The figure above represents any of the four trihalides of nitrogen, $$NF_{3}, NCl_{3}, NBr_{3}$$, and $$NI_{3}$$, where the X represents the halogen atoms. Here is some additional information regarding the elements involved. Element Electronegativity Atomic Radius $$N$$ $$3.0$$ $$70 pm$$ $$F$$ $$4.0$$ $$64 pm$$ $$Cl$$ $$3.0$$ $$99 pm$$ $$Br$$ $$2.8$$ $$114 pm$$ $$I$$ $$2.5$$ $$133 pm$$ Of the four nitrogen trihalides, only one is not a molecular dipole. Which one is the nonpolar molecule, and why is it nonpolar?

$$NF_{3}$$. The electron density in the three N-F bonds is drawn toward the base of the trigonal pyramidal structure, while the lone pair place electron density at the top of the pyramid. The effects off-set.

$$NI_{3}$$. The bond polarities off-set effect of the lone pair.

$$NBr_{3}$$. The bonds are virtually nonpolar, and the size of the bromine atoms cause the bond angles to be larger than the normal $$109.5^{\circ}$$ for tetrahedral orbital geometry.

$$NCl_{3}$$. Nitrogen and chlorine each have electronegativities of $$3.0$$, so the bonds are nonpolar, which means the molecule will be nonploar

You have three solids. One is a metal; one is an ionic compound; one is a molecular/covalent compound. You test all three solid samples to identify their bonding. Sample Appearance Melting Point $$( ^{\circ}C)$$ Conductivity A Shiny $$962$$ conductive B Crystalline $$801$$ conductive in aqueous solution C Powdery $$186$$ not conductive Based on this data, identify the type of bonds present in each sample.

Sample A is a metal. Sample B is a salt. Sample C is a molecule.

Sample A is a molecule Sample B is a metal. Sample C is a salt.

Sample A is a salt. Sample B is a metal. Sample C is a molecule.

Sample A is a molecule. Sample B is a salt. Sample C is a metal.

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

Which of the following contains maximum number of lone pairs on the central atom?

0%
$\mathrm{C}\mathrm{l}\mathrm{O}_{4}^{-}$
0%
$\mathrm{X}\mathrm{e}\mathrm{F}_{4}$
0%
$\mathrm{S}\mathrm{F}_{4}$
0%
$\mathrm{I}_{3}^{-}$

Explanation

From figure we get option

$D$ is correct.

Which of the following orbitals can not undergo hybridisation amongst themselves?

$(I)\ 3d, 4s$

$(II)\ 3d, 4d$

$(III)\ 3d, 4s$ &

$4p$

$(IV) \ 3s, 3p$ &

$4s$

0%
$II$
0%
$II$ and $III$
0%
$I, II$ and $IV$
0%
$II$ and $IV$

Explanation

$(II) \ 3d$ and

$4d$ can not undergo hybridization without involving

$4s$ .

$(IV) \ 3s, 3p$ and

$4s$ can not undergo hybridization without involving

$3d$ .

Which of the following statements is incorrect?

0%
Out of trimethylamine and trimethylphosphine, trimethylamine has higher dipole moment.
0%
Out of $(SiH_3)_2O$ and $(CH_3)_2O$ , $(SiH_3)_2O$ is more basic.
0%
The critical temperature of the water is higher than that of $O_2$ because of $H_2O$ molecule has the dipole moment.
0%
Intermolecular hydrogen bonding increases the enthalpy of vaporization of a liquid due to the increase in the attraction between molecules.

Explanation

(A) Nitrogen is more electronegative than phosphorus.
So, dipole moment of trimethylamine is greater than trimethy phosphine.
(B) In trisilyl ether the lone pair of electron on oxygen atom is less easily available for donation because of

$p \pi - d \pi$ delocalisation due to presence of the vacant d-orbital with Si. This however is not possible with carbond in CH

$_3$ - O - CH

$_3$ due to the absence of d-orbital making it more basic.
(C) The critical temperature depends on the magnitude (of strength) of intermolecular force of attraction between the molecules. If a molecule has dipole moment it means there is dipole attraction between the molecules and thus it will affect the critical temperature.
(D) Hydrogen bond acts as a bridge between atoms which holds one atom by covalent bond and the other by hydrogen bond and this increase in the attraction between molecules.

Which of the following types of bonds are present in

$CuSO_4\cdot 5H_2O$ ?
I: Electrovalent II: Covalent III: Coordinate

0%
I, II
0%
II, III
0%
I, III
0%
I, II, III

Explanation

Here,

$CuSO_{4}$ is ionic compound. When it is hydrated, it forms hydrogen bonding with one of the water molecule and coordinate bonds with the other 4 water molecules. It has covalent bonding in

$SO_4^{2-}$ .

∴CuSO4.5H2O contains all the three types of bond

i. Electrovalent (Ionic interaction between

$Cu^{2+}$ &

$SO_4^{2-}$ )

ii.Covalent (with

$SO_4^{2-}$ )

iii.Co ordinate (H2O molecules to

$Cu^{2+}$ )

Maximum covalency shown by

$Be$ and

$B$ is:

0%
2 and 3
0%
2 and 4
0%
4 and 3
0%
4 and 4

$XeF_2, XeF_4$ and

$XeF_6$ , the number of lone pairs of

$Xe$ is respectively:

0%
$2, 3, 1$
0%
$1, 2, 3$
0%
$4, 1, 2$
0%
$3, 2, 1$

Explanation

number of lone pairs of

$Xe$ are

$3,2$ and

$1$ .

In which of the following pairs, indicated bond is of greater strength?

0%
I
0%
II
0%
Both are equal
0%
None of the above

Which among the following have two lone pair of electrons on center atom?

0%
$CO_2$
0%
${ClF}_3$
0%
$SO_3^{2-}$
0%
$XeF_4$

Explanation

$XeF_4$ and

$ClF_3$ , four electrons are present in the nonbonding orbitals of

$Cl$ and

$Xe$ .
Also, from the structure it can be seen that

$XeF_4$ and

$ClF_3$ have two lone pair of electrons on center atom.

The ratio of lone pairs in

$\displaystyle XeF_{2}$ molecule and the lone pairs on its central atom is:

0%
$3$
0%
$4$
0%
$5$
0%
$6$

Explanation

Total no. of lone pairs molecule:

$9$

Ratio of lone pairs in molecule to that of central atom is

$9:3$

$\Rightarrow 3:1$

$\therefore$ correct answer is option A.

In a molecule of phosphorus (V) oxide, there are:

0%
$4P-P, 10P-O$ and $4P = O$ bonds
0%
$12P-O$ and $4P = O$ bonds
0%
$2P-O$ and $4P = P$ bonds
0%
$6P-P, 12P-O$ and $4P= P$ bonds

Explanation

As can be seen from the image, it contains

$12$

$P-O$ bonds and

$4 \:P=O$ bonds.

Rotation around the bond (between the underlined atoms) is restricted in:

0%
$\displaystyle\:\underline{C}_{2}H_{4}$
0%
$\displaystyle\:\underline{C}_{2}O_{2}$
0%
$\displaystyle\:\underline{C}_{2}H_{2}$
0%
$\displaystyle\:\underline{C}_{2}H_{6}$

Explanation

$H_2C=CH_2$ ,

$HC\equiv CH$ and

$O=C=C=O$ have double/triple bond. Thus, they have pi bonds and in pi bond, the rotation is not possible as the bond is formed by lateral overlapping.

so rotation is restricted in these compounds.

$CH_3-CH_3$ have a single bond. Thus it contains sigma bond which can be rotated on its axis. Thus, in

$CH_3-CH_3$ the free rotation is possible.

The total number of valence electrons in

$4.2\ gm$ of

${N}_{3}^{-}$ ion are :

0%
$2.2\ N_A$
0%
$4.2\ N_A$
0%
$1.6\ N_A$
0%
$3.2\ N_A$

Explanation

As each molecule of

${N}_3^-$ has

$16$ valence electrons.

Therefore, one mole will have

$16$ mole of electrons.

Number of valence electrons = No. of moles

$\times N_A\times$ valence electrons.

Here,

$N_A$ is the Avagadro number.

$= \dfrac{4.2}{42}\times N_A\times 16$

$= 1.6N_A$

Hence, option C is correct.

The correct statement regarding

$SO_2$ molecule is:

0%
two $p\pi-d\pi$ bonds
0%
molecule has 2 lone pair, $2\sigma$ bonds and $2\pi$ bonds
0%
two $p\pi-p\pi$ bonds
0%
one $p\pi-p\pi$ bond and one $p\pi-d\pi$ bond

Explanation

$SO_2$ molecule has one

$p\pi-p\pi$ bond and one

$p\pi-d\pi$ bond.

${ SO }_{ 2 }=6+6(2)=18e-16e=2e\rightarrow 1$ lone pair

$\rightarrow$ G.S.=[Ne]

$3{s}^{2}$

$3{p}^{4}$

2 Unpaired electron

${1}^{st}$ E.S.

$= [Ne] 3{s}^{2}3{p}^{3}3{d}^{1}$

4 unpaired electron

$O\rightarrow 1{ s }^{ 2 }2{ s }^{ 2 }2{ p }^{ 4 }$

$\rightarrow$ Geometry of

$SO_2$ is BENT/ANGULAR

$\rightarrow$ one

$p\pi-p\pi$ and one

$p\pi-d\pi$ Bond

$\rightarrow$

$2\pi$ Bonds

$\rightarrow$

$1$ lone pair on the central atom.

Out of given reaction which show change in hybridization of central atom__________.

0%
${ H }_{ 2 }{ BO }_{ 3 }$ dissolve in water
0%
${ H }_{ 2 }{ SO }_{ 4 }$ dissolve in water
0%
${ N }_{ 2 }{ O }_{ 5(g) }\longrightarrow { N }_{ 2 }{ O }_{ 5(s) }$
0%
$P{ Br }_{ 5(g) }\longrightarrow P{ Br }_{ 5(s) }$
0%
${ C }_{ 2 }{ H }_{ 6 }\xrightarrow [ bond\ cleavage\ of\ C-C\ bond ]{ Homolytic }$

Explanation

Correct Option

$- A$

Reason:-

$H_{3}BO_{3}\longrightarrow HBO_{2} + H_{2}O$

$H_{3}BO_{3}$ has

$sp^{3}$ hybridisation at the central atom and

$HBO_{2}$ has

$sp^{2}$ hybridisation at the central atom.

Assuming the bond direction to the z-axis, which of the overlapping of atomic orbitals of two atoms (

$A$ ) and (

$B$ ) will result in bonding?
(I)

$s$ -orbital of

$A$ and

$p_x$ orbital of

$B$
(II)

$s$ -orbital of

$A$ and

$p_z$ orbital of

$B$
(III)

$p_y$ orbital of

$A$ and

$p_z$ orbital of

$B$
(IV)

$s$ -orbital of both (

$A$ ) and (

$B$ )

0%
I and IV
0%
I and II
0%
III and IV
0%
II and IV

Explanation

As s is nondirectional so it will form bond in all directions.
Also

$p_z$ is directed on z-axis so it will form bond on z-axis.
So II and IV are correct.

The distance between two neutral atoms in equilibrium is :

0%
$\displaystyle \sqrt\frac{2A}{B}$
0%
$\displaystyle \left(\frac{2A}{B}\right)^\dfrac{1}{6}$
0%
$\displaystyle \frac{2A}{B}$
0%
$\displaystyle \left(\frac{2A}{B}\right)^\dfrac{1}{2}$

Explanation

The distance between two neutral atoms in equilibrium is

$\displaystyle \left(\frac{2A}{B}\right)^\dfrac{1}{6}$ .
This is obtained when the value of the force F is substituted as zero in the expression for the force which is

$F=\dfrac {12A}{r^{13}}-\dfrac {6B}{r^7}$ .

Dipole moment of

$\displaystyle NH_{3}$ is ________ than

$\displaystyle NF_{3}.$

0%
medium
0%
less
0%
more
0%
none of the above

Explanation

Dipole moment of

$NH_3$ is more than

$NF_3$ . Because in

$NH_3$ , the Nitrogen is more electronegativity than the hydrogen, so electron density shift towards nitrogen. On the other hand, in

$NF_3$ , the electron equally distributed to the three fluorine atoms as it has more electronegative than nitrogen.

The correct option is C.

Match list I (Complex ions) with list II (number of unpaired electrons) and select the correct answers:

	List I		List II
A	$[CrF_6]^{4-}$	1	One
B	$[MnF_6]^{4-}$	2	two
C	$[Fe(CN)_6]^{4-}$	3	zero
D	$[Mn(CN)_6]^{4-}$	4	four
		5	five

0%
$A-4, B-1, C-2, D-5$
0%
$A-2, B-5, C-3, D-1$
0%
$A-4, B-5, C-3, D-1$
0%
$A-2, B-1, C-3, D-5$

Explanation

(A) The complex ion

$[CrF_6]^{4-}$ contains four unpaired electrons.
(B) The complex ion

$[MnF_6]^{4-}$ contains five unpaired electrons.
(C) The complex ion

$[Fe(CN)_6]^{4-}$ contains zero unpaired electrons.
(D) The complex ion

$[Mn(CN)_6]^{4-}$ contains one unpaired electrons.

An ionic compound is expected to have tetrahedral structure if r+ / r- lies in the range of:

0%
0.414 to 0.732
0%
0.225 to 0.414
0%
0.115 to 0.225
0%
0.732 to 1

Which of the following combination of orbitals does not form any type of covalent bond (if z-axis is molecular axis)?

0%
$p_z+p_z$
0%
$p_y+p_y$
0%
$s+p_y$
0%
$s+s$

Explanation

$s+p_y$ doesnt form covalent bond. as

$p_y$ is oriented along

$y$ axis and

$s$ is non-directional. So according to

$VBT$ they will not form a covalent bond.

Consider the given lewis dot structure:

The formal charges on

$B$ and

$F$ are respectively.

0%
$-1 , 0$
0%
$-1 , -1$
0%
$0 , -1$
0%
$-1 , +1$

Explanation

Formal Charge = [valence electrons on atom] - [non-bonded electrons + number of bonds].

$B=3-[0+4]=-1$

$F=7-[1+6]=0$

Which plot best represents the potential energy $(E)$ of two hydrogen atoms as they approach one another to form a hydrogen molecule?

Which of the following orbital combination does not form

$\pi$ -bond?

0%
$p_x+p_x$ sideways overlapping
0%
$d_{\displaystyle x^2-y^2}+p_y$ sideways overlapping
0%
$d_{xy}+d_{xy}$ sideways overlapping
0%
$d_{yz}+p_y$ sideways overlapping

Explanation

$d_{\displaystyle x^2-y^2}+p_y$ sideways overlapping does not form

$\pi$ -bond as the two orbitals do not have proper orientation for the overlap.

Which species has the maximum number of lone pair of electron on the central atom?

0%
${ \left( Cl{ O }_{ 3 } \right) }^{ - }$
0%
$Xe{F}_{4}$
0%
$Sn{F}_{4}$
0%
${ \left( { I }_{ 3 } \right) }^{ - }$

Explanation

${ I }_{ 3 }^{ - }$ has the maximum number of lone pair of electron on the central atom.

${ I }_{ 3 }^{ - },Xe{F}_{4},S{F}_{4}$ and

$Cl{ O }_{ 3 }^{ - }$ have

$3,2,1,1$ lone pair of electron respectively.

What is the formal charge on the chlorine atom in the oxyacid

$HOClO_2$ if it contains single bonds ?

0%
$-2$
0%
$-1$
0%
$+1$
0%
$+2$

Explanation

formal charge= valence electrons-(lone electrons+bonds)

$=7-(2+3)=+2$

Assertion: In dilute benzene solutions, equimolar addition of

$R_3N$ and

$HCl$ produce a substance with a dipole moment. In the same solvent, equimolar addition of

$R_3N$ and

$SO_3$ produce a substance having an almost identical dipole moment.
Reason:(

$N-S$ ) bond is a more polar bond.

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct
0%
Both Assertion and Reason are incorrect

Explanation

In dilute benzene solutions, equimolar addition of

$R_3N$ and

$HCl$ produce a substance with a dipole moment. In the same solvent, equimolar addition of

$R_3N$ and

$SO_3$ produce a substance having an almost identical dipole moment.

This is because, both

$HCl$ and

$SO_3$ are Lewis acids and can react with the amine base to form polar substances which undergo ionic dissociation in a solvent sufficiently more polar than benzene.

Moreover, (

$N-S$ ) bond is a more polar bond.

The correct order of dipole moment is :

0%
$C{ H }_{ 4 }< N{ F }_{ 3 }< N{ H }_{ 3 }< { H }_{ 2 }O$
0%
$N{ F }_{ 3 }< C{ H }_{ 4 }< N{ H }_{ 3 }< { H }_{ 2 }O$
0%
$N{ H }_{ 3 }< N{ F }_{ 3 }< C{ H }_{ 4 }< { H }_{ 2 }O$
0%
${ H }_{ 2 }O< N{ H }_{ 3 }< N{ F }_{ 3 }< C{ H }_{ 4 }$

Explanation

The correct order of dipole moment is

$C{ H }_{ 4 }< N{ F }_{ 3 }< N{ H }_{ 3 }< { H }_{ 2 }O$ . Dipole moment of

${CH}_{4}$ = 0,

$N{ F }_{ 3 }$ =0.235 D,

$N{ H }_{ 3}$ = 1.46 D and that of

${ H }_{ 2 }O = 1.85 D.$

Assuming the additivity of covalent radii in the

$C - I$ bond, what would be the iodine-iodine distance in each of the diiodobenzene? Assume that the ring is regular hexagon and that each

$C - I$ bond lies on a line through the centre of the hexagon. (Take

$C- C$ bond-length

$= 1.40\mathring{A}$ radius of iodine atom

$=1.33\mathring{A}$ , radius of carbon atom

$= 0.77\mathring{A}$ )
In

$o -, m -$ and

$p -$ isomers, bond angles are respectively

$60^o, 120^o$ and

$180^o$ . Hence, the distance between two-atoms is:

0%
$3.5\mathring{A}$
0%
$4.0\mathring{A}$
0%
$7.0\mathring{A}$
0%
$8.0\mathring{A}$

Explanation

(a)

$I-$ atom are at

$C$ and

$D$ positions.

$CD = ?$
Also,

$\triangle OCD$ is equilateal triangle.

$CD = OC = OA + AC$

$= 1.40 + 1.33 + 0.77$

$= 3.50\mathring{A}$
(

$AC$ is the sum of radii of carbon and iodine and

$OA = AB$ )
(b)

$CD = 2DE$
But

$\displaystyle \frac{DE}{OD} = sin \,60^o$

$DE = OD \, sin \, 60^o$

$= (OB+BD)sin \, 60^o$

$= (1.40 + 0.77 +1.33)sin \, 60^o$

$=\displaystyle 3.50\times \frac{\sqrt{3}}{2} = 3.03\mathring{A}$

$\therefore CD = 6.06\mathring{A}$
(

$BD = AC$ as in (a))
(c)

$OC = OD = 3.50\mathring{A}$

$CD = 2\times 3.50=7.0\mathring{A}$

$XeF_2$ ,

$XeF_4$ and

$XeF_6$ the number of lone pairs on Xe is respectively__________.

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

Explanation

$XeF_2$ has 2 bond pairs and 3 lone pairs.

$XeF_4$ has 4 bond pairs and 2 lone pairs.

$XeF_6$ has 6 bond pairs and 1 lone pairs.

For the type of interactions:

$(I)$ Covalent bond,

$(II)$ van der waal's forces,

$(III)$ Hydrogen bonding,

$(IV)$ Dipole-dipole interaction, which represents the correct order of increasing stability?

0%
$(I)< (III)< (II)< (IV)$
0%
$(II)< (III)< (IV)< (I)$
0%
$(II)< (IV)< (III)< (I)$
0%
$(IV)< (II)< (III)< (I)$

The

$C- C$ single-bond distance is

$1.54\mathring {A}$ . What is the distance between the terminal carbons in propane? Assume that the Four bonds of any carbons atom are pointed towards the corners of a regular tetrahedron.
Plan structure of propane is assumed to be tetrahedral with bond angle of

$\theta = 109^o28'$ ,. thus, form the value of

$\left(\frac{\theta}{2}\right)$ and sin values, distance between terminal C-atoms can be determind.

0%
$6.52\mathring{A}$
0%
$4.52\mathring{A}$
0%
$2.52\mathring{A}$
0%
None of these

Explanation

Two terminal carbons can be assumed to be at A and B, while the central carbon at O. Then,

$AB = 2AP$
BUt,

$\displaystyle \frac{AP}{AO} = sin \left(\frac{\theta}{2}\right)$

$= \displaystyle sin \left(\frac{109^o28'}{2}\right) \because$ (in tetrahedral structure

$\theta = 109^o28'$ )

$= sin (54^o 44')$

$\therefore AP = AO\,sin (54^o 44')$

$=1.54 \times 0.82 = 1.26 \mathring{A}$

$\therefore AB = 2AP = 2.52\mathring{A}$

The hybridizations of atomic orbitals of nitrogen in

$NO_2^+ ,\ N{ O }_3^-$ and

$NH_4^+$ respectively are :

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

Explanation

$NO_2^{+}$ contains 2 sigma bonds so hybridization is

$sp$ and shape is linear.

$N{ O }_3^{-}$ contains 3 sigma bonds so hybridization is

$sp^{2}$ and shape is trigonal.

$NH_4^{+}$ contains 4 sigma bonds so hybridization is

$sp^{3}$ and shape is tetrahedral.

Hence, option A is correct.

What is hybridization?

0%
The mixing of two or more atomic orbitals to form new orbitals that describe the covalent bonding in molecules.
0%
The mixing of electrons in the electron sea model to account for the number of bonds in a molecule.
0%
The mixing of two or more atomic orbitals to form new orbitals that describe the location of electrons in atoms.
0%
The mixing of two or more molecules
0%
The mixing of two or more atoms to make a new compound or molecule.

The formal charges of

$N_{(1)}, N_{(2)}$ and

$O$ atoms in

$: \overset {. .}{N_{(1)}} = N_{(2)} = \overset {. .}{O}:$ are respectively

0%
$+1, -1, 0$
0%
$-1, +1, 0$
0%
$+1, +1, 0$
0%
$-1, -1, 0$

Explanation

You can write this two ways:

$:N:::N:O:::$
That has a (+) on the middle nitrogen and a

$(-)$ on the oxygen
Or you can write it as

$::N::N::O::$
That has a

$(+)$ on the middle nitrogen and a

$(-)$ on the left nitrogen. The first way is preferred because the oxygen (the more electronegative element) gets the negative charge.

What is a non-bonding electron domain?

0%
A lone pair of electrons
0%
An area of electron density
0%
A double bond
0%
The area of electron density in an anion

What is the first thing you need to know to use the hybridization theory?

0%
The number of valence electrons in the participating atoms.
0%
The number of total electrons in the participating atoms.
0%
Where the element lies on the periodic table.
0%
How many of each molecule are being made.
0%
The valence bond theory.

Which of the following best describes the term dipole moment?

0%
When electrons are unevenly distributed within a molecule
0%
When several atoms in a compound are electronegative
0%
A non polar molecule
0%
Having more than four dipoles

Calculate the formal charges on oxygen atoms 1, 2, 3 respectively.

0%
0, -1, +1
0%
0, +1, -1
0%
1, 0, -1
0%
1, 0, +1

Explanation

The formula of formal charge,

$FC = V - N - \dfrac B2$

where V the number of valence electrons of the neutral atom in isolation (in its ground state); N is the number of non-bonding valence electrons on this atom in the molecule, and B is the total number of electrons shared in bonds with other atoms in the molecule.

The atoms have been numbered as 1, 2, and 3. The formal charge on:
The central O atom marked 1:

$= 6 - 2 - \dfrac 12 (6) = +1$

The end O atom marked 2:

$= 6 - 4 - \dfrac 12 (4) = 0$

The end O atom marked 3:

$= 6 - 6 - \dfrac 12 (2) = -1$

The figure above represents any of the four trihalides of nitrogen,

$NF_{3}, NCl_{3}, NBr_{3}$ , and

$NI_{3}$ , where the X represents the halogen atoms. Here is some additional information regarding the elements involved.

Element	Electronegativity	Atomic Radius
$N$	$3.0$	$70 pm$
$F$	$4.0$	$64 pm$
$Cl$	$3.0$	$99 pm$
$Br$	$2.8$	$114 pm$
$I$	$2.5$	$133 pm$

Of the four nitrogen trihalides, only one is not a molecular dipole. Which one is the nonpolar molecule, and why is it nonpolar?

0%
$NI_{3}$ . The bond polarities off-set effect of the lone pair.
0%
$NBr_{3}$ . The bonds are virtually nonpolar, and the size of the bromine atoms cause the bond angles to be larger than the normal $109.5^{\circ}$ for tetrahedral orbital geometry.
0%
$NCl_{3}$ . Nitrogen and chlorine each have electronegativities of $3.0$ , so the bonds are nonpolar, which means the molecule will be nonploar
0%
$NF_{3}$ . The electron density in the three N-F bonds is drawn toward the base of the trigonal pyramidal structure, while the lone pair place electron density at the top of the pyramid. The effects off-set.

Explanation

$NF_3$ the net dipole moment is zero as the dipole towards

$F$ is cancelled by the dipole of the lone pair. Hence, cancelling each other's effect.

Dipole moment of

$CH_2 X_2 > CH_3 X,$ then

$X$ is :

0%
$F$
0%
$Cl$
0%
$Br$
0%
$I$

The number of

$\sigma-bond$ and

$\pi-bond$ in

$HCP$ are respectively:

0%
$2$ and $2$
0%
$1$ and $3$
0%
$2$ and $1$
0%
none of these

The

$H-F$ bond and the

$H-I$ bond have different values of their bond polarity. Which of the following statements correctly identifies the bond that is more polar and gives the correct reason for this difference?

0%
The $H-I$ bond is more polar than the $H-F$ bond because iodine is more electronegative than fluorine.
0%
The $H-F$ bond is more polar than the $H-I$ bond because fluorine is more electronegative than iodine.
0%
Depends on the temperature.
0%
None of these.

Explanation

$H-F$ is more polar then

$H-I$ due to the high electronegativity of

$F$ , it attracts the shared pair of

${ e }^{ - }$ towards itself more, thereby acquiring

$S-$ charge on

$F$ and

${ S }^{ + }$ on

$H$ .

${ H }^{ \delta + }-{ F }^{ \delta - }$

You have three solids. One is a metal; one is an ionic compound; one is a molecular/covalent compound. You test all three solid samples to identify their bonding.

Sample	Appearance	Melting Point $( ^{\circ}C)$	Conductivity
A	Shiny	$962$	conductive
B	Crystalline	$801$	conductive in aqueous solution
C	Powdery	$186$	not conductive

Based on this data, identify the type of bonds present in each sample.

0%
Sample A is a metal.

Sample B is a salt.

Sample C is a molecule.
0%
Sample A is a molecule

Sample B is a metal.

Sample C is a salt.
0%
Sample A is a salt.

Sample B is a metal.

Sample C is a molecule.
0%
Sample A is a molecule.

Sample B is a salt.

Sample C is a metal.

Explanation

Sample A:- Shiny Appearance

${ 962 }^{ 0 }C$ melting point

conducts electricity

Sample A is a metal because of a very high melting point and as it conducts electricity. As metals are lustrous, malleable, ductile and good conductors of heat and electricity. As it is shiny, it is likely to be solid, it is a metallic solid.

Sample B:- Crystalline appearance

${ 801 }^{ 0 }C$ melting point

conducts electricity in aqueous solution.

Sample B is a salt as it is a crystal and conducts electricity only in aqueous solution as it forms its ions in aqueous solution.

Sample C:- Powdery appearance

${ 186 }^{ 0 }C$ melting point

do not conduct electricity.

In addition reactions of aldehydes, the carbonyl carbon atom changes from:

0%
$sp^3$ to $sp^2$
0%
$sp^2$ to $sp^3$
0%
$sp$ to $sp^3$
0%
$sp^3$ to $sp$

Explanation

In addition reactions of aldehyde,carbonyl carbon atom changes from

$sp^2 \rightarrow sp^3$ as the

$C=O$ bond breaks to

$C-O^-$ bond.

Among the following, the molecule of highest dipole moment is:

0%
$C{ Cl }_{ 4 }$
0%
$N{ H }_{ 3 }$
0%
${ H }_{ 2 }O$
0%
$CH{ Cl }_{ 3 }$
0%
$B{ F }_{ 3 }$

Explanation

As the s-character increases, dipole moment increases.

Therefore, percentage s-character in

$B{F}_{3} \; \& \; {H}_{2}O$ is highest among all given compounds.

$B{F}_{3}$ , all the three bonds are in the same plane and dipole moments cancel one another resulting net dipole moment equal to zero.

Hence,

${H}_{2}O$ has highest dipole moment among all the given compounds.

In case of water molecule the two O—H bonds are oriented at an angle of 104.5° and have a bent structure. The dipole moment of

${H}_{2}O$ is 1.84 D, i.e., the resultant of the dipole moments of two bonds.

Find the incorrect option with respect to the relevant characteristic.

0%
$NH_3> NF_3$ (Dipole moment)
0%
$CO_2< CO_3^-2$ (CO bond length)
0%
$NH_2^- < NH_4^+$ (Bond Angle)
0%
$I^->I^+$ (Ionization energy)

If element

$Mn^{y+}$ has spin only magnetic moment 1.732 BM, then:

0%
Number of unpaired electron in $Mn^{+y}=1$
0%
$y=4$
0%
$y=6$
0%
Both $(A) \& (C)$

Hybridisation of carbon in

${ CH }_{ 2 }=CH-\overset { \oplus }{ { CH }_{ 2 } }$ is:

0%
${ sp }^{ 2 }$
0%
$sp$
0%
${ sp }^{ 3 }$
0%
all of these

Explanation

$S{ N }_{ 1 }=\dfrac { 4+2 }{ 2 } =\dfrac { 6 }{ 2 } =3-s{ p }^{ 2 }\\ S{ N }_{ 2 }=\dfrac { 4+2 }{ 2 } =\dfrac { 6 }{ 2 } =3-s{ p }^{ 2 }\\ S{ N }_{ 3 }=\dfrac { 4+3-1 }{ 2 } =3-sp^{ 2 }$

Methanoic acid,

$HCOOH$ has two different

$C-O$ bond lengths.

0%
True
0%
False

Explanation

$2$

$C-O$ bonds are of different bond length. One is

$1.23 \overset {o} {A}$ and other is

$1.32 \overset {o} {A}$ .

Complete the balanced and following reaction:

$Na_2HPO_4 + X \rightarrow Y + NaH_2PO_4$

0%
X = $NaHCO_3$

Y = NaCl
0%
X = $HCO_{3}^-$

Y = $NaHCO_3$
0%
X = $NaHCO_3$

Y = $H_2CO_3$
0%
X = $H_2CO_3$

Y = $NaHCO_3$

Radius ratio	Structural arrangement
Above 0.732	Cubic
0.414 – 0.732	Octahedral
0.225 – 0.414	Tetrahedral
0.155 – 0.225	Triangular

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

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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