The electron of one atom are shared between two atoms

MCQ Questions for Class 11 Medical Chemistry Chemical Bonding And Molecular Structure Quiz 6

The bond length in

$CO$ is

$1.128 \dot A$ . What will be the bond length of

$CO$ in

$Fe{ \left( CO \right) }_{ 5 }$ ?

0%
$1.158\dot A$
0%
$1.128\dot A$
0%
$1.078\dot A$
0%
$1.118\dot A$

Explanation

In $Fe(CO)_5$ there exists a synergic bond between $Fe$ and $CO$ . Due to synergic bond formation between metal and $CO$ , the bond order of $CO$ decreases.
As bond order decreases the Bond length increases.Here the only option having greater bond length than $1.128A^{0}$ is option A- $1.158A^{0}$ .
Hence, option $A$ is the correct answer.

Which of the following compound has zero dipole moment?

0%
$1, 1-$ Dichloro ethylene
0%
Cis $- 1, 2-$ dichoroethylene
0%
Trans $-1, 2 -$ dichoroethylene
0%
Both $2$ and $3$

Explanation

Trans-1,2 - dichloroethylene has

$C=C$ and the two

$Cl$ on opposite sides of the double bond. Since the two

$C-Cl$ bonds have equal and opposite dipole moments they cancel. So no dipole moment.

So, correct answer is option

$C$ .

974548_1020547_ans_b775903dae484660b4c43de0b47d69fe.png

Order of the bond strength of

$C-H$ bonds involving sp,

$sp^2$ and

$sp^3$ hybridized carbon atoms is:

0%
$sp \, > \, sp^2 \, > \, sp^3$
0%
$sp^3 \, > \, sp^2 \, > \, sp$
0%
$sp^2 \, > \, sp^3 \, > \, sp$
0%
$sp^2 \, > \, sp \, > \, sp^3$

Explanation

$\equiv \overset {sp}{C}-H$

$=\overset {sp^2}{CH}-H$

$-\overset {sp^3}{CH_2}-H$

$sp, sp^2$ and

$sp^3$ hybridized central carbon atoms, the percentage of

$s$ character in bonding orbitals (i.e.

$C-H$ bonds) are

$50$ %,

$33.3$ % and

$25$ % respectively. We know that percentage of

$s$ character increases the bond strength because the shape of

$s$ orbitals is spherical and it is more closer to nucleus of an atom. Therefore order of bond strength of

$C-H$ bonds will be:

$sp >sp^2 > sp^3$

The magnetic moment of a complex ion is

$2.83$ BM. The complex ion is

0%
$[V(H_2O)_6]^{3+}$
0%
$[Cr(H_2O)_6]^{3+}$
0%
$[Cu(CN)_4]^{2-}$
0%
$[MnCl_4]^{2-}$

Which has maximum dipole moment?

Explanation

Charge separation creates an aromatic ring only in case of

$C$ .
This separation leads to more stable aromatic ring. Also, the positive charge is more stabilized in structures

$B$ and

$C$ due to the presence of benzene rings. Hence, this separation will be favored leading to high dipole moment.

1053468_989400_ans_950c05a1b64c4d3fb30bc378d92ae19f.png

1053468_989400_ans_950c05a1b64c4d3fb30bc378d92ae19f.png

Which of the following does not contain co-ordinate bond?

0%
$\overset { + }{ PH_ 4 }$
0%
$NO_2$
0%
$O_3$
0%
$CO^{2-}_{3}$

Which of the following has zero dipole moment?

0%
I & II
0%
II & III
0%
II & IV
0%
III & IV

In covalency:

0%
Electrons are transferred
0%
Electrons are equally shared
0%
The electron of one atom are shared between two atoms
0%
None of the above

Explanation

The covalency of an atom is the number of electron it shares with the other atoms when forming chemical bonds. Usually but not always, it is the number of covalent bonds it forms with the other bonds.

Example: In

$CH_4$ , carbon shares its four electrons with four electrons of four different

$H-$ atoms to form

$4$ covlent bonds. Hence covalency of carbon is

$4$ .

Which among the following have zero dipole moment?

0%
cis- $C_2H_2Cl_2$
0%
trans- $C_2H_2Cl_2$
0%
$PCl_3$
0%
$2-$ Pentene

Explanation

(A)

$\mu\neq 0$

(B)

$\mu= 0$

(C)

$\mu \neq 0$

(D)

$\mu \neq 0$

1381147_1066877_ans_9a4e518ff1354ed0acec659d23258db4.png

The molecule which has zero dipole moment is:

0%
$CH_2Cl_2$
0%
$BF_3$
0%
$NF_3$
0%
$ClO_2$

Explanation

The

$BF_3$ molecule has a zero dipole moment.

The boron atom in

$BF_3$ show

$sp^2$ hybridization with the bond angle of

$120^{\circ}$ . Thus, it has a symmetrical structure and zero dipole moment.

Which of the following molecule is having shortest bond length of

$C-O$ bond?

0%
$CH_3OH$
0%
$H_2CO$
0%
$CO$
0%
$Na_2CO_3$

Explanation

$CO, C-O$ bond gets triple bond character in one of the resonating structures. So it has shortest bond length of

$C-O$ bond. But in

$A$

$C-O$ has single bond character. In

$B$ ,

$C-O$ bond has double character. In

$D$ ,

$C-O$ bonds have bond length greater than double bond but less than single

$C-O$ bond.

983135_1043604_ans_280a2ff7f6fa4e11a78406ab62c11bc0.JPG

The correct order of

$C-N$ bond length in the given compounds is :

$P:CH_{3}CN \quad Q:HNCO \quad R:CH_{3}CONH_{2}$

0%
$P>Q>R$
0%
$P=Q=R$
0%
$R>Q>P$
0%
$R>P>Q$

Among

${ KO }_{ 2 },{ AIO }_{ 2 }^{ - },{ BaO }_{ 2 }\ and\ { NO }_{ 2 }^{ + }$ , unpaired electron is present in :

0%
${ NO }_{ 2 }^{ + }\quad and\quad { BaO }_{ 2 }$
0%
${ KO }_{ 2 }\quad and\quad AI{ O }_{ 2 }^{ - }$
0%
${ KO }_{ 2 }$ only
0%
$Ba{ O }_{ 2 }$ only

Explanation

This can be done by counting number of valence electrons

$KO_2=1+2\times 6=17$ -Odd number of electrons and hence unpaired.

$AlO^{2-}_2=3+2\times 6+1=14$ -Even number of electrons and hence paired.

$BaO_2=2+2\times6=14$ -Even number of electrons and hence paired.

$NO^{+}_{2}=5+2\times6 -1=16$ -Even number of electrons and hence paired.

Hence option C is the correct answer.

Which of the following molecules has dipole moment greater than zero?

0%
3-3-Diethylpentane
0%
trans-2-pentene
0%
2,2,3,3-tetramethyl butane
0%
Neopentane

Among the alkaline earth metals, the element forming predominantly covalent compound is :

0%
Be
0%
Mg
0%
Sr
0%
Ca

Explanation

$Be$ to

$Ba,$ ionic character increasing due to increasing size and covalent character decreases. So, the element forming predominantly covalent compound is

$Be.$

Which statement is correct :-

0%
All the compounds having polar bonds, have dipole moment
0%
$SO_{2}$ is non-polar
0%
$H_{2}O$ molecule is non polar, having polar bonds
0%
$PH_{3}$ is polar molecule having non polar bonds

Which of the following is the strongest bond?

0%
$Na - Cl$
0%
$Cs - Cl$
0%
Both $(A)$ and $(B)$
0%
None of these

Explanation

According to Fazan's rule, a large cation with small positive charge has less polarizability there by forming strong ionic compound. between

$NaCl$ &

$CsCl$ ,

${ Na }^{ + }$ &

${ Cs }^{ + }$ have same charge of

$+1$ unit.

And,

${ Cs }^{ + }$ is bigger than

${ Na }^{ + }$ .

So,

${ Na }^{ + }$ will cause more polarization. Higher polarization of anion, greater is the covalent character & lower ionic character.

Hence,

$CsCl$ has more ionic character than

$NaCl$ .

Higher the ionic character lower is the bond strength.

Hence, the correct answer is A.

Which of the following compounds has dipole moment approximately equal to that of chlorobenzene?

0%
o-dichlorobenzene
0%
m-dichlorobenzene
0%
p-dichlorobenzene
0%
p-chloronitrobenzene

Which of the following pair of molecules posses have permanent dipole moment?

0%
${ NO }_{ 2 }\quad and\quad { CO }_{ 2 }$
0%
${ NO }_{ 2 }\quad and\quad { O }_{ 3 }$
0%
${ SiF }_{ 4 }\quad and\quad { CO }_{ 2 }$
0%
${ SiF }_{ 4 }\quad and\quad { NO }_{ 2 }$

Number of

$\sigma$ and

$\pi$ bonds in

$C_{2}$ molecule is/are:

0%
$1\sigma$ and $1\pi$
0%
$1\sigma$ and $2\pi$
0%
$2\pi$ only
0%
$1\sigma$ and $3\pi$

Explanation

$C_{ 2 }$ molecules have been found in vapour state. Their double bonds are made of two pi bonds because four electrons need to be accomodated in each bond. So this is against the rule that there must be a sigma bond before a pi bond is formed in the double bond.
Conclusion: hence the answer option (c) is correct.

Select correct statement (s) regarding

$\sigma$ and

$\pi-$ bonds.

0%
$\sigma-$ bond lies on the line joining the nuclei of bonded atoms
0%
$\pi-$ electron cloud lies on either side to the line joining the nuclei of bonded atoms
0%
$(2p_{\pi}-3d_{\pi})$ pi-bond is stronger than $(2p_{\pi}-3d_{\pi})$ pi-bond.
0%
$\sigma-$ bond has primary effect of decide direction of covalent bond, while $\pi-$ bond has no primary effect in direction of bond

Arrange the compounds in decreasing order of their dipole moment.

0%
IV> III> I> II
0%
IV> I> III> II
0%
I> III> IV> II
0%
I> IV> III> II

Explanation

More the charge separation more will be dipole moment and more

$EN$ electronegativity difference, more will be dipole moment.

${ CH }_{ 3 }{ CH }_{ 2 }{ NO }_{ 2 }>{ CH }_{ 3 }{ CH }_{ 2 }CN>{ CH }_{ 3 }{ CH }_{ 2 }-F>{ CH }_{ 3 }{ CH }_{ 2 }-{ NH }_{ 2 }$

Dipole moment

$\propto$ diff.in

$e^{-}$ unity is

$HF > HCl > HBr > HI$

0%
True
0%
False

Explanation

Dipole moment is a measure of bond polarity which in turn affected by the electronegativity of atoms in the chemical bond. When there is a greater the electronegativity of atom higher will be the dipole moment.

Electronigativity of fluorine is highest. So it has more tendency to pull the electrons towards itself than other atoms. In periodic table,down the group electronegativity decreases from top to bottom. So fluorine is most electronegative and iodine is the least electronegative element. So charge polarisation is most in HF and least in HI and hence HF has the highest dipole moment than other element.

electronegativity of halogen decreases in the order of:

$HF>HCl>HBr>HI$

Numbers of

$\sigma$ and

$pi$ bonds in

$C_{2}$ molecule is/are:

0%
$1\sigma$ and $1\pi$
0%
$1\sigma$ and $2\pi$
0%
$2\pi$ only
0%
$1\sigma$ and $3\pi$

Explanation

Configuration of

$C₂$ molecule is

$σ1s² , σ*1s² , σ2s², σ*2s² , π2py² ≈ π2pz²$

Here, you see that outermost electrons of

$C₂$ molecule { e.g.,

$π2py² ≈ π2pz²$ } is

$4$ . And it belongs to

$π$ molecular orbitals. Also, you know in bond formation only valence electrons or outermost electrons participate. Hence, in

$C₂$ molecule only

$π$ bond exists and number of pi bond =

$2$

Conclusion: hence the answer option(C) is correct.

Which of the following compounds has dipole moment approximately equal to half of chlorobenzene?

0%
$o-$ dicholorobenzene
0%
$m-$ dicholorobenzene
0%
$p-$ dicholorobenzene
0%
$p-$ choloronitrobenzene

Which (have) zero value of dipole moment?

0%
$XeF_4$
0%
$CHCl_3$
0%
$CO_2$
0%

Explanation

$XeF_4$ has square bypyramidal structure with all

$F$ atoms in one place. Hence, its dipole moment is zero. For

$CO_2$ , the two dipoles are in opposite direction.

(Refer to Image 1)

Hence its dipole moment is zero.

(Refer to Image 2)

In this case also, dipole moment is zero.

1104153_1076116_ans_1fbf5e0a2c5d4096874bb15f61cc1a30.png

The number of

$C - C$ sigma bonds present in the given compound:

0%
$16$
0%
$14$
0%
$18$
0%
$11$

$S-S$ bond is present in:

0%
$\alpha - (SO_3)_n$
0%
$\gamma (S_3O_9)$
0%
$H_2S_2O_3$
0%
$H_2S_2O_8$

Which compound has non-zero dipole moment?

The correct order of increasing bond length of

$C-H$ ,

$C-O$ ,

$C-C$ and

$C=C$ is :

0%
$C-H\ <\ C-O\ <\ C-C\ <\ C=C$
0%
$C-H\ <\ C=C\ <\ C-O\ <\ C-C$
0%
$C-C\ <\ C=C\ <\ C-O\ <\ C-H$
0%
$C-O\ <\ C-H\ <\ C-C\ <\ C=C$

Explanation

The

$C-H$ bond is the shortest due to the small size of both atoms & large electronegativity difference.

The double bonds are shorter than single bonds.

Hence

$C=C < C-O$ and

$C-C$

Due to electronegativity difference between

$C$ and

$O$

$C-O < C-C$

$\therefore$ Correct order of increasing bond length is :

$C-H < C=C < C-O < C-C$

Hence, the correct option is B.

Which of the following molecule does not possess a permanent dipole moment?

0%
$CS_2$
0%
$SO_{3}^{2-}$
0%
$H_2S$
0%
$SO_2$

Explanation

Structure of given molecules.

${ CS }_{ 2 }$ does not passes a permanent dipole moment.

1380475_1098000_ans_f47f880ed94e4ef9a7faf88c971a20f5.jpg

Select the molecule that has one

$pi\ bond$ ?

0%
$CH=CH$
0%
${CH}_{2} =CH-CHO$
0%
${CH}_{3}-{CH}={CH}_{2}$
0%
${CH}_{3}-{CH}={CH}-{COOH}$

Explanation

(A)

$CH_2\overset {1}{=}CH_2$

$O$

(B)

$CH_2\overset {1}{=}CH-\overset {2||}{C}-H$

(C)

$CH_3-CH\overset {1}{=}CH_2$

$O$

(D)

$CH_3-CH\overset {1}{=}CH-\overset {2||}{C}-OH$

Compound in option

$(A)$ &

$(B)$ have one

$Pi-$ bond in their structure.

Which of the following is not correct?

0%
$\text{A sigma bond is weaker than } \pi \ bond$
0%
$\text{A sigma bond is stronger than } \pi \ bond$
0%
$\text{A double bond is stronger than a single bond}$
0%
$\text{A double bond is shorter than a single bond}$

Explanation

The sigma bond is formed by the co-axial overlapping of the orbitals while pi bond is formed by the lateral overlapping of the orbitals.

Thus, the extent of overlapping is more in sigma bond than in pi bond.

Thus $\sigma$ bond is stronger than the $\pi$ bond.

987381_1094561_ans_71702128952b4fe2ad03f91f2a633059.PNG

A: Tetracyanomethane
B: Carbondioxide
C: Benzene
C:

$1,3-buta-di-ene$
Ratio of

$\sigma$ and

$\pi$ :

0%
in A and B are equal
0%
in C and D are equal
0%
is same in A, B, C, and D
0%
in B and D are equal

Explanation

$a:$ Tetracyanomethane, see the figure i. where

$\sigma$ bond

$=8$ , ratio

$=1$ ,

$\pi$ bond

$=8.$

$b:$ Carbondioxide,

$O=C=O\\ \sigma=2\\ \pi=2$ , ratio

$=1.$

$c:$ Benzene, see the figure ii, where

$\sigma$ bond

$=12,$

$\pi$ bond

$=3$ , ratio

$=4.$

$d:$

$1,3-$ buta-di-ene, see the figure iii, where

$\sigma$ bond

$=9,$

$\pi$ bond

$=2,$ ratio

$=\cfrac{9}{2}.$

$\therefore\text{ }A=B$

1167969_1098912_ans_3c847adfe51040acbf5103374acbcc3f.png

From the given list of compounds whose proton bonding is highest?

0%
$H_2O$
0%
$H_2S$
0%
$NH_3$
0%
$PH_3$

Explanation

1. $H_2O$ has the highest proton bonding.

2.(a) In a water molecule, the electrons of the hydrogen atoms are strongly attracted to the oxygen atom.

3. Therefore, the part of the water molecule containing the oxygen atom has a slightly negative charge, and the part containing hydrogen atoms has a slightly positive charge.

4.In $NH_3, PH_3 , H_2S$ hydrogen bonding is comparably low.

Which of the following has the maximum value of dipole moment?

Explanation

Here, Chlorine is more electronegative than

$NO_2$ .

Therefore, for chlorine the dipole moment will be greater than that for

$NO_2$ .

Also,

$\mu=e\times d$

The greater the distance between the atoms, the higher will be the dipole moment.

Therefore, the distance is greater in IMAGE

$01$ and IMAGE

$02$ .

Therefore, IMAGE

$01$ has the maximum value of dipole moment.

1057839_1108581_ans_c22fb8d825f9488c9ce4aa7ac7af0588.PNG

In which of the following molecules, bonding is not taking place in excited state:

0%
$CH_4$
0%
$BF_3$
0%
$IF_7$
0%
$PCl_3$

Explanation

In the case of

$PCl_3$ , only outer electrons are going to be shown for simplicity.

Each atom in $PCl_3$ has $2,8$ electrons in inner layers of orbit.
Every atom has a noble gas structure.

So, in

$PCl_3$ bonding is not taking place in excited state.

The molecular that will have dipole moment IS ZERO

0%
2,2-Dimethylpropane
0%
Cis-2-butane
0%
Trans-2-pentene
0%
2,2,3,3-Tereathylpentane

Peroxide linkage is present in :

0%
$H_2O$
0%
$BaO_2$
0%
$OF_2$
0%
$H_2SO_4$

Explanation

Peroxide linkage

means (0-0) linkage

this linkage is present in

$BaO_2$ which is barium oxidation .

refer to image

B is the correct answer.

1052768_1113649_ans_95c6a78430ca4c9fbd8e5db10bfe8306.PNG

The correct order of decreasing bond length of

$C = C$ in the above compounds is:

0%
$II>I>IV>III$
0%
$III>II>I>IV$
0%
$IV>II>I>III$
0%
none of the above

Explanation

The bond length of

$C=C$ is affected by presence or absence of electron pushing group like

$-CH_3$ . Presence of methyl groups decrease the electron deficiency and the length of double bond thus increases. Similarly, presence of more double bonds has more electron attracting tendency of the bonds, due to which more double bonds cause shorter

$C=C$ .

Hence the decreasing bond length of

$C=C$ will be:

$II > I > IV >III$

In which of the following dimer empty atomic orbital of central atom of monomer does not involve in hybridisation?

0%
$Ga_2H_6$
0%
$Al_2Br_6$
0%
$Be_2H_4$
0%
$Cl_2O_6$

Explanation

Dichlorine hexoxide is the chemical compound with the molecular formula

$Cl_2O_6$ , which is correct for its gaseous state. However, in liquid or solid form, this chlorine oxide ionizes into the dark red ionic compound chloryl perchlorate

$[ClO_2]^+[ClO_4]^−$ , which may be thought of as the mixed anhydride of chloric and perchloric acids.

$Ga_2H_6,Al_2Br_6,Be_2H_4$ have no empty orbital only in

$Cl_2O_6$

$Cl$ has empty orbital and

$Cl_2O_6$ dimer empty atomic orbital of central atom of monomer does not involve in hybridisation.

The dipole moment of a bond

$AB$ is

$5\ D$ and its bond distance is

$150\ pm$ . The percentage ionic charactor of this bond is?

0%
$78.5\%$
0%
$83\%$
0%
$24\%$
0%
$69.5\%$

Explanation

Dipole moment= charge

$\times$ distance between the dipoles

$=1.6 \times 10^{-19}\times 150 \times 10^{-12}$ (

$1$ pm=

$10^{-12}$ m)

$=\cfrac {240 \times 10^{-31}}{3.336}=7.190$ (

$10=3.336 \times 10^{-30}$ cm)

% of ionic character=

$\cfrac {5}{7.19}\times 100=69.54$ %

A diatomic molecule has a dipole moment of 1.2D. If the bond distance is

$1.0 \mathring{A}$ , fraction of an electronic charge on each atom is:

0%
$10\%$
0%
$20\%$
0%
$25\%$
0%
$50\%$

Explanation

Electronic charge

$'e'=\dfrac { Dipole\quad moment }{ Bond\quad length } =\dfrac { 1.2\times { 10 }^{ -18 }esucm }{ 1\times { 10 }^{ -18 }cm } =1.2\times { 10 }^{ -10 }esu$

Now fraction of electronic charge

$=\dfrac { 1.2\times { 10 }^{ -10 } }{ 4.8\times { 10 }^{ -10 } } =0.25$

$\because$ charge of electron

$=4.8\times { 10 }^{ -10 }esu=25$ % of

${ e }^{ - }$

Ans :- Option C.

Which of the following pair of molecules will have permanent dipole moments for both members?

0%
$SiF_{4}$ and $NO_{2}$
0%
$CO_{2}$ and $NO_{2}$
0%
$CO_{2}$ and $O_{3}$
0%
$SiF_{4}$ and $CO_{2}$

Explanation

$CO_2$ is a linear molecule, thus its net dipole moment is zero.

(Refer to Image 1)

$SiF_4 \longrightarrow$ tetrahedral geometry (Refer to Image 2)

Net dipole moment is not zero.

$NO_2 \longrightarrow$ bent structure, because of lone pair of Nitrogen

(Refer to Image 3)

Thus net dipole moment is not zero.

Thus

$SiF_4$ &

$NO_2$ both contain permanent dipole moment.

1162232_1117287_ans_feca4b410f324bd68400e7be9faf6949.png

Which of the following has zero dipole moment?

0%
$BF_3$
0%
$BeClBr$
0%
$CH_2Cl_2$
0%
COS

Explanation

$BF_3$ the shape of the molecule is trigonal planar. The three

$F$ are so attached to

$B$ that the resultant dipole due to two

$B-F$ bonds cancels the third dipole. Hence the overall dipole moment of the molecule is

$0$ . Hence the correct option is

$A$ .

1108110_1117711_ans_4a28abe8cc7248a6b4fa13f7c5d8292b.webp

If dipole moment of HI is

$0.38$ D and bond length of HI is

$1.6 AA$ , then percentage covalent characted in HI bond is?

0%
$4.95 \%$
0%
$5.9 \%$
0%
$95.05 \%$
0%
$47.5 %$

The structural formula of a compound is

$CH_3-CH=C=CH_2$ . The type pf hybridization at the four carbon atoms from left to right is?

0%
$sp^2, sp, sp^2, sp^3$
0%
$sp^2, sp^3, sp^2, sp$
0%
$sp^3, sp^2, sp, sp^2$
0%
$sp^3, sp^2, sp^2, sp^2$

The Hybridisation and magnetic behavior of complex

$[Ni(CO)_{4}]$ is

0%
$dsp^{2}$ , paramagnetic
0%
$dsp^{2}$ , diamagnetic
0%
$sp^{3}$ , paramagnetic
0%
$sp^{3}$ , diamagnetic

The dipole moment of

$HBr$ is

$0.78\ \times 10^{-18}$ esu cm and interatomic spacing is

$1.41\ \mathring {A}$ The

$\%$ ionic character of

$HBr$ is?

0%
$7.5$
0%
$11.$
0%
$15$
0%
$27$

Explanation

$D=$ charge

$\times$ distance

$=1.6 \times 10^{-19}\times 1.41\times 10^{-8}$

$=\cfrac {2.256\times 10^{-27}}{3.33\times 10^{-30}}$

$=6.76 \times 10^2$

% of ionic charge=

$0.78 \times 10^{-18} esu$

$=0.78$

$=\cfrac {0.78}{6.76}=11$ % .

Which of the following molecule stabilizes by removal of electron?

0%
$C_2$
0%
$O_2$
0%
$N_2$
0%
$H_2$

Explanation

In all these molecules that is,

$C_2,\ O_2, \ , N_2$ and

$H_2$ .

$O_2$ , is stabilized by removal of electrons because it is having electrons in the anti-bonding orbital. And thus removal of electron from the anti bonding orbital , increases the bond order of

$O_2$ . And thus is becomes more stable after removal of electrons. Bond order of

$O_2{^+}$ is 2.5 than that of

$O_2$ which is having bond order as 2. Therefore the answer is option

$B$ .

CBSE Questions for Class 11 Medical Chemistry Chemical Bonding And Molecular Structure Quiz 6 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Chemical Bonding And Molecular Structure Quiz 6 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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