In $$Fe(CO)_5$$, the $$Fe-C$$ bond posses:

bond $$\pi$$ and $$\sigma$$ characters

MCQ Questions for Class 11 Medical Chemistry Chemical Bonding And Molecular Structure Quiz 9

Among the following molecules in increasing order of

$\sigma$ to

$\pi$ bond ratio.

0%
$2 < 3 < 4 < 1$
0%
$2 < 4 < 3 < 1$
0%
$3 < 2 < 1 < 4$
0%
$2 < 3 < 1 < 4$

The enolic form of butanone contains?

0%
$12\sigma$ bonds, $1\pi$ bond and $2$ lone pairs of electrons
0%
$11\sigma$ bonds, $1\pi$ bond and $2$ lone pairs of electrons
0%
$12\sigma$ bonds, $1\pi$ bond and $1$ lone pair of electrons
0%
$10\sigma$ bonds, $2\pi$ bonds and $2$ lone pairs of electrons
0%
$13\sigma$ bonds, $1\pi$ bonds and $2$ lone pairs of electrons

The dipole moment of

$CH_3 F$ is greater than that

$CH_3 Cl$

0%
True
0%
False

Explanation

The dipole moment of

$CH_3 F$ is greater than that

$CH_3 Cl$ because electronegativity of Fluorine is greater than that of Chlorine.

Hence, The given statement is True.

In the group the carbonyl carbon is joined to other atoms by:

0%
two sigma and one pi bond
0%
three sigma and one pi bond
0%
two sigma and two pi bonds
0%
three sigma and two pi bonds

A compound with molecular formula

$C_{4}H_{6}$ may contain.

0%
two double bond
0%
Two double bonds or a triple bond
0%
A triple bond
0%
All single bonds

Which cholrine atom is more electronegative in the following ?

0%
$CH_3CI$
0%
$CH_3CH_2CI$
0%
$(CH_3)_2CHCI$
0%
$CH_3 CH_2 - \underset{Cl}{\underset{|}{C}} - (CH_3)_2$

Explanation

$3^0$ carbon is more electronegative

hence option D is correct because of the halide attach carbon is

$3^0$

Find the incorrect match.

0%
$Al_2Cl_6$ : $3C-4e$ bond is present
0%
$Al_2(CH_3)_6$ : All carbon atoms are $sp^3$ - hybridized
0%
$I_2Cl_6$ : Nonplanar
0%
$Al_2Br_6$ : Nonpolar

Explanation

Square planar geometry for individual

$I$ -atoms, and to minimize the repulsion of lone pair on

$I$ -atoms overall molecule must be planar.

$d_{x^2-y^2}$ orbital is involved in which of the following hybridization?

0%
$sp^3d$ (sq. pyramidal)
0%
$dsp^2$
0%
$sp^3d^2$
0%
$sp^3d^3$

Explanation

$sp^3d\Rightarrow sp_xp_yp_zd_{x^2-y^2}$

$dsp^3\Rightarrow d_{x^2-y^2}sp_xp_yp_z$

$sp^3d^2\Rightarrow sp_xp_yp_zd_{x^2-y^2}d_{z^2}$

$sp^3d^3\Rightarrow sp_xp_yp_zd_{x^2-y^2}d_{z^2}d_{xy}$ .

Which of the following does not contain three electron bond?

0%
$ClO_2$
0%
$CO_2$
0%
$O_2^-$
0%
$NO$

Explanation

$ClO_{2},O_{2}^{-},NO$ all have three electron bonds. But

$CO_{2}$ has two carbon-oxygen double bonds in the regular structure. So the correct option is B.
1885607_1722021_ans_037b10efa860495ea2bdb17bc8a22db9.png

1885607_1722021_ans_037b10efa860495ea2bdb17bc8a22db9.png

Which of the following molecule/ion has a zero dipole moment?

0%
$ClF_3$
0%
$ICl_2^-$
0%
$SF_4$
0%
None of these

Explanation

Iodine is

$sp^{3}d$ hybridised in the complex

$ICl_{2}^{-}$ . Here two chlorine atoms occupy the axial positions. So the compound has a linear structure. Hence dipole moment of one

$I-Cl$ bond gets cancelled by the other

$I-Cl$ bond. So the ion has zero dipole moment. Option B is correct.
1890873_1722190_ans_ad86b9f726eb4f3ea7d3414ce7f8c6b1.png

Which is/are not correct for

$B_2H_6$ structure ?

0%
It has $4B-H$ terminal bonds and two $3C-2e$ bonds
0%
It has six $B-H$ terminal bonds and one $3C-2e$ bond
0%
It has four $B-H$ terminal bonds two $3C-2e$ bonds and one B-B bond
0%
It has interaction between $[BH_2]^+$ and $[BH_4]^-$

Explanation

The structure of diborane (B2H6) contains:

In diborane there are two bridging hydrogens which form the 3C-2e bonds (banana bond) where three atoms(two boron atoms and a hydrogen atom) share two electrons. It has four

$B-H$ terminal bonds where 2C-2e bond takes place. So statement B and C are incorrect.

Linear combination of two hybridized orbitals, belonging to two atoms and each having one electron, leads to:

0%
Sigma-bond
0%
Double bound
0%
Coordinate covalent bond
0%
Pi-bond

Explanation

Linear combination of two hybridized orbitals, belonging to two atoms and each having one electron, leads to

$\sigma$ - bond formation.

1584417_1722152_ans_d8d954bb35f84d6eb4a64576346aedee.gif

Select the correct order :

0%
Bond Strength : $NO^- < NO < NO^+$
0%
N-O bond angle : $NO^+_2 < NO^-_2 < NO^-_3$
0%
Thermal stability : $LiF > NaF >KF >RbF > CsF$
0%
Hydrated size : $Be^{2+} (aq) < Mg^{2+} (aq) < Ca^{2+} (aq) < Sr^{2+} (aq) < Ba^{2+}(aq)$

Explanation

The order of bond strength is

$NO^{-}<NO<NO^{+}$ The bond order for the ions can be calculated using the molecular orbital theory. Bond order of

$NO^{+},NO,NO^{-}$ are 3, 2.5, 2 respectively. The bond strength is directly proportional to bond order, hence greater the bond order of a ion greater is the bond strength.

The size of

$F^{-}$ is comparatively small than other common anions. So if the size of the cation is also small then the packing efficiency of that salt will be higher than the salt consisting of large size cation and fluoride ion. Higher the packing efficiency higher will be the thermal stability.

This factor overcomes the effect of Fajan's rule.

Hence option A and C are correct.

The bond between two identical non metal atoms has a pair of electron

0%
unequally shared betweem the two
0%
transferred fully from one atom to another
0%
with identical spins
0%
equally shared between them

The bond between

$B$ and

$C$ will be

0%
Ionic
0%
Covalent
0%
Hydrogen
0%
Cordiante

Explanation

Correct Answer: Option B

Explanation of Correct Option:

A covalent bond is formed by sharing of electrons.
The elements in which the last electron enters the $p$ −subshell belong to the $p$ block.
The electronegativity difference between $p$ − block elements is very less. Therefore, the bond is formed by sharing of electrons and is covalent in nature.

Explanation of Incorrect Options:

An ionic bond is formed by the complete exchange of electrons between the atoms of metal and non-metal, which results in the formation of cations and anions when the bond is broken.
A hydrogen bond is formed by a hydrogen atom and an electronegative atom and is weaker than ionic and covalent bonds.
A coordinate bond is a type of covalent bond in which the electrons are shared by only one atom.

Final Step: Correct answer - (B) Covalent.

Among the following species, identify the isostructural pairs:

$NF_3,\ NO^-_3,\ H_3O^+ ,\ N_3H,\ BF_3$

0%
$[NF_3,\ NO^-_3]$ and $[ BF_3, \ H_3O^+]$
0%
$[ NF_3,\ N_3H]$ and $[ NO^-_3,\ BF_3 ]$
0%
$[NF_3 ,\ H_3O^+]$ and $[ NO^-_3, \ BF_3]$
0%
$[NF_3,\ H_3O^+ ]$ and $[N_3H,\ BF_3 ]$

Explanation

$NF_ 3$ and

$H_3O^+$ are pyaramidal structures with three bond pairs and two lone paird due to

$sp^3$ hybridization.
1547446_1723955_ans_bdcc59d8447c4e759ef6132f58c3fdaf.PNG

All molecules with polar bonds have dipole moment.

0%
True
0%
False

Which of the following compounds has a zero dipole moment?

0%
1,1 - dichlorethylene
0%
cis -1,2 - dichlorocthylene
0%
trans -1,2 - dichloroethylene
0%
none of the above

On hybridization of one s- and one p-orbital we get :

0%
Two mutually perpendicular orbitals
0%
Two orbitals at $180^0$
0%
Four orbitals directed tetrahedrally
0%
Three orbitals in a plane

The number of lone pairs of electrons in

$XeF_2, XeF_4, XeF_6$ and

$XeOF_2$ are respectively:

0%
$3,2,1,2$
0%
$2,1,1,2$
0%
$3,2,1,1$
0%
$3,2,2,2$

The number and type of bonds between two carbon atoms in

$CaC_2$ are :

0%
One $\sigma$ and one $\pi$ bond
0%
One $\sigma$ and two $\pi$ bond
0%
One $\sigma$ and one-half $\pi$ bond
0%
One $\sigma$ bond

Explanation

$CaC_2\rightarrow Ca^{2+} + C\equiv C^{2-}$

There is a triple bond between two carbon atoms which means that there is one

$\sigma$ bond and two

$\pi$ bonds .

Acetylene molecule contains:

0%
5 sigma bonds
0%
4 sigma and $1 \pi - bond$
0%
3 sigma and $2\pi - bonds$
0%
2 sigma and $3 \pi-bonds$

A molecule possessing dipole moment is:

0%
$CH_4$
0%
$H_2 O$
0%
$BF_3$
0%
$CO_2$

The number and type of bonds in

$C^{2-}_2$ in

$CaC_2$ are:

0%
Two $\sigma$ bond and one $\pi$ bond
0%
Two $\sigma$ bond and two $\pi$ bond
0%
One $\sigma$ bond and two $\pi$ bond
0%
One $\sigma$ bond and one $\pi$ bond

Explanation

$Ca^{+2} [ C \equiv C]^{-2}$

A single bond between two atoms is always considered as sigma bond.

A double bond between two atoms is always considered as one sigma and one pi bond.

A triple bond between two atoms is always considered as one sigma bond and two pi bonds.

The types of bonds present in

$NH_4Cl$ are:

0%
electrovalent, covalent and coordinate
0%
only ionic
0%
only covalent
0%
covalent and coordinate

Explanation

$NH_4Cl$ molecule, ionic or electrovalent bond is formed between

$NH^{4+}$ and

$Cl^-$ ions, 3 covalent bonds are formed between N and three H atoms and one coordinate bond is formed between N and 1 H atom.
1682471_1725463_ans_d77c757e55b04ff49d2b6fdfd2b264f5.png

1682471_1725463_ans_d77c757e55b04ff49d2b6fdfd2b264f5.png

The type of bond present in

$CuSO_4 .5H_2O$ are only.

0%
Electrovalent and covalent
0%
Electrovalent and coordinate covalent
0%
Electrovalent,covalent and coordinate covalent
0%
Covalent and coordinate covalent

On hybridization of one s-and one p-orbitals, we get:

0%
Two mutually perpendicular orbitals
0%
Two orbitals at $180^o$
0%
four orbitals directed tetrahedrally
0%
Three orbitals in the plane

Explanation

On hybridization of one s-and one p-orbital, we get two orbitals at

$180^o$ . The number of hybrid orbitals formed is equal to the number of participating atomic orbitals, which in present case is 2. The angle between these two is equal to

$360^o$ divided by the number of orbitals.

$\dfrac {360^o}{2} =180^o$

The hybrid state and oxidation state of two oxygen atoms in

$H_2O_2$ are respectively:

0%
$sp^2, -1$
0%
$sp^2, +1$
0%
$sp^3, -1$
0%
$sp^3, +1$

The hybridization of I in ICl

$_2^+$ is :

0%
$sp$
0%
$sp^2$
0%
$sp^3$
0%
$dsp^2$

The hybridisation of oxygen atom in

$H_2O_2$ is:

0%
$sp^3d$
0%
$sp$
0%
$sp^2$
0%
$sp^3$
0%
$sp^3d^2$

Explanation

each oxygen atom in peroxide is

$sp^3$ hybridized. There is a covalent

$sp^3-sp^3$ bond between the two oxygens. Each oxygen atom is covalently linked to its respective hydrogen via.

$\mathbf{Hence\ the\ correct\ answer\ is\ option\ (D)}$

Which statement(s) is are correct?

0%
A sigma bond is stronger than $\pi-bond$
0%
A sigma bond is weaker than $\pi - bond$
0%
Hydrogen bonding is weaker than covalent bonding
0%
A triple is weaker than double bond

In which of the following the central atom has maximum number of lone piar of electrons?

0%
$SF_4$
0%
$XeF_4$
0%
$ClO_3^-$
0%
$I_3^-$

Which of the following statements is true about

$[Cu(NH_3)_4] SO_4$ ?

0%
It has coordinate as well as covalent bonds
0%
It has only coordinate bonds
0%
It has only electrovalent bonds
0%
It has electrovalent,covalent as well as coordinate bonds

Which of the following does not have

$S-S$ linkage?

0%
$S_{2}O_{8}^{2-}$
0%
$S_{2}O_{6}^{2-}$
0%
$S_{2}O_{5}^{2-}$
0%
$S_{2}O_{3}^{2-}$

The correct order of

$O-O$ bond length in

$O_{2}, H_{2}O_{2}$ and

$O_{3}$ is:

0%
$O_{3}>H_{2}O_{2}>O_{2}$
0%
$O_{2}>H_{2}O_{2}>O_{3}$
0%
$O_{2}>O_{3}>H_{2}O_{2}$
0%
$H_{2}O_{2}>O_{3}>O_{2}$

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

0%
HCl
0%
HF
0%
HI
0%
HBr

Explanation

We know dipole moment generally vary with separated charge and bond distance. The electronegativity order in halogens

$F>Cl>Br>I$ , but the bond distance order between

$HX\ (X=F,Cl,Br,I)$ is

$HI>HBr>HCl>HF$ .

In this case, bond distance and electronegativity are contradictory factors. But we also know 'F' is highly electronegative in the periodic table for which' F' can attract bond pair electron in

$H-F$ more than iodine in

$H-I$ .

So in

$H-F$ bond 'H' becomes highly electropositive and 'F' becomes highly electronegative, in this case, the charge factor is more effective than bond distance.

$OF_{2}$ molecule, the total number of bond pairs and line pairs of electron present respectively are:

0%
$2,6$
0%
$2,8$
0%
$2,10$
0%
$2,9$

In which of the following

$S-S$ link is present?

0%
Caro's acid
0%
Dithionic acid
0%
Thiosulphuric acid
0%
Chlorosulphonic acid

Explanation

From figures, we can see that, Dithionic acid and Thiosulphuric acid have

$S-S$ bond.

Hence, option

$B$ and

$C$ are the correct options.

1672502_1728667_ans_88cddd24c649447ba064b8647bd06bd2.JPG

Number of lone pairs of electrons on Xe atoms in

$XeF_2, XeF_4$ and

$XeF_6$ moleculaes are respectively :

0%
3, 2 and 1
0%
4, 3 and 2
0%
2, 3 and 1
0%
3, 2 and 0

Explanation

$XeF_2$ is linear in shape having 3 lone pairs and 2 bond pair electrons.

$XeF_4$ is square planar in shape having 2 lone pairs and 4 bond pair electrons.

$XeF_6$ is pentagonal pyramidal in shape having 1 lone pair and 6 bond pair electrons.

How many

$P-O$ bonds and how many lone pairs respectively are present in

$P_4O_6$ molecule?

0%
$12, 4$
0%
$8, 8$
0%
$12, 16$
0%
$12, 12$

Explanation

From the structure of

$P_4O_6$ , we can count the number of

$P-O$ bonds and the total number of lone pairs in the molecule. Each

$P$ atom is bonded to 3 oxygen atom, hence there are 12

$P-O$ bonds are present.

Each oxygen has 2 lone pairs and each phosphorus has one lone pair. Hence, total 16 lone pairs are present in

$P_4O_6$ ,

Hence, the answer is option

$C$ .

1623314_1728905_ans_63bbdfdb0cdc4f5a8c1cbbb3e1a1e454.JPG

The number of sigma and pi bonds in peroxydisulphuric acid are respectively:

0%
$9$ and $4$
0%
$11$ and $4$
0%
$4$ and $8$
0%
$4$ and $9$

Which of the following bonds is the weakest?

0%
$Si-Si$
0%
$Si-O$
0%
$Si-Cl$
0%
$Si-H$

Explanation

The

$Si-Si$ bond is weaker than

$Si-O$ ,

$Si-Cl$ ,

$Si-H$ is simply that because the Si atom is bigger than the

$O, Cl, H$ atom. The electrons forming the bond are farther from the nuclei and don't lower their energy as much by forming the bond.

So, the answer is option

$A$ .

The number of lone pairs and the number of

$S-S$ bonds in

$S_{8}$ molecules are respectively:

0%
$8,8$
0%
$16,8$
0%
$8,16$
0%
$8,4$

In which of the following ions, there is no

$S-S$ bond?

0%
$S_{2}O_{4}^{2-}$
0%
$S_{2}O_{6}^{2-}$
0%
$S_{2}O_{2}^{2-}$
0%
$S_{2}O_{3}^{2-}$
0%
$S_{2}O_{7}^{2-}$

Which of the following molecules of possess a permanent dipole moment?

0%
$H_{2}S$
0%
$SO_{2}$
0%
$NH_{3}$
0%
$CS_{2}$

Explanation

The molecular shape of $H_2s$ is $bent.$
The electron geometry of $SO_2$ is formed in the shape of a $trigonal\ planner.$
The molecular geometry of $NH_3$ is $trigonal \ pyramidal$

1690752_1733939_ans_6d0dfbc409bd40e6bbdff76027cbf209.jpg

$Fe(CO)_5$ , the

$Fe-C$ bond posses:

0%
$\pi$ character only
0%
$\sigma$ character only
0%
ionic character only
0%
bond $\pi$ and $\sigma$ characters

Which of the following ions does not have S-S linkage?

0%
$S_{2}O_{8}^{2-}$
0%
$S_{2}O_{6}^{2-}$
0%
$S_{2}O_{5}^{2-}$
0%
$S_{2}O_{3}^{2-}$

$M-L$ bonds in carbonyl possess:

0%
only sigma character
0%
only pi $(\pi)$ character
0%
both sigma $(\sigma)$ and pi $(\pi)$ characters
0%
none of the above

Explanation

The formation of a metal-carbon σ bond takes place because of the donation of electrons by the carbonyl molecules.They donate it to the vacant orbitals of the metal. This is one way of formation of the metal carbonyls.

The other form is the creation of a metal-carbon

$\pi$ bond because of the donation of a pair of electrons from a filled d orbital metal into the vacant antibonding

$\pi^{*}$ orbital of carbonyl ligand.

1575223_1730987_ans_efa5a36c438b4dc3a53246250a545f1f.png

The cation containing maximum number of unpaired electrons is:

0%
$Co^{2+}$
0%
$Ni`^{2+}$
0%
$Fe^{3+}$
0%
$Fe^{2+}$

B.P of

$H_2O (100^{\circ}C)$ and

$H_2S (-42^{\circ}C)$ explained by:

0%
Vander Waal's forces
0%
Covalent bond
0%
Hydrogen bond
0%
Ionic bond

Explanation

B.P of

$H_2O$ is more than that of

$H_2S$ because

$H_2O$ forms hydrogen bonding while

$H_2S$ does not.

CBSE Questions for Class 11 Medical Chemistry Chemical Bonding And Molecular Structure Quiz 9 - 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 9 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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