MCQ Questions for Class 11 Medical Chemistry Chemical Bonding And Molecular Structure Quiz 8

Covalancy of nitrogen in molecule is:

0%
$1$
0%
$3$
0%
$6$
0%
$4$

Explanation

$N=1s^22s^22p^3=3$

Nitrogen has two

$2s$ and three

$2p$ valence electrons. By sharing the three

$2p$ electrons,nitrogen can form three covalent bonds.So the co valency of nitrogen atom is 3.

Hybridization of sulphur in

$SF$

$_{4}$ molecules are :

0%
$sp$ $^{3}d^{3}$
0%
$sp$ $^{3}$ d
0%
$sp$ $^{3}d^{2}$
0%
$dsp$ $^{2}$

${ PO }_{ 4 }^{ 3- }$ ion the formal charge on the oxygen atom of P-O bond is:

0%
+1
0%
-1
0%
-0.75
0%
+0.75

Explanation

Formal charge on an atom

$=$ total no. of values electron no. of lone pair

$+\dfrac{1}{2}$ ( bonding electrons )

Formal charge on oxygen

$=6- \left[ 6+ \dfrac{1(2)}{2 \times 2} \right]$

$=6-7$

$=-1$

We know that there are

$3$ oxygen having charge than formal charge will be

$-1 \times 3 =-3$

Now the formal charge on oxygen

$= -3/4 =-0.75$

Which of the following posses two lone pairs of electrons on the cebtral atom and square planar in shape?
(i)

$SF _ 4$ (ii)

$XeO _ 4$ (iii)

$XeF _ 4$ (iv)

${ ICl }_{ 4 }^{ - }$

0%
I, III
0%
III, IV
0%
II, IV
0%
All

The main factor responsible for weak acidic nature of

$B-F$ bonds in

$BF_3$ is:

0%
large electronegativity of F
0%
three centred two electron bonds in $BF_3$
0%
$p \pi$ - $d \pi$ back bonding
0%
$p \pi$ - $p \pi$ back bonding

Explanation

As refer to the above image, in $BF_3$ boron atom accepts the lone pairs of fluorine to form $p^{\pi}-p^{\pi}$ back bonding. Hence, its Lewis acidic character is decreased.

1100225_1224595_ans_c713831d55aa4d13a3771884456d90f0.PNG

In which of the following molecule, the central atom is used

$sp^{3}d$ hybridized orbitals?

0%
$SF$ $_{4}$
0%
$NH$ $_{3}$
0%
$XeF$ $_{2}$
0%
$XeF$ $_{4}$

Explanation

A) The hybridization that is involved in

$SF_{4}$ is

$sp^{3}d$ type.

When the central metal atom is surrounded with five pairs of electrons the electronic geometry is trigonal bi pyramidal. If all the electron pairs are shared, the molecular geometry will also be trigonal bi pyramidal. In this molecule

$SF_{4}$ , there are five electron pairs which should lead to trigonal bi pyramidal structure.

B) In

$NH_{3}$ , the central atom

$NH_{3}$ is

$sp^3$ hybridized.

$\rightarrow$ The geometry of such a molecule is pyramidal.

$\rightarrow$ The bond angle is

$107^{0}$ .

C)The hybridization of

$XeF_{2}$ (Xenon Difluoride) is an

$sp^{3}d$ type.

$\rightarrow$ It involves

$Sp^{3}$ hybridization of xenon atoms.

$\rightarrow$ The molecule has trigonal bipyramidial structure with 3 position occupied by lone pair. So,

$XeF_{2}$ is linear.

$XeF_{4}$

$\rightarrow$ The central atom has 6 electron pairs out of which two are lone pair electrons. The hybridization in Xenon is

$sp^{3}d^{2}$ because there is a migration of two electrons of p to d orbital which results in the formation of sigma bond with F.

$\rightarrow$

$XeF_{4}$ is

$AB_{4}L_{2}$ in which the Xenon is bonded with four fluorine atoms with 4 sigma bonds and xenon has two lone pairs (4 sigma bonds and 2 lone pairs).

So, the hybridisation is

$sp^{3}d^{2}$ .

Hence, The correct option is A .

$H_{2}O$ is polar but

$BeF_{2}$ is not. The reason of this is

0%
$H_{2}O$ is linear, $BeF_{2}$ is angular
0%
$H_{2}O$ is angular, $BeF_{2}$ is linear
0%
$H_{2}O$ involves hydrogen bonding whereas, $BeF_{2}$ does not
0%
Electronegativitiy of $F$ is greater then that of $O$

Explanation

${H}_{2}O$ is polar because it angular and net dipole moment is not equal to 0.

${Be}{F}_{2}$ is not polar because it is linear and the net dipole moment is 0.

How many sigma and pi bonds are there in tetracyano ethylene molecule?

0%
$9\sigma$ and $9\pi$
0%
$9\sigma$ and $7\pi$
0%
$5\sigma$ and $9\pi$
0%
$5\sigma$ and $7\pi$

The number of anti-bonding electron pairs in

${O}_2^-$ molecular ion on the basis of molecular orbital theory is (atomic no. of

$O$ is

$8$ ) :

0%
$5$
0%
$2$
0%
$7$
0%
$6$

Two

$\pi$ and half

$\sigma$ bonds are present in:

0%
$N_2^+$
0%
$N_2$
0%
$O_2^+$
0%
$O_2$

Explanation

$N_2^{\oplus} \Longrightarrow BO = 2.5 \Longrightarrow \left[ \pi - Bond = 2 \& \sigma - Bond = \dfrac{1}{2}\right]$

$N_2 \Longrightarrow B.O. = 3.0 \Longrightarrow [\pi - Bond = 2 \& \sigma - Bond = 11]$

$O_2^{\oplus} = B.O. \Longrightarrow 2.5 \Longrightarrow [\pi-Bond = 1.5 \& \sigma - Bond = 11]$

$O_2 \Longrightarrow 2 \Longrightarrow [\pi - Bond \Longrightarrow 1 \& \sigma - Bond = 11]$

Identify the incorrect statement :

0%
Electronic theory of valency was presented by kossel and Lewis
0%
Covalent bond is chemical bond
0%
Hydrogen bond is chemical bond
0%
Covalent bond is strong bond

Among

$KO_2, KAlO_2, CaO_2, CaO_2 and NO_2^+$ unpaired electron is present in :

0%
$NO_2^+ , CaO_2$
0%
$KO_2 and KAlO_2$
0%
$KO_2 only$
0%
$CaO_2 only$

The valency of carbon is four. On what principle it can be explained in a better way?

0%
Resonance
0%
Hybridization
0%
Electron transfer
0%
None of the above

The hybridisation of

$S$ in

$SF_6$ molecule is

0%
$sp^3$
0%
$sp^3d^2$
0%
$sp^3d$
0%
None

Explanation

Sulphur forms

$6$ bond pairs. It thus needs

$6$ hybrid orbitals and therefore, hybridization is

$sp^3d^2$ this molecule has octahedral structure.

1921361_1370880_ans_151c4db47c7a457fb4595b7191d43a8f.png

Which one of the following compounds has permanent dipoles?

0%
$CH_4$
0%
$CO_2$
0%
$HCl$
0%
$H_2$

The dipole moment of

$HBr$ is

$2.6 \times {10^{ - 30}}cm$ and the interatomic spacing

$1.41\,A^\circ$ . The percentage of ionic in

$HBr$ is

0%
$10.5$
0%
$13.5$
0%
$12.5$
0%
$11.5$

Which of the following is not

$sp^3$ hybridise

0%
$BH_3$
0%
$BH^-_4$
0%
$NH_4^+$
0%
$NH_3$

Explanation

$BH_3 \to Sp^2$

$NH_4^+ \to Sp^3$

$BH_4^- \to Sp^3$

$NH_3 \to Sp^3$

The molecule which does not exibit dipole moment is :

0%
$N{H_3}$
0%
$CHC{l_3}$
0%
${H_2}O$
0%
$CC{l_4}$

What type of carbon-carbon bonds are present in vanaspati ghee?

0%
single bonds
0%
double bonds
0%
triple bonds
0%
none of these

Which of the following compounds has a zero dipole moment?

Choose the correct option for the following molecule in view of chemical bonding:

0%
Non-planar
0%
$\mu \ne 0$
0%
A & B both
0%
$\mu = 0$

Heat of sublimation of

$Na = x J$
Ionization energy of

$Na =yJ$
Bond dissociation energy of

$Cl_2 = a J$

$e^-$ gain enthalpy of

$Cl = -b J$
Enthalpy of formation of

$NaCl = -c J$
Then what will be the lattice energy of

$NaCl(s)$ ?

0%
$b + \left ( x+y+c-\frac{a}{2} \right )$
0%
$b - \left ( x+y+c-\frac{a}{2} \right )$
0%
$b - \left ( x+y+c+\frac{a}{2} \right )$
0%
$b -\frac{a}{2} +x+y+c$

Number of lone pairs in Xe

$OF_{ 4 }$ is

0%
0
0%
1
0%
2
0%
3

Which of the following hydrocarbons has the lowest dipole moment?

0%
cis-2-butene
0%
2-butyne
0%
I=butyne
0%
$H_2C=CH-C\equiv CH$

In a polar molecule, the ionic charge is

$4.8\times 10^{-10}$ e.s.u. If the inter ionic distance is one

$\overset {\circ}{A}$ unit, then the dipole moment is

0%
$41.8\ debye$
0%
$4.18\ debye$
0%
$4.8\ debye$
0%
$0.48\ debye$

What is the proportion of sigma

$(\sigma)$ , and pi

$(\pi)$ bond in

$1, 4- Benzoquinone$ ?

0%
$1 : 3$
0%
$3 :2$
0%
$2 : 1$
0%
$2 : 3$

Which of the following elements can show covalencey greater than 4?

0%
Be
0%
Li
0%
P
0%
B

The dipole moment of a given compound is .............than that of cyclohexyl chloride.
Choose the most appropriate word fill the blank is more

0%
more
0%
lower
0%
same
0%
may be more of lower depends upon

The molecule with the highest dipole moment is

0%
$CH_{2}Cl_{2}$
0%
$CH_{3}Cl$
0%
$CHCl_{3}$
0%
$CCl_{4}$

Which of the following name is incorrectly matched ?

0%
$H_{2}O$ - dihydrogen monoxide
0%
$SO_{2}$ - sulphur dioxide
0%
$HOCl$ - chlorous acid
0%
$H_{3}PO_{3}$ - phosphorous acid

The correct order of dipole moment is

0%
${ CH }_{ 4 }<{ NF }_{ 3 }<{ NH }_{ 3 }<{ H }_{ 2 }O$
0%
${ NF }_{ 3 }<{ CH }_{ 4 }<{ NH }_{ 3 }<{ H }_{ 2 }O$
0%
${ NH }_{ 3 }<{ NF }_{ 3 }<{ CH }_{ 4 }<{ H }_{ 2 }O$
0%
${ H }_{ 2 }O<{ NH }_{ 3 }<{ NF }_{ 3 }<{ CH }_{ 4 }$

Explanation

order of dipole movement is

$H_2O$ >

$NH3$ >

$NF_3$ >

$CH_4$

The number of ion pairs formed from one mole of

$Ca{(OH)}_{2}$ are

0%
$3$ mol ion
0%
$2$ mol ion
0%
$1$ mol ion
0%
$4$ mol ion

The atomic weight of an element is double its atomic number. If there are four electrons in

$2p$ orbital, the element is:

0%
$C$
0%
$N$
0%
$O$
0%
$Ca$

Which of the following molecule have both

$P \pi - p \pi$ and

$p \pi -d \pi$ bonding?

0%
$CIO^+_ 2$
0%
$NO^+_2$
0%
$SO^{2-}_3$
0%
$CIO^-_4$

The number of

$\sigma$ bonds and

$\pi$ bonds in the following structure respectively are:-

0%
$8,5$
0%
$8,6$
0%
$3,8$
0%
$5,13$

Dipole-dipole interaction possible ONLY if the molecules interacting are ______

0%
non-polar with zero dipole moment
0%
polar with permanent non-zero dipole moment
0%
non-polar with induced dipole moment
0%
ionic

Which of the following exhibit minimum number of oxidation states :

0%
Mn
0%
Np
0%
Th
0%
Cr

In which of the following shape is same but hybridization is different :

0%
$ICl_2^-, XeF_2$
0%
$SO_2, NO_2^+$
0%
$SO_2, NH_2^-$
0%
$CO_2, SO_2$

Explanation

Geometry for

$SO_2$ &

$NH_2^-$ is bent, but hybridization are

$sp^2$ and

$sp^3$ respectively.

% s-character of N-H bond is maximum in :

0%
$N_2H_2$
0%
$N_2H_4$
0%
$NH_3$
0%
$NH_4^+$

Which of the following is the correct order of radius?

0%
$H^{-} > H > H^{+}$
0%
$Na^{+} > F^{-} > O^{2-}$
0%
$F^{-} > O^{2-} > Na^{+}$
0%
$Al^{3+} > Mg^{2+} > N^{3-}$

Explanation

H has one electron and one proton. If one more electron is added to the shell, the effective nuclear charge per electron decreases and hence electron cloud tends to move away from the attractive positive charge and hence, the radius increases as compared to the parent atomic radius.

On the other side, when the only electron of H is removed, a single proton is left in the atom solely having a positive charge. There is no electron cloud as in to move away. So it tends to contract and therefore it's radius decreases as compared to the parent atomic radius.

Hence, the correct order is

$H^- > H > H^+$ .

Therefore, option A is correct.

The major binding force in graphite is

0%
ionic bond
0%
covalent bond
0%
hydrogen bond
0%
London force

Solid

$CO_{2}$ form:

0%
Ionic bond
0%
Van der Waal's bond
0%
Chemical bond
0%
Covalent bond

Explanation

The correct option is D.

A covalent bond is a chemical bond, involving the exchange of pairs of electrons between atoms. In case of

$CO_2$ , two oxygen and one carbon atoms share their outermost electrons between them which forms a strong covalent bond. Therefore, solid

$CO_2$ form covalent bond by sharing of electrons among them.

Option A is incorrect because ionic requires transfer of electrons.

Options B is incorrect because the van der Waals force is caused by temporary attractions between electron rich region of one molecules and electron poor regions of another.

Option C is incorrect because chemical bonding may form between two oppositely charge atom.

Consider the following species:
(i)

$[O_2]^{2-}$
(ii)

$[CO]^+$
(iii)

$[O_2]^+$
Among these sigma bond alone is present in?

0%
(i) alone
0%
(ii) alone
0%
(iii) alone
0%
(i) and (ii)
0%
(i), (ii) and (iii)

Explanation

Only

$[O_2]^{2−}$ contains one σ bond between two bonded atoms.

$O_2^{2-} =\ \ ^-O-O^-$

∴ (i) alone is the correct option.

_______ bond is the most polar.

0%
$C-O$
0%
$C-Br$
0%
$C-S$
0%
$C-F$

Explanation

$C-F$ bond is more polar since fluorine

$(F)$ is more electronegative than other elements given.

The order of electronegativity is

$F>O>Br>S$ .

Hence the correct answer is option D.

Allyl cyanide contains _____

$\sigma$ and ____

$\pi$ -bonds.

0%
$5\sigma, 7\pi$
0%
$9\sigma, 3\pi$
0%
$3\sigma, 4\pi$
0%
$9\sigma, 9\pi$

Explanation

There are 9

$\sigma$ and 3

$\pi$ bonds.
1550882_1717156_ans_4a684af8e2744a1bb2994306ccf98472.png

In which different forms does nitrogen exist in compounds?

$1$ . bonded by a triple covalent bond

$2$ . as part of a cation

$3$ in a oxidation state of

$+5$ .

0%
$1, 2$ and $3$ are correct
0%
$1$ and $2$ only are correct
0%
$2$ and $3$ only are correct
0%
$1$ only is correct

Explanation

$1$ . Di-Nitrogen is bonded by a triple covalent bond

$2$ . Nitrogen act as part of a cation as shown in the figure.

$3$ Nitrogen can have oxidation state of

$+5$ in

$N_2O_5$ .
1360889_1649244_ans_37b14f59d5944731baca1c9c5efd8ee6.png

The hybrid states of carbon atoms in

${(\overset{B}CN)_4}\overset{A}{C_2}$ are A and B and number of

$\pi$ bonds in compound is C. Then?

0%
$A=sp, B=sp^2$ and $C=9$
0%
$A=sp^2, B=sp$ , and $C=9$
0%
$A=sp^3, B=sp$ and $C=9$
0%
$A=sp^2, B=sp^2$ and $C=9$

Explanation

From the above shown strucutre we can observe that-
The hybrid state B is

$sp$ and that of A is

$sp^2$ and the total no. of

$\pi$ bond,

$C=9$ .

Option B is correct.
1552776_1717885_ans_5650fd72ff15474fac995d33331ae4c5.jpg

In the hydrocarbon,

$H_3\underset{6}{C}-\underset{5}{C}H=\underset{4}{C}H-\underset{3}{C}H_2-\underset{2}{C}\equiv \underset{1}{C}H$

The state of hybridization of carbons

$1, 3, 5$ are in the following sequence is:

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

Explanation

$H_3\underset{6}{\overset{sp^3}C}-\underset{5}{\overset{sp^2}C}H=\underset{4}{\overset{sp^2}C}H-\underset{3}{\overset{sp^3}C}H_2-\underset{2}{\overset{sp}C}\equiv \underset{1}{\overset{sp}C}H$

The state of hybridisation of carbon in

$1, 3$ and

$5$ position are

$sp, sp^3, sp^2$

Conjugated double bond is present in_______

0%
Propylene
0%
Isobutylene
0%
Butylene
0%
Butadiene

Explanation

Butadiene

$CH_{2}=CH-CH=CH_{2}$

A single bond separated by two double bonds is known as conjugated double bond.

Order of arrangement of the following compounds with increasing dipole moment is?

0%
(i) $<$ (iv) $<$ (ii) $<$ (iii)
0%
(iv) $<$ (i) $<$ (ii) $<$ (iii)
0%
(iv) $<$ (i) $<$ (iii) $<$ (ii)
0%
(iv) $<$ (ii) $<$ (i) $<$ (iii)

CBSE Questions for Class 11 Medical Chemistry Chemical Bonding And Molecular Structure Quiz 8 - 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 8 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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