Chemical Bonding and Molecular Structure Test 37

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Chemical Bonding and Molecular Structure Test 37

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Question 1
1 / -0

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?

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

B
$$NBr_{3}$$. 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.

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

D
$$NF_{3}$$. 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.

Solution

$$NF_3$$ 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.
Question 2
1 / -0

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

A
Number of unpaired electron in $$ Mn^{+y}=1$$

B
$$y=4$$

C
$$y=6$$

D
Both $$(A) \& (C)$$

Solution
Question 3
1 / -0

Dipole moment of $$CH_2 X_2 > CH_3 X, $$ then $$X$$ is :

A
$$F$$

B
$$Cl$$

C
$$Br$$

D
$$I$$

Solution
Question 4
1 / -0

The number of $$\sigma-bond$$ and $$\pi-bond$$ in $$HCP$$ are respectively:

A
$$2$$ and $$2$$

B
$$1$$ and $$3$$

C
$$2$$ and $$1$$

D
none of these

Solution
Question 5
1 / -0

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?

A
The $$H-I$$ bond is more polar than the $$H-F$$ bond because iodine is more electronegative than fluorine.

B
The $$H-F$$ bond is more polar than the $$H-I$$ bond because fluorine is more electronegative than iodine.

C
Depends on the temperature.

D
None of these.

Solution

$$H-F$$ 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 $$H$$.
$${ H }^{ \delta + }-{ F }^{ \delta - }$$
Question 6
1 / -0

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.

A
Sample A is a metal.

Sample B is a salt.

Sample C is a molecule.

B
Sample A is a molecule

Sample B is a metal.

Sample C is a salt.

C
Sample A is a salt.

Sample B is a metal.

Sample C is a molecule.

D
Sample A is a molecule.

Sample B is a salt.

Sample C is a metal.

Solution

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.
Question 7
1 / -0

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

A
$$sp^3$$ to $$sp^2$$

B
$$sp^2$$ to $$sp^3$$

C
$$sp$$ to $$sp^3$$

D
$$sp^3$$ to $$sp$$

Solution

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.
Question 8
1 / -0

Correct sequence of repulsion is:

A
Lone pair (LP)-Lone pair $$> $$ Lone pair-bond pair (BP) $$> $$ B.P $$-$$ B.P

B
B.P-B.P$$> $$ B.P-L.P $$> $$ L.P-L.P

C
B.P-B.P $$> $$ L.P.-L.P. $$> $$ BP-LP.

D
double bond-single bond $$< $$ single bond-single bond $$< $$ LP.-LP.

Solution

A) Correct
Lone pair-lone pair / lone pair-bond pair > B.P - B.P. Since lone pairs are not involved in bond formation, they are free to make more repulsion. But bond pairs are bond between two atoms and they are very close to nucleus. As a result they make lower repulsion.
Question 9
1 / -0

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

A
$$C{ Cl }_{ 4 }$$

B
$$N{ H }_{ 3 }$$

C
$${ H }_{ 2 }O$$

D
$$CH{ Cl }_{ 3 }$$

E
$$B{ F }_{ 3 }$$

Solution

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.
In $$B{F}_{3}$$, 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.
Question 10
1 / -0

Among the following, the molecules with the lowest dipole moment is

A

B

C

D

Solution

Amine group is an electron donating group nitro group is an electron withdrawing group.
Therefore, in para- position,the dipole moment of both the group act in the same direction and add up increasing the overall dipole moment.

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

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

Chemical Bonding and Molecular Structure Test 37

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Chemical Bonding and Molecular Structure Test 37

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Question 1

$$NI_{3}$$. 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

$$NF_{3}$$. 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.

Solution

Question 2

Number of unpaired electron in $$ Mn^{+y}=1$$

$$y=4$$

$$y=6$$

Both $$(A) \& (C)$$

Solution

Question 3

$$F$$

$$Cl$$

$$Br$$

$$I$$

Solution

Question 4

$$2$$ and $$2$$

$$1$$ and $$3$$

$$2$$ and $$1$$

none of these

Solution

Question 5

The $$H-I$$ bond is more polar than the $$H-F$$ bond because iodine is more electronegative than fluorine.

The $$H-F$$ bond is more polar than the $$H-I$$ bond because fluorine is more electronegative than iodine.

Depends on the temperature.

None of these.

Solution

Question 6

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

Sample A is a moleculeSample 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.

Solution

Question 7

$$sp^3$$ to $$sp^2$$

$$sp^2$$ to $$sp^3$$

$$sp$$ to $$sp^3$$

$$sp^3$$ to $$sp$$

Solution

Question 8

Lone pair (LP)-Lone pair $$> $$ Lone pair-bond pair (BP) $$> $$ B.P $$-$$ B.P

B.P-B.P$$> $$ B.P-L.P $$> $$ L.P-L.P

B.P-B.P $$> $$ L.P.-L.P. $$> $$ BP-LP.

double bond-single bond $$< $$ single bond-single bond $$< $$ LP.-LP.

Solution

Question 9

$$C{ Cl }_{ 4 }$$

$$N{ H }_{ 3 }$$

$${ H }_{ 2 }O$$

$$CH{ Cl }_{ 3 }$$

$$B{ F }_{ 3 }$$

Solution

Question 10

Solution

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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.