Chemical Bonding and Molecular Structure Test 39

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Chemical Bonding and Molecular Structure Test 39

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Weekly Quiz Competition

Question 1
1 / -0

The number $$sp^2 - sp^2$$ sigma bonds in the compound given below is:

A
1

B
3

C
4

D
5

Solution

There are 8 $$sp^2$$ hybridized carbon atoms in the given compound which form 4 $$sp^2-sp^2$$ sigma bond with each other.
Thus, the correct answer is 4.
Question 2
1 / -0

Which of the following overlap provides strongest bond?

A
$$s{ p }^{ 3 }-s{ p }^{ 3 }$$

B
$$p-p$$

C
$$s-s$$

D
$$s-p$$

Solution

$$sp^{3}$$ has high quantity of $$p-$$ character and we know that $$p-$$orbital is directiona and hence, it would more overlap than $$s-$$orbital.
$$\therefore$$ Have more bond strength than others.
Question 3
1 / -0

Which of the following compound has highest dipole moment?

A

B

C

D

Solution
Question 4
1 / -0

The percentage ionic character of a bond having $$1.257A^0$$ its length and $$1.03$$D its dipole moment is:
($$q = 4.8 \times 10^{-10}$$ esu)

A
$$10\%$$

B
$$15\%$$

C
$$16.83\%$$

D
$$18.88\%$$

Solution

$$\mu_{ionic}(ionic\,dipole)=4.8\times 10^{-10}\times 1.275\times 10^{-8}$$
$$\mu_{ionic}=6.12\times10^{-18} esu\, cm$$=$$6.12\,D$$
Observed dipole moment $$\mu_{observed}=1.032\,D$$
% ionic character =$$\frac{\mu_{observed}}{\mu_{ionic}}\times 100$$
=$$\frac{1.032\,D}{6.12\,D}\times 100$$
=$$16.38\%$$
Question 5
1 / -0

Which of the following will show least dipole character?

A
Water

B
Ethanol

C
Ethane

D
Ether

Solution

Ethane has a symmetrical structure. Also, it does not contain any lone pair of electrons, unlike water, ethanol, and ether.
Thus, it shows the least dipole moment.
Question 6
1 / -0

Isostructural species are those which have the same shape and hybridisation. Among the given species, identify the isostructural pairs.

A
[$$NF_3$$ and $$BF_3$$]

B
[$$BF_4^-$$ and $$NH_4^+$$]

C
[$$BCl_3$$ and $$BrCl_3$$]

D
[$$NH_3$$ and $$NO^-_3$$]

Solution

From the structural point of the view, we see that,
$$NF_{3}$$ is pyramidal whereas $$BF_{3}$$ is planar triangular in structure.
$$BF^{-}_{4}$$ and $$NH^{+}_{4}$$ ions both are tetrahedral in structure.
$$BCl_{3}$$ is triangular planar whereas $$BrCl_{3}$$ is pyramidal in structure.
$$NH_{3}$$ is pyramidal whereas $$NO_{3}$$ is triangular planar structure.
So as we can see the only option B has both the elements same in structure.
The correct answer is option B.
Question 7
1 / -0

If a molecule $$MX_3$$ has zero dipole moment, the sigma bonding orbitals used by $$M$$ (atomic number $$< 21$$) are:

A
pure $$p$$

B
$$sp$$ hybrid

C
$$sp^2$$ hybrid

D
$$sp^3$$ hybrid

Solution

If dipole moment is zero in $${ MX }_{ 3 }$$, then the bond angle is $$120°$$. There should be no lone pair on central atom and $$3$$ bounds surrounded to it. One $$'s'$$ and $$2'p'$$ orbitals are included. So hybridisation is $${ sp }^{ 2 }$$.
Example, $${ BF }_{ 3 }$$.
Question 8
1 / -0

The dipole moment of HBr is $$1.6\times 10^{-30}$$ c.m. and inter atomic spacing is $$1\overset{o}{A}$$. The percentage ionic character of HBr is:

A
$$10$$

B
$$0$$

C
$$15$$

D
$$27$$

Solution

Percentage of ionic character = $$\dfrac { Observed\ dipole\ moment }{ Calculated\ dipole\ moment } \times 100$$
Observed dipole moment $$=1.6\times { 10 }^{ -30 }cm$$
Calculated dipole moment = electron charge $$\times $$ interatomic distance
$$=1.6\times { 10 }^{ -19 }\times { 10 }^{ -10 }\\ =1.6\times { 10 }^{ -29 }cm$$
% of ionic character $$=\dfrac { 1.6\times { 10 }^{ -30 } }{ 1.6\times { 10 }^{ -29 } } \times 100=10$$
$$\therefore $$ Percentage of ionic character in $$HBr$$ is $$10$$
Question 9
1 / -0

$$\triangle $$H of which of the following reaction is zero?

A
$${ H }_{ 2 }\left( g \right) \rightarrow { 2H }^{ + }\left( g \right) +{ 2e }^{ - }$$

B
$$2H\left( g \right) +aq\rightarrow { 2H }^{ + }\left( aq \right) +{ 2e }^{ - }$$

C
$$2H\left( g \right) \rightarrow { 2H }^{ + }\left( g \right) +{ 2e }^{ - }$$

D
$${ H }_{ 2 }\left( g \right) +aq\rightarrow { 2H }^{ + }\left( aq \right) +{ 2e }^{ - }$$

Solution

For electron displacement reactions of native state elements, $$\triangle H=0$$ ,
Here $${ H }_{ 2 }$$ is a native state element.
$${ H }_{ 2 }(g)+aq\longrightarrow 2{ H }^{ + }(aq)+{ 2e }^{ - }$$
Question 10
1 / -0

The dipole moments of halo compounds are in the order:

A
$$CHCl_3 > CCl_4 > CH_2Cl_2 > CH_3Cl$$

B
$$CH_3Cl > CHCl_3 > CH_2Cl_2 > CCl_4$$

C
$$CH_3Cl > CH_2Cl_2 > CHCl_3 > CCl_4$$

D
$$CHCl_3 > CH_2Cl_2 > CH_3Cl > CCl_4$$

Solution