contains one unpaired electron in *2py1 molecular orbital so it is paramagnetic. Rest all species
have paired electrons so diamagnetic.
gy- O2– , O2 dk O;qRiUu gksrk gS o 17 e– j[krk gSA blfy, bldk v.kq d{kd bysDVªkWfud foU;kl gSaA
 1s2  *1s 2  2s 2  *2s 2  2 p 2  2 p 2  2p 2  *2 p 2  *2p 1
z x y x y
pawfd ;g *2py1 v.kqd{kd esa ,d v;qfXer bysDVªkWu j[krk gS blfy, ;g vuqpEq cdh; gksrk gSA ckdh lHkh ;qfXer
bysDVªkWu j[krs gS blfy, izfrpqEcdh; gksrs gSA
3. According to molecular orbital theory, which one of the following statements about the molecular
species O2+ is correct ? [JEE–2004(S), 3/144]
(A) It is paramagnetic and has less bond order than O2
(B*) It is paramagnetic and more bond order than O2
(C) It is diamagnetic and has less bond order than O2
(D) It is diamagnetic and has more bond order than O2
vkf.od d{kd fl)kar ds vuqlkj Lih'kht O2+ ds fy, fuEu esa ls dkSulk dFku lgh gSa \ [JEE–2004(S), 3/144]
(A) ;g vuqpqEcdh; gSa rFkk bldk cU/k Øe O2 ls de gksrk gSA
(B*) ;g vuqpqEcdh; gSa rFkk bldk cU/k Øe O2 ls vf/kd gksrk gSA
(C) ;g izfrpqEcdh; gSa rFkk bldk cU/k Øe O2 ls de gksrk gSA
(D) ;g izfrpqEcdh; gSa rFkk bldk cU/k Øe O2 ls vf/kd gksrk gSA
Sol. Molecular orbital electronic configuration is  1s2  *1s 2  2s 2  *2s 2  2 p 2  2 p 2  2p 2  *2 p 1  *2p 0
z x y x y
As it contains one unpaired electron it is paramagnetic and bond order = (10 – 5) / 2 = 2.5 (O2 = 2.0).
gy- vkf.od d{kd bysDVªkWfud foU;kl 1s2  *1s 2  2s 2  *2s2  2 p 2  2 p 2  2p 2  *2 p 1  *2p 0 gSA
z x x
y y
blesa ,d v;qfXer bysDVªkWu gksrk gSA vr% ;g vuqpqEcdh; gS vkSj cU/k Øe = (10 – 5) / 2 = 2.5 (O2 = 2.0).
4. Arrange the following three compounds in terms of increasing O–O bond length :
O2, O2[AsF6], K[O2]
Justify your answer based on the ground state electronic configuration of the dioxygen species in these
three compounds. [JEE–2004(M), 2/144]
fuEu rhu ;kSfxdksa dks O–O ca/k yEckbZ ds c<+rs gq, Øe esa O;ofLFkr dhft,A
O2, O2[AsF6], K[O2]
viuk mÙkj bu rhu ;kSfxdksa es a MkbvkWDlhtu ds vk| voLFkk bysDVªkWfud vfHkfoU;kl ds vk/kkj ij le>kb;sA
[JEE–2004(M), 2/144]
Sol. The electronic configuration of O2 will be:
O2 = 1s2 *1s2 2s2 *2s2 2p2z 2p2x = 2p2y *2p1x = *2p1y
N  Na
Now bond order = b
2
Where, Nb = Number of electrons in bonding orbitals
Na = Number of electrons in antibonding orbitals
ADVCBO - 20
Chemical Bonding-IV
10  6
bond order = =2
2
Similarly electronic configuration of O2– (in KO2) will be
1s2 *1s2 2s2 *2s2 2p2z 2p2x = 2p2y *2p2x = *2p1y
10  7 3
Bond order = = = 1.5
2 2
In O2 [AsF4]–, O2 is O2+ .
The electronic configuration of O2+ will be
 1s2 *1s2 2s2 *2s2 2p2z 2p2x, = 2p2y *2p1x
10  5
bond order  =2.5
2
1
Hence bond length order will be O+2 < O2 < O–2 because Bond order  .
Bond length
gy. O2 dk bysDVªksfud foU;kl gksxkA
O2 = 1s2 *1s2 2s2 *2s2 2p2z 2p2x = 2p2y *2p1x = *2p1y
N N
vc cU/k Øe = b a
2
tgkW] Nb = cfU/kr d{kdksa esa bysDVªksuks dh la[;k] Na = vcfU/kr d{kdksa esa bysDVªksuks dh la[;k
10  6
cU/k Øe = =2
2
O2– bysDVªksfud foU;kl (KO2 esa) lkekU;r% gksxkA
1s2 *1s2 2s2 *2s2 2p2z 2p2x = 2p2y *2p2x = *2p1y
10  7 3
cU/k Øe = = = 1.5
2 2
O2 [AsF4]– esa O2 ] O2+ esa gSA
O2+ dk bysDVªksfud foU;kl gksxkA
1s2 *1s2 2s2 *2s2 2p2z 2p2x, = 2p2y *2p1x
10  5
cU/k Øe  =2.5
2
1
vr% cU/k nwjh dk Øe gksxk O+2 < O2 < O–2 D;ksafd cU/k Øe  .
cU/k nwjh
5. The species having bond order different from that in CO is : [JEE–2007, 3/162]
buesa ls fdl Lih'kht (species) dk ca/kØe CO ls vyx gS \ [JEE–2007, 3/162]
(A*) NO¯ (B) NO+ (C) CN¯ (D) N2
Nb – Na
Sol. Bond order =
2
CO = 1s2  *1s2  2s2  *2s2  2p2 2p 2  2p 2
x y z
Bond order of CO = 10 – 4 / 2 = 3
NO–  1s2  *1s2  2s2  *2s2  2p 2  2p 2 2p 2  *2p 1  *2p 1 
z x y x y
Bond order of NO– = 10 – 6 / 2 = 2

NO+  1s2  *1s2  2s2  *2s2  2p 2  2p 2 2p 2
z x y
Bond order of NO+ = 10 – 4 / 2 = 3

CN–  1s2  *1s2  2s2  *2s2  2p 2 2p 2  2p 2 
x y z
Bond order of CN– = 10 – 4 / 2 = 3

N2  1s2  *1s2  2s2  *2s2  2p 2 2p 2  2p 2 
x y z
Bond order of N2 = 10 – 4 / 2 = 3.
ADVCBO - 21
Chemical Bonding-IV
Nb – Na
gy- cU/k Øe =
2
CO = 1s2  *1s2  2s2  *2s2  2p2 2p 2  2p 2 
x y z
CO esa cU/k Øe = 10 – 4 / 2 = 3
NO–  1s2  *1s2  2s2  *2s2  2p 2  2p 2 2p 2  *2p 1  *2p 1
z x y x y
–
NO esa cU/k Øe = 10 – 6 / 2 = 2
NO+  1s2  *1s2  2s2  *2s2  2p 2  2p 2 2p 2
z x y
NO+ esa cU/k Øe = 10 – 4 / 2 = 3

CN–  1s2  *1s2  2s2  *2s2  2p 2 2p 2  2p 2 
x y z
–
CN esa cU/k Øe = 10 – 4 / 2 = 3
N2  1s2  *1s2  2s2  *2s2  2p 2 2p 2  2p 2 
x y z
N2 esa cU/k Øe = 10 – 4 / 2 = 3
6. Among the following, the paramagnetic compound is : [JEE–2007, 3/162]

buesa ls dkSu lk ;kSfxd vuqpqEcdh; gS \ [JEE–2007, 3/162]
(A) Na2O2 (B) O3 (C) N2O (D*) KO2
Sol. KO2 exists as K+ & O2– .
In O2–, superoxide ion there are total number of electrons = 16 + 1 = 17.
MOT configuration =  1s2  *1s2  2s2  *2s2  2p 2  2p 2 2p 2  *2p 2  *2p 1
z x y x y
O2 has one unpaired electron in antibonding *2py . So it is paramagnetic.

– 1
gy- KO2 , K+ & O2– esa jgrk gSA

O2–, lqij vkWDlkbM vk;u esa] dqy bysDVªkWu la[;k = 16 + 1 = 17.
vkf.od d{kd vfHkfoU;kl = 1s2  *1s2  2s2  *2s2 2p 2  2p 2 2p 2  *2p 2  *2p 1
z x y x y
O2– ds izfrcU/kh vkf.od d{kd *2py1 esa ,d v;qfXer bysDVªkWu gSA vr% ;g vuqpqEcdh; gSA

7. Statement-1 : Band gap in germanium is small, because
Statement-2 : The energy spread of each germanium atomic energy level is infinitesimally small.
[JEE–2007, 3/162]
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C*) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True
oDrO;& 1 : tesZfu;e esa cSaM vUrjky (Band gap) NksVk gS] D;ksafd
oDrO;& 2 : tesZfu;e ds izR;sd ijek.oh; ÅtkZ Lrj esa ÅtkZ foLrkj (energy spread) vuUr% lw{e gksrk gSA
[JEE–2007, 3/162]
(A) oDrO;&1 lR; gS] oDrO;&2 lR; gS ; oDrO;&2, oDrO;&1 dk lgh Li"Vhdj.k gSA
(B) oDrO;&1 lR; gS] oDrO;&2 lR; gS ; oDrO;&2, oDrO;&1 dk lgh Li"Vhdj.k ugha gSA
(C*) oDrO;&1 lR; gS] oDrO;&2 vlR; gSA
(D) oDrO;&1 vlR; gS] oDrO;&2 lR; gSA
Sol. Statement-1 : Germanium is a semiconductor, where the energy gap between adjacent bands is
sufficiently small for thermal energy to be able to promote a small number of electrons from the full
valence band to the empty conduction band. This leaves both bands partially filled, so the material can
conduct electricity.
Statement-2 : Incorrect statement.
gy- oDrO;-1 : teksZfu;e ,d v)Zpkyd gS] ftles nks lehiorhZ cS.M ds e/; ÅtkZ vUrjky ml rkih; ÅtkZ dh rqyuk
esa] i;kZIr NksVk gksrk gS tks fd dqN la[;k esa bysDVªkWu dks] iw.kZ iwfjr la;kstdrk cS.M ls] fjDr pkydrk cS.M esa
LFkkukUrfjr dj ldrh gSA bl izdkj nksuksa cS.M vkaf'kd :i ls Hkjs gksrs gS] vr% inkFkZ fo|qr dk pkyu djrk gSA
oDrO;-2 : dFku lgh ugha gSA
ADVCBO - 22
Chemical Bonding-IV
8. Statement-1 : Boron always forms covalent bond, because
Statement-2 : The small size of B3+ favours formation of covalent bond. [JEE–2007, 3/162]
(A*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-
1.
(C) Statement-1 is True, Statement-2 is False.
(D) Statement-1 is False, Statement-2 is True.
oDrO;& 1 : cksjku (B) lnSo lgla;kstd vkca/k cukrk gS] D;ksafd
oDrO;& 2 : lgla;kstd vkca/k cukus esa B3+ dk NksVk vkdkj lgk;d gSA [JEE–2007, 3/162]
(A*) oDrO;&1 lR; gS] oDrO;&2 lR; gS ; oDrO;&2, oDrO;&1 dk lgh Li"Vhdj.k gSA
(B) oDrO;&1 lR; gS] oDrO;&2 lR; gS ; oDrO;&2, oDrO;&1 dk lgh Li"Vhdj.k ugha gSA
(C) oDrO;&1 lR; gS] oDrO;&2 vlR; gSA
(D) oDrO;&1 vlR; gS] oDrO;&2 lR; gSA
Sol. Due to small size of B, it is very difficult to remove the electrons from boron to form ionic bond as it will
require very high energy. On the other hand, due to its very small size having high polarising power
causes greater polarisation and eventually significant covalent character according to Fajan's rule.
gy- B ds NksVs vkdkj ds dkj.k] vk;fud ca/k dk fuekZ.k ds fy, cksjksu ls rhu bySDVªkWu fudkyuk vf/kd dfBu gksrk gS]
D;ksafd blds fy, vf/kd ÅtkZ dh vko';drk gksrh gSA nwljs 'kCnksa esa] blds cgqr NksVs vkdkj ds dkj.k] ;g mPp
/kqzo.k {kerk j[krk gSA bl dkj.k ;g cgqr vf/kd /kzqo.krk mRiUu djrk gS] vr% QkW;ku fu;e ds vuqlkj] blesa
lgla;kstd xq.k ik;k tkrk gSA
9. Match each of the diatomic molecules in Column I with its property/properties in Column II.
Column I Column II [JEE–2009, 8/160]
(A) B2 (p) Paramagnetic
(B) N2 (q) Undergoes oxidation
(C) O2– (r) Undergoes reduction
(D) O2 (s) Bond order  2
(t) Mixing of 's' and 'p' orbitals
dkWye I esa fy[ks izR;sd f}ijekf.od v.kq dks dkWye II esa fn;s x;s blds xq.k @ xq.kksa ds lkFk lqesy dhft,A
dkWye I dkWye II [JEE–2009, 8/160]
(A) B2 (p) vuqpEq cdh; (Paramagnetic)
(B) N2 (q) vkWDlhdj.k (Oxidation)
–
(C) O2 (r) vip;u (Reduction)
(D) O2 (s) vkcU/k Øe (Bond order)  2
(t) 's' vkSj 'p' d{kdksa (orbitals) dk feJ.k (mixing)
Ans. (A) - p, q, r, t ; (B) - q, r, s, t ; (C) - p, q, r ; (D) - p, q, r, s
Sol. (A) B2 1s2 *1s 2 2s 2 *2s 2 2px1 = 2py1
64
Bond order = =1 Paramagnetic with two unpaired electrons.
2
It undergoes oxidation as well as reduction which can be explained by taking the
following reactions.
2B + 3Cl2 2BCl3 ; 2B + 3Ca Ca3B2 (boride)
Mixing of 's' and 'p' orbitals takes place.
(B) N2 1s2 *1s 2 2s 2 *2s 2 2px2 = 2py2 2pz2
10  4
Bond order = =3 Diamagnetic
2
N2 + O2  2NO ; 6Li + N2  2Li3N
Mixing of 's' and 'p' orbitals takes place.
ADVCBO - 23
Chemical Bonding-IV
(C) O2– 1s2 *1s 2 2s 2 *2s 2 2pz2 2px2 = 2px 2 *2px2 = *2py1
10  7
Bond order = = 1.5 Paramagnetic with one unpaired electron.
2
O2–  O2 + e– ; O2– + e–  O22–
Mixing of 's' and 'p' orbitals does not take place.
(D) O2 1s2 *1s 2 2s 2 *2s 2 2pz2 2px2 = 2px2 *2px1 = *2py1
10  6
Bond order = =2 Paramagnetic with two unpaired electrons.
2
O2 O2+ + e– ; O2 + e–  O2–
Mixing of 's' and 'p' orbitals does not take place.
Sol. (A) B2 1s2 *1s 2 2s 2 *2s 2 2px1 = 2py1
64
vkcU/k Øe = =1 nks v;qfXer bysDVªkWu ds lkFk vuqpqEcdh;
2
bldk vkWDlhdj.k rFkk vip;u gksrk gS tks fd fuEu vfHkfØ;k }kjk le>k;k tk ldrk gSA
2B + 3Cl2 2BCl3 ; 2B + 3Ca  Ca3B2 (boride)
blesa 's' rFkk 'p' d{kdks dk feJ.k gksrk gSA
(B) N2 1s2 *1s 2 2s 2 *2s 2 2px2 = 2py2 2pz2
10  4
vkcU/k Øe = =3 izfrpqEcdh;
2
bldk vkWDlhdj.k rFkk vip;u gksrk gS tks fd fuEu vfHkfØ;k }kjk le>k;k tk ldrk gSA
N2 + O2  2NO ; 6Li + N2 2Li3N
blesa 's' rFkk 'p' d{kdks dk feJ.k gksrk gSA
(C) O2– 1s2 *1s 2 2s 2 *2s 2 2pz2 2px2 = 2px 2 *2px2 = *2py1
10  7
vkcU/k Øe = = 1.5 ,d v;qfXer bysDVªkWu ds lkFk vuqpqEcdh;
2
bldk vkWDlhdj.k] vip;u gksrk gS tks fd fuEu vfHkfØ;k }kjk le>k;k tk ldrk gSA
O2–  O2 + e– ; O2– + e–  O22–
blesa 's' rFkk 'p' d{kdks dk feJ.k ugha gksrk gSA
(D) O2 1s2 *1s 2 2s 2 *2s 2 2pz2 2px2 = 2px2 *2px1 = *2py1
10  6
vkcU/k Øe = =2 nks v;qfXer bysDVªkWu ds lkFk vuqpqEcdh;A
2
bldk vkWDlhdj.k] vip;u gksrk gS tks fd fuEu vfHkfØ;k }kjk le>k;k tk ldrk gSA
O2 O2+ + e– ; O2 + e– O2–
blesa 's' rFkk 'p' d{kdks dk feJ.k ugha gksrk gSA
10. Assuming that Hund's rule is violated, the bond order and magnetic nature of the diatomic molecule B 2
is :
(A*) 1 and diamagnetic (B) 0 and diamagnetic [JEE–2010, 5/163]
(C) 1 and paramagnetic (D) 0 and paramagnetic
gqUM fu;e (Hund's rule) ds mYya?ku gksus ij f}ijek.kqd v.kq B2 dk cU/k Øe vkSj pqEcdh; voLFkk gS %
(A*) 1 vkSj izfrpqEcdh; (B) 0 vkSj izfrpqEcdh;
(C) 1 vkSj vuqpqEcdh; (D) 0 vkSj vuqpqEcdh;
ADVCBO - 24
Chemical Bonding-IV
Sol. B2 ; total number of electrons = 10. The MOT electron configuration violating the Hund's rule will be
thus :
1s2  *1s2 2s2  *2s2 2px2  2py 2pz0 
0
64
So, bond order = =1
2
As all electrons are paired, the molecule is diamagnetic.
Sol. B2 ; dqy bysDVªksuksa dh la[;k = 10. MOT ds vuqlkj bysDVªksfud foU;kl tks gq.M fu;e dk ikyu ugha djrk og gS
1s2  *1s2 2s2  *2s2 2px2  2py 2pz0 
0
64
vr%] ca/k Øe = =1
2
D;ksafd lHkh bysDVªksu ;qfXer gSa vr% v.kq izfrpqEcdh; gSA
11. Assuming 2s-2p mixing is NOT operative, the paramagnetic species among the following is :
;g ekurs gq, fd 2s-2p dk feJ.k fØ;kdkjh ugha gS] fuEu esa vuqpEq cdh; fLih'kht gS %
[JEE(Advanced) 2014, 3/120]
(A) Be2 (B) B2 (C*) C2 (D) N2
Sol. If 2s-2p mixing is not operative, the increasing order of Molecular orbitals will be :
 2p1y 
1s2, *1s2, 2s2, *2s2, 2Px2  1

 2pz 
Considering this Be2 & B2 become diamagnetic, so does N2. Only C2 would be paramagnatic with
electronic configuration as above
gy ;fn 2s-2p feJ.k fØ;kfUor ugha gksrk gS] rks vk.kfod d{kdksa dk vkjks gh Øe fuEu gksxk %
 2p1y 
 1s2, *1s2, 2s2, *2s2, 2Px2  1

 2pz 
Be2 o B2 izfrpqEcdh; gS] ,slk gh N2 ds lkFk gksrk gSA dsoy mijksDr bysDVªkWfud foU;kl ds lkFk C2 vuqpqEcdh;
gksxkA
12. Match the orbital overlap figures shown in List-I with the description given in List-II and select the
correct answer using the code given below the lists. [JEE(Advanced) 2014, 3/120]
List-I List-II
P. 1. p–d  antibonding
Q. 2. d–d  bonding
R. 3. p–d  bonding
S. 4. d–d  antibonding
lwph-I esa n'kkZ;s d{kh; vfrO;kiu (orbital overlap) vkd`fr dks lwph-II esa n'kkZ;s o.kZu ls lqesy dhft, rFkk lwfp;ksa
ds uhps fn;s dksM dk iz;ksx djds lgh mÙkj pqfu;s %
lwph-I lwph-II
P. 1. p–d  izfrvkcU/ku (antibonding)
Q. 2. d–d  vkcU/ku (bonding)
R. 3. p–d  vkcU/ku (bonding)
S. 4. d–d  izfrvkcU/ku (antibonding)
ADVCBO - 25
Chemical Bonding-IV
Code :
dksM :
P Q R S P Q R S
(A) 2 1 3 4 (B) 4 3 1 2
(C*) 2 3 1 4 (D) 4 1 3 2
Sol.  It is d–d axial overlap in same phase, so d–d  bonding.
 It is p & d lateral overlap in same phase, so it is p–d  bonding.
 It is p and d lateral overlap in opposite phase, so it is p–d  antibonding.
 It is d–d axial overlap in opposite phase, so it is d–d  antibonding.
gy.  ;g leku izkoLFkk esa d–d v{kh; vfrO;kiu j[krs gS] blfy, d–d  cU/ku j[krs gSA
 ;g leku izkoLFkk esa p o d leikf'oZd vfrO;kiu j[krs gS] blfy, p–d  cU/ku j[krs gSA
 ;g foijhr izkoLFkk esa p o d leikf'koZd vfro;kiu j[krs gS] blfy, p–d  izfrcfU/kr j[krs gSA
 ;g foijhr izkoLFkk esa d–d v{kh; vfrO;kiu j[krs gS blfy, ;g d–d  izfrcfU/kr j[krs gSA
13.* According to Molecular orbital Theory, [JEE(Advanced) 2016, 4/124]

(A*) C22– is expected to be diamagnetic
(B) O22+ is expected to have a longer bond length than O2
(C*) N2+ and N2– have the same bond order
(D) He2+ has the same energy as two isolated He atoms
v.kq d{kd fl}kUr (Molecular orbital Theory) ds vuqlkj [JEE(Advanced) 2016, 4/124]
(A*) C2 izR;kf'’kr :i ls izfrpqEcdh; (diamagnetic) gSA
2–
(B) O22+ dh vkca/k yEckbZ (bond length) izR;kf’'kr :i ls O2 dh vkca/k yEckbZ ls yEch gSA
(C*) N2+ rFkk N2– dh vkca/k dksfV (bond length) leku gSA
(D) He2+ dh ÅtkZ nks ,dy (isolated) He ijek.kqvksa dh ÅtkZ ds leku gSA
Sol. (A) C22 Total no. of electrons = 14 so it is diamagnetic
(B) O22 Bond order = 3; O 2 Bond order = 2
 Bond length in O22 is less than bond length in O2.
(C ) Bond order of N2 = 2.5
Bond order of He2 = 1/2
 Some energy is released during the formation of He2 from two isolated He atoms.
Sol. (A) C22 bysDVªkWuksa dh dqy la[;k = 14 blfy, ;g izfrpqEcdh; gSA
(B) O22 cU/k Øe = 3; O 2 cU/k Øe = 2
 O22 esa ca/k yEckbZ O2 dh rqyuk esa de gksrh gSA
(C ) N2 dk cU/k Øe = 2.5
He2 dk cU/k Øe = 1/2
 nks He ijek.kqvksa ls He2 ds fuekZ.k ds nkSjku dqN ÅtkZ eqDr gksrh gSA
ADVCBO - 26
Chemical Bonding-IV
PART - II : JEE (MAIN) / AIEEE PROBLEMS (PREVIOUS YEARS)
Hkkx - II : JEE (MAIN) / AIEEE ¼fiNys o"kksZ½ ds iz'u
JEE(MAIN) OFFLINE PROBLEMS
1. Increasing order of bond strength of O2, O2–, O22– and O2+ is : [AIEEE-2002, 3/225]
O2, O2–, O22– rFkk O2+ dh ca/k lkeF;Z dk c<+rk gqvk Øe gS % [AIEEE-2002, 3/225]
(1) O2+ < O2 < O2– < O22– (2) O2 < O2+ < O2– < O22–
(3) O2– < O22– < O2+ < O2 (4*) O22– < O2– < O2 < O2+
–
Sol. O2 bond order = 2 ; O2 bond order = 1.5 ; O2 bond order = 1.0 ; O2+ bond order = 2.5.
2–
gy- O2 dk ca/k Øe = 2 ; O2– dk ca/k Øe = 1.5 ; O22– dk ca/k Øe = 1.0 ; O2+ dk ca/k Øe = 2.5.
2. The bond order in NO is 2.5 while that in NO + is 3. Which of the following statements is true for these
two species? [AIEEE-2004, 3/225]
(1) Bond length in NO+ is greater than in NO (2*) Bond length in NO is greater than in NO+
(3) Bond length in NO+ is equal to that in NO (4) Bond length is unpredictable
NO esa cU/k Øe 2.5 gSa tcfd NO+ esa 3 gSA bu nks iztkfr;ksa ds fy, dkSulk dFku lR; gS \ [AIEEE-2004, 3/225]
(1) NO+ esa cU/k yEckbZ] NO dh vis{kk vf/kd gSA (2*) NO esa cU/k yEckbZ NO+ dh vis{kk vf/kd gSA
(3) NO esa cU/k yEckbZ] NO ds leku gSA
+
(4) cU/k yEckbZ dks crk;k ugh tk ldrk gSA
+
Sol. NO and NO are derivative of O2.
NO(isoelectronic with O2+) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px1 = *2py)
Bond order = 1/2(10 – 5) = 2.5.
NO+(isoelectronic with O22+) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px = *2py)
Bond order = 1/2(10 – 4) = 3.
Bond order  1/bond length.
So NO+ has shorter bond length.
gy % NO rFkk NO+ nksuksa O2 O;qRiUu gSA
NO (O2+ ls lebysDVªkWfud) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px1 = *2py)
ca/k Øe = 1/2(10 – 5) = 2.5.
NO+(O22+ ls lebysDVªkWfud) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px = *2py)
ca/k Øe = 1/2(10 – 4) = 3.
ca/k Øe  1/ca/k yEckbZ
blfy, NO+ esa NksVh ca/k yEckbZ gksrh gSaA
3. Which one of the following species is diamagnetic in nature ? [AIEEE-2005, 1½/225]
 izd`fr esa fuEu esa ls dkSulh iztkfr izfrpqEcdh; gSa \ [AIEEE-2005, 1½/225]
(1) He2+ (2*) H2 (3) H2+ (4) H2–.
Sol. He2   (1s) * (1s) , one unpaired electron.
+ 2 1
H2  (1s)2 , * (1s)0 , no unpaired electron.

H2+  (1s)1 , * (1s)0, one unpaired electron.
H2– (1s)2, * (1s)1, one unpaired electron.
Sol. He2+   (1s)2 * (1s)1, ,d v;qfXer bysDVªkWuA
H2  (1s)2 , * (1s)0 , dksbZ Hkh v;qfXer bysDVªkWu ugha gSA
H2+  (1s)1 , * (1s)0, ,d v;qfXer bysDVªkWu gSA
H2– (1s)2, * (1s)1, ,d v;qfXer bysDVªkWu gSA
4. Which of the following molecules/ions does not contain unpaired electrons? [AIEEE-2006, 3/165]
 fuEu esa ls dkSulk v.kq @ vk;u v;qfXer bysDVªkWu ugha j[krk gS \ [AIEEE-2006, 3/165]
2– +
(1*) O2 (2) B2 (3) N2 (4) O2
Sol. The electronic configuration of oxygen atom is 1s2, 2s2, 2p4. Each oxygen atom has 8 electrons. Hence
in each O22– ion there are 16 + 2 = 18 electrons. The electronic configuration of O 22– ion is
KK (2s)2 * (2s)2  (2pz)2  (2px) p (2py)2*(2px)2 * (2py)2
Here KK represents non-bonding molecular orbital of 1s orbital. O22– contains no unpaired electrons.
ADVCBO - 27
Chemical Bonding-IV
The electronic configuration of B2 molecule is
KK (2s)2 *(2s)2 (2px)1 (2pz)1
it contains 2 unpaired electrons. The electronic configuration of N2+ ion is
KK (2s)2 *(2s)2 (2px)2 (2py)2 (2pz)1
it contains one unpaired electron.
The electronic configuration of O2 molecule is
KK (2s)2 *(2s)2 (2px)2 (2py)2 (2pz)2 *(2px)1 *(2py )1
It contains 2 unpaired electron.
Sol. vkWDlhtu ijek.kq dk bysDVªkWfud foU;kl 1s2, 2s2, 2p4 gSA çR;sd vkWDlhtu ijek.kq ds ikl 8 bysDVªkWu gSA
vFkkZr~ O22– ds ikl 16 +2 = 18 bysDVªkWu gSA O22– vk;u dk bysDVªkWfud vfHkfoU;kl gSA
KK (2s)2 * (2s)2  (2pz)2  (2px) p (2py)2*(2px)2 * (2py)2
tgk¡ KK, 1s d{kd ds foijhr ca/kh vkf.od d{kdksa dks n'kkZrk gSA O22– esa dksbZ Hkh v;qfXer bysDVªkWu ugha gSA
B2 v.kq dk bysDVªkWfud foU;kl
KK (2s)2 *(2s)2 (2px)1 (2pz)1
;g 2 v;qfXer bysDVªkWu j[krk gSA N2+ vk;u dk bysDVªkWfud foU;kl
KK (2s)2 *(2s)2 (2px)2 (2py)2 (2pz)1
;g ,d v;qfXer bysDVªkWu j[krk gSA
O2 v.kq dk bysDVªkWfud foU;kl
KK (2s)2 *(2s)2 (2px)2 (2py)2 (2pz)2 *(2px)1 *(2py )1
;g 2 v;qfXer bysDVªkWu j[krk gSA
5. Which of the following species exhibits the diamagnetic behaviour? [AIEEE-2007, 3/120]
fuEu esa ls dkSulh Lih'kht çfrpqEcdh; O;ogkj çnf'kZr djrk gS \ [AIEEE-2007, 3/120]
(1*) O22– (2) O2+ (3) O2 (4) NO
Sol. O2 (8 + 8 + 2 = 18) : 1s *1s 2s2 *2s2 2pz2 2px2
2– 2 2
2py2 *2px2 *2py2 ; all electrons are
paired. So diamagnetic.
gy- O22– (8 + 8 + 2 = 18) : 1s2 *1s2 2s2 *2s2 2pz2 2px2 2py2 *2px2 *2py2 ; lHkh bysDVªkWu
;qfXer gSA vr% çfrpqEcdh; gSA
6. In which of the following ionization processes, the bond order has increased and the magnetic
behaviour has changed ? [AIEEE-2007, 3/120]
fuEu esa ls fdl vk;fud izØe esa cU/k Øe c<+rk gSa vkSj pqEcdh; O;ogkj ifjofrZr gksrk gS \[AIEEE-2007, 3/120]
(1) O2   O2+ (2) N2   N2+ (3) C2 
 C2+ (4*) NO 
 NO+
Sol. Molecular orbital configuration of
vkf.od d{kd dk foU;kl
O2  1s2* 1s2 2s2 *2s2 2pz2 2py2 *2p1x *2p1y
 Paramagnetic vuqpqEcdh;
10 – 6
ca/k dksfV Bond order = =2
2
O 2+  1s2* 1s 2s2* 2s2 2p2z 2px2 2py2 * 2p1x
10 – 5
ca/k dksfV Bond order = = 2.5
2
N2  1s2* 1s2 2s2 *2s2 2p2x 2p2y 2p2z
 paramagnetic vuqpqEcdh;
10 – 4
2
N2+  1s2 * 1s2 2s2 * 2s2 2p2x 2p2y 2p1z
9–4
2
ADVCBO - 28
Chemical Bonding-IV
C2  1s2 * 1s2 2s2 * 2s2 2p2x 2p1y
8–4
2
C2+  1s2 * 1s2 2s2 * 2s2 2p2x 2p1y
7–4
2
NO  1s2 * 1s2 2s2 * 2s2 2p2z 2p2x 2p2y * 2p1x
10 – 5
2
NO+  1s2 * 1s2 2s2 * 2s2 2p2z 2p2x 2p2y
 Diamagnetic çfrpqEcdh;
10 – 4
ca/k dksfV Bond order = = 3.
2
7. Which one of the following pairs of species has the same bond order? [AIEEE-2008, 3/105]
(1) CN– and CN+ (2) O2– and CN– (3) NO+ and CN+ (4*) CN– and NO+
Lih'kht ds fuEu ;qXeksa esa ls fdlds cU/k Øe ,d leku gS ? [AIEEE-2008, 3/105]
– – – –
(1) CN vkSj CN +
(2) O2 vkSj CN (3) NO vkSj CN
+ +
(4*) CN vkSj NO+
Sol. NO is derivative of O2 ; NO (isoelectronic with O2 ) 1s * 1s  2s * 2s2 2p2z 2p2x 2p2y
+ + 2+ 2 2 2
10 – 4
Bond order = = 3.
2
CN– is derivative of N2 ; CN– (isoelectronic with N2) 1s2 * 1s2 2s2 * 2s2 2p2x 2p2y 2p2z
10 – 4
Bond order = = 3.
2
NO+ , O2 dk O;qRiUu gS ; NO+ ( O22+ ds lkFk lebysDVªkWfud gS) 1s2 * 1s2 2s2 * 2s2 2p2z 2p2x 2p2y
10 – 4
ca/k Øe = = 3.
2
CN– , N2 dk O;qRiUu gS ; CN– (N2 ds lkFk lebysDVªkWfud gS) 1s2 * 1s2 2s2 * 2s2 2p2x 2p2y 2p2z
10 – 4
ca/k Øe = = 3.
2
8. Using MO theory predict which of the following species has the shortest bond length ?
[AIEEE-2009, 4/144]
MO fl)kUr dk iz;ksx djds Kkr dhft, dh fuEu esa ls fdl Lih'kht esa vkcU/k yEckbZ U;wure gksxh \
[AIEEE-2009, 4/144]
(1) O2+ (2) O2– (3) O22– (4*) O22+
Sol. Bond Order (cU/k Øe)
10  5
(1) O2+ = 2.5
2
10  7
(2) O2– = 1.5
2
10  8
(3) O22– =1
2
10  4
(4) O22+ =3
2
1 1
Bond order  (cU/k Øe  )
bond length cU/k yEckbZ
So, O22+ has the shortest bond length. vr%] O22+ dh cU/k yEckbZ lcls de gksxhA
ADVCBO - 29
Chemical Bonding-IV
9.* Which one of the following molecules is expected to exhibit diamagnetic behaviour ?
[JEE(Main)-2013, 4/120]
fuEu v.kqvksa esa ls fdlls izfrpqEcdh; O;ogkj dh vis{kk dh tkrh gS \ [JEE(Main)-2013, 4/120]
(1*) C2 (2*) N2 (3) O2 (4) S2
Sol. N2 and C2 both are diamagnetic
Ans is (1) and (2)
N2 rFkk C2 nksuksa izfrpqEcdh; gSA
vr% mRrj (1) vkSj (2) gksxkA
10. In which of the following pairs of molecules/ions, both the species are not likely to exist ?
[JEE(Main)-2013, 4/120]
v.kq@vk;uksa ds fuEu ;qXeksa esa ls fdlesa nksuksa Lih'kht ds gksus dh laHkkouk ugha gSa \ [JEE(Main)-2013, 4/120]
(1) H2+, He22– (2) H2–, He22– (3*) H22+, He2 (4) H2–, He22+
Sol. H22+ : Bond order = 0
22
He2 : Bond order = 0
2
So, both H22+ & He2 do not exist.
gy- H22+ : ca/k Øe = 0
22
He2 : ca/k Øe = 0 vr%] H22+ rFkk He2 nksuksa vfLrRo ugha j[krs gSA
2
11. Stability of the species Li2, Li2– and Li2+ increases in the order of : [JEE(Main)-2013, 4/120]
Lih'kht Li2, Li2– vkSj Li2+ dh fLFkjrk dk c<+rk Øe gS % [JEE(Main)-2013, 4/120]
(1) Li2 < Li2+ < Li2– (2*) Li2– < Li2+ < Li2 (3) Li2 < Li2– < Li2+ (4) Li2– < Li2 < Li2+
Sol. Li2 1s2  1s2 2s2 Bond order = 1
Li2+ 1s2 1s2 2s1 Bond order = 0.5
Li2– 1s2  1s2 2s2  2s1  Bond order = 0.5
Stability order Li2 > Li2+ > Li2–
Sol. Li2 1s2  1s2 2s2 ca/k Øe = 1
Li2+ 1s2 1s2 2s1 ca/k Øe = 0.5
Li2 1s  1s 2s
– 2  2 2
 2s  ca/k Øe = 0.5
 1
LFkkf;Ro dk Øe Li2 > Li2+ > Li2–

12. Which of the following species is not paramagnetic? [JEE(Main)-2017, 4/120]
fuEu es ls dkSu lh Lih’'kht vuqpqEcdh; ugha gS\ [JEE(Main)-2017, 4/120]
(1*) CO (2) O2 (3) B2 (4) NO
Sol. No of electron in CO = 6 + 8 = 14
(i) CO 1S2, *1S2,  2S2,* 2S2, 2Px  2Py  2PZ
2 2 2
All electrons are paired so diamagnetic

(ii) O2 1S2, *1S2,  2S2, * 2S2, 2PZ2 2Px2  2Py2  ,  * 2Px1  * 2Py1 
   
Unpaired electron = 2 (Paramagnetic)
(iii) B2 1S2, *1S2, 2S2, *2S2  2Px  2Py  (Paramagnetic)
1 1
2  * 0
(iv) NO 1S2,*1S2, 2S2, *2S2, 2PZ2  2Px  2Py  ,   2Px   2Py  (Paramagnetic)
2 1 *
 
Sol. CO esa bysDVªksuks dh la[;k = 6 + 8 = 14
(i) CO 1S2, *1S2,  2S2, *2S2, 2Px  2Py  2PZ

2 2 2
lHkh bysDVªkWu ;qfXer gSA vr% izfrpqEcdh; gSA

(ii) O2 1S2, *1S2,  2S2, *2S2, 2PZ2 2Px  2Py  ,   2Px   2Py 
2 2 * 1 * 1
v;qfXer bysDVªkWu = 2 (vuqpqEcdh;)
ADVCBO - 30
Chemical Bonding-IV
(iii) B2 1S2, *1S2, 2S2, *2S2  2Px  2Py  (vuqpqEcdh;)

1 1
2  * 0
(iv) NO 1S2, *1S2, 2S2, * 2S2, 2PZ2  2Px  2Py  ,   2Px   2Py  (vuqpqEcdh;)
2 1 *
 
13. According to molecular orbital theory, which of the following will not be a viable molecule ?
[JEE(Main)-2018, 4/120]
v.kqd{kd fl)kUr ds vuqlkj] fuEu esa ls dkSu lk v.kq O;ogk;Z ugha gksxk \ [JEE(Main)-2018, 4/120]
(1) H2– (2*) H2–
2 (3) He22 (4) He2
Sol. H22– have bond order zero  do not exist
H22– 'kqU; cU/k Øe j[krk gSA vr% vfLrRo ugh gSA
2–2
H22–1S2, *1S2, B.O. = 0
2
JEE(MAIN) ONLINE PROBLEMS
1. Which of the following has unpaired electron(s) ? [JEE(Main) 2014 Online (09-04-14), 4/120]
bu esa ls fdl esa v;qfXer bySDVªkWu gksrk gS ;k gksrs gSa \ [JEE(Main) 2014 Online (09-04-14), 4/120]
(1) N2 (2*) O2– (3) N22+ (4) O22–
2. The correct order of bond dissociation energy among N2, O2, O2– is shown in which of the following
arrangements? [JEE(Main) 2014 Online (11-04-14), 4/120]
fuEu O;oLFkkvksa esa ls fdl esa N2, O2, O2– dh vkcU/k fo;kstu ÅtkZ ds lgh Øe dks fn[kk;k x;k gS \
[JEE(Main) 2014 Online (11-04-14), 4/120]
(1) N2 > O2– > O2 (2) O2– > O2 > N2 (3*) N2 > O2 > O2– (4) O2 > O2– > N2
3. Which one of the following molecules is paramagnetic ? [JEE(Main) 2014 Online (19-04-14), 4/120]
fuEu v.kqvksa esa ls dkSu vuqpqEcdh; gS \ [JEE(Main) 2014 Online (19-04-14), 4/120]
(1) N2 (2*) NO (3) CO (4) O3
4. After understanding the assertion and reason, choose the correct option.
Assertion : In the bonding molecular orbital (MO) of H2, electron density is increased between the
nuclei.
Reason : The bonding MO is A + B, which shows destructive interference of the combining electron
waves.
[JEE(Main) 2015 Online (10-04-15), 4/120]
(1*) Assertion is correct, reason is incorrect.
(2) Assertion is incorrect, reason is correct.
(3) Assertion and reason are correct, but reason is not the correct explanation for the assertion.
(4) Assertion and reason are correct and reason is the correct and reason is the correct explanation for
the assertion.
dFku o dkj.k dks le>us ds i'pkr~ lgh fodYi dk p;u dhft,A
dFku % H2 ds ca/kh vkf.od d{kd (MO) esa] ukfHkdksa ds e/; bysDVªkWu ?kuRo c<+rk gSA
dkj.k % cU/k vkf.od d{kd A + B gksrs gS] tks la;ksth (combining) bysDVªkWu rjaxksa dk fouk'kh O;frdj.k n'kkZrk
gSaA
(1*) dFku lgh gS] dkj.k xyr gSA [JEE(Main) 2015 Online (10-04-15), 4/120]
(2) dFku xyr gS] dkj.k lgh gSA
(3) dFku rFkk dkj.k lgh gS] ysfdu dkj.k dFku dh lgh O;k[;k ugha djrk gSA
(4) dFku rFkk dkj.k lgh gS rFkk ysfdu dkj.k dFku dh lgh O;k[;k djrk gSA
Sol. Bonding molecular orbital results in increased electron density between nuclei due to constructive
interference of combining electron waves.
Sol. la;ksth bysDVªkWu rjaxksa dk leiks"kh; O;frdj.k ds dkj.k ca/kh vkf.od d{kd (MO) esa] ukfHkdksa ds e/; bysDVªkWu
?kuRo c<+rk gSA
ADVCBO - 31
Chemical Bonding-IV
5. In the molecular orbital diagram for the molecular ion, N2 , the number of electrons in the 2p molecular
orbital is : [JEE(Main) 2018 Online (15-04-18), 4/120]

vkf.od vk;u] N2 ds fy, vkf.od d{kd Mk;xzke esa] 2p vkf.od d{kd esa bysDVªkWuksa dh la[;k gS %
[JEE(Main) 2018 Online (15-04-18), 4/120]
(1) 0 (2) 2 (3) 3 (4*) 1

Sol. N2  1s2 , 1s2 , 2s2 ,  2s2 , [  2p x = 2p ]  2p z
* * 2 2
y
1
Number of electron in 2pz is 1

2pz esa bysDVªksuksa dh la[;k = 1
6. Which of the following best describes the diagram below of a molecular orbital ?
[JEE(Main) 2018 Online (15-04-18), 4/120]
+ –
– +
(1) A non-bonding orbital (2) An antibonding  orbital

(3) A bonding  orbital (4*) An antibonding  orbital
v.kq d{kd ds fn;s x;s fp=k dks] fuEu esa ls dkSu loksZÙke <ax ls le>krk gS \
[JEE(Main) 2018 Online (15-04-18), 4/120]
+ –
– +
(1) ,d vukca/kh d{kd (2) ,d izfrvkca/kh  d{kd

(3) ,d vkca/kh  d{kd (4*) ,d izfrvkca/kh  d{kd
+ –
Sol.
– +
ABMO Anti bonding molecular orbital (* bond)

ABMO izfrvkca/kh vkf.od d{kd (* ca/k)
7. According to molecular orbital theory, which of the following is true with respect to Li 2+ and Li2– ?
[JEE(Main) 2019 Online (09-01-19), 4/120]
(1) Li2+ is unstable and Li2– is stable (2) Li2+ is stable and Li2– is unstable
(3*) Both are stable (4) Both are unstable
vkf.od d{kd fl)kUr ds vuqlkj Li2+ rFkk Li2– ds laca/k esa fuEufyf[kr esa ls dkSu lR; gS ?
[JEE(Main) 2019 Online (09-01-19), 4/120]
(1) Li2+ vLFkk;h gS rFkk Li2– LFkk;h gSA (2) Li2+ LFkk;h gS rFkk Li2– vLFkk;h gSA
(3*) nksuksa LFkk;h gaSA (4) nksuksa vLFkk;h gSaA
 Li2–  7e –
Sol. Li2  5e –
1s2 * 1s22s1 1s2 * 1s22s2 * 2s1
ADVCBO - 32
Chemical Bonding-IV
3–2 1 4–3 1
B.O.   B.O.  
2 2 2 2
Li2 & Li2– have same bond order since no. of bonding electrons are more than antibonding electrons
both are stable.
Li2 rFkk Li2– leku ca/k Øe j[krs gS D;ksafd cfU/kr bysDVªkWuksa dh la[;k izfrcfU/kr bysDVªkWuksa dh la[;k ls vf/kd
gksrh gSA nksuksa LFkk;h gSA
8. In which of the following processes, the bond order has increased and paramagnetic character has
changed to diamagnetic? [JEE(Main) 2019 Online (09-01-19), 4/120]
uhps fn, x, fdl izØe esa] vkca/k dksfV c<+ x;h vkSj vuqpacq dh; xq.k izfrpqacdh; esa cny x;k\
[JEE(Main) 2019 Online (09-01-19), 4/120]
(1*) NO  NO+ (2) O2  O22– (3) O2  O2+ (4) N2  N2+
Sol. NO 15e paramagnetic bond order = 2.5
NO+ 14e diamagnetic bond order = 3
NO 15e vuqpqEcdh; vkca/k dksfV = 2.5
NO+ 14e izfrpqEcdh; vkca/k dksfV = 3
9. Two pi and half sigma bonds are present in : [JEE(Main) 2019 Online (10-01-19), 4/120]
nks ikbZ rFkk vk/kk flXek vkcU/k fuEu esa ls fdlesa mifLFkr gSa \ [JEE(Main) 2019 Online (10-01-19), 4/120]
(1*) N2+ (2) O2 (3) O2+ (4) N2
Sol. N2+ = kk 2s2 *2s2  2p2x p2x 2p2y  2p1z
1
BO = 2.5 = 2  & 
2
ADVCBO - 33

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Chemical Bonding-IV

 Marked questions are recommended for Revision.

Section (B) : Applilcation of MOT

PART - II : ONLY ONE OPTION CORRECT TYPE

B-6. S1  The HOMO in F2– is *2px = *2py molecular orbitals.

PART - III : MATCH THE COLUMN

 Marked questions are recommended for Revision.

1. Number of antibonding electrons in N2 is :

2. Following is the molecular orbital configuration of a diatomic molecule

Its bond order is : (bldk ca/k Øe gS%)

3. The bond order of He2 molecule ion is :

NO2+ > NO2 > NO2– (ca/k dks.k)

gy. He2+ cU/k Øe = 2  1  1 ; O2– cU/k Øe = 10  7  1.5

Sol. N2 : (1s) (*1s) (2s) (*2s) (2p x = 2p2y) 2pz)2

The bond order of N2 is 1/2(10 – 4) = 3.

O2–, NO2 and ClO2 are odd electron species.

11. Which of the following is observed in metallic bonds ?

2. Find out the no. of correct statements :

3. In how many conversions, the bond length increases ?

4. Which of the following have bond order less than two ?

PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE

2. The species which are paramagnetic is/are :

Bond order  stability (i.e., bond strength)

vkca/k Øe LFkkf;Ro (vFkkZr~ vkca/k lkeF;Z)

9. Which is incorrect combination?

10. Which is correct combination for Diamagnetic species ?

* Marked Questions may have more than one correct option.

PART - I : JEE (ADVANCED) / IIT-JEE PROBLEMS (PREVIOUS YEARS)

Bond order of NO– = 10 – 6 / 2 = 2

Bond order of NO+ = 10 – 4 / 2 = 3

Bond order of CN– = 10 – 4 / 2 = 3

NO+ esa cU/k Øe = 10 – 4 / 2 = 3

6. Among the following, the paramagnetic compound is : [JEE–2007, 3/162]

O2 has one unpaired electron in antibonding *2py . So it is paramagnetic.

gy- KO2 , K+ & O2– esa jgrk gSA

Q. 2. d–d  vkcU/ku (bonding)

R. 3. p–d  vkcU/ku (bonding)

S. 4. d–d  izfrvkcU/ku (antibonding)

 It is p & d lateral overlap in same phase, so it is p–d  bonding.

 It is p and d lateral overlap in opposite phase, so it is p–d  antibonding.

 It is d–d axial overlap in opposite phase, so it is d–d  antibonding.

13.* According to Molecular orbital Theory, [JEE(Advanced) 2016, 4/124]

H2  (1s)2 , * (1s)0 , no unpaired electron.

LFkkf;Ro dk Øe Li2 > Li2+ > Li2–

All electrons are paired so diamagnetic

(i) CO 1S2, *1S2,  2S2, *2S2, 2Px  2Py  2PZ

lHkh bysDVªkWu ;qfXer gSA vr% izfrpqEcdh; gSA

v;qfXer bysDVªkWu = 2 (vuqpqEcdh;)

(iii) B2 1S2, *1S2, 2S2, *2S2  2Px  2Py  (vuqpqEcdh;)

Number of electron in 2pz is 1

(1) A non-bonding orbital (2) An antibonding  orbital

(1) ,d vukca/kh d{kd (2) ,d izfrvkca/kh  d{kd

ABMO Anti bonding molecular orbital (* bond)

1s2 * 1s22s1 1s2 * 1s22s2 * 2s1

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B-6. S1  The HOMO in F2– is 2px = 2py molecular orbitals.

Sol. N2 : (1s) (1s) (2s) (2s) (2p x = 2p2y) 2pz)2

(i) CO 1S2, 1S2,  2S2, 2S2, 2Px  2Py  2PZ

(iii) B2 1S2, 1S2, 2S2, 2S2  2Px  2Py  (vuqpqEcdh;)