Chemistry Test 172

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Chemistry Test 172

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Weekly Quiz Competition

Question 1
4 / -1

NO₂ can be represented as

Formal charge on each oxygen atom is

A
-2

B
-1

C
0

D
+1

Solution
Question 2
4 / -1

Directions For Questions

Fluorine nitrate, is an oxidising agent, used as a rocket propellant.
this oxygen is different from other two)

...view full instructions

Which is/are correct a -bonding scheme?

A

B

C
O(p)— N(sp²)

D
All are correct σ -bonding schemes

Solution
Question 3
4 / -1

In compound X, all bond angles are exactly 109° 28’, X is:

A
Chloroform

B
Chloromethane

C
Carbon tetrachloride

D
Iodoform

Solution

In compound X, all the bond angles are exactly 109^∘28′X is carbon tetrachloride.

CCl has sp³ - hybridization with 4 bond pairs and 0 lone pairs of electrons giving regular tetrahedron geometry as all the four C−Cl bonds are equivalent. In others the geometry is little distorted inspite of sp³ - hybridization due to different atoms on the vertices of tetrahedron.
Question 4
4 / -1

The most stable shape of ClF₃ is shown by

A
(i) only

B
(i) and (iii)

C
(ii) only

D
(iii) only

Solution

In (i), the Ip are atequatorial position so there are less Ip - bp repulsions as compared to other positions.

Hence, T-shape is moststable.
Question 5
4 / -1

Which of the following statements is correct regarding the structure of PCI₅?

A
Three P—Cl bonds lie in one plane and two P—Cl bonds lie above and below the equatorial plane

B
Five P—Cl bonds lie in the same plane

C
The bond angle in all P—Cl bonds is 90°

D
The bond length of all P—Cl bonds is same

Solution

In PCI₅, three axial and two equatorial bonds are present.
Question 6
4 / -1

Correct decreasing order of dipole moment of CH₃F, CH₃CI and CH₃Br is

A
CH₃F > CH₃CI > CH₃Br

B
CH₃F > CH₃Br > CH₃C

C
CH₃CI > CH₃F > CH₃Br₂

D
CH₃CI > CH₃Br > CH₃F

Solution

Bond length and (EN) oppose each other, with the larger halogens having longer bonds but weaker (EN). Thus CH₃CI > CH₃F > CH₃Br.
Question 7
4 / -1

Ratio of fraction of electric charge existing on each atom in HCI and HI is

A
1.2:1

B
2.7:1

C
3.3:1

D
1:3.3

Solution
Question 8
4 / -1

PCI_x F_{5 - x} shows non-polar behaviour when x is

A
0, 5

B
0

C
2, 1,4

D
0, 4, 1

Solution

The Phosphorus Pentafluoride (PF₅) is a nonpolar molecule because the configuration of PF₅ has trigonal bipyramidal, and PF₅ consists of a central phosphorus atom, surrounded by five fluorine atoms. So the three planar fluorine’s electronegativity cancels out each other, which resulting PF₅ is a nonpolar molecule.

similarly, in PCl₅case chlorine's electronegativity cancels out each other.

In case of PCl₃F₂, the molecule is non-polar, even though the fluorine atoms are slightly more electro-negative than the chlorine. This could be explain in a couple ways. The more intuitive approach is to consider how the presence of two fluorine atoms would create symmetry. The less intuitive, but more accurate, process would be to realize that the structure of the molecule is divided in to two non-polar structure(trigonal planer and linear). With other molecules, like PF₃Cl₂ it's not possible to divide it in to non-polar shapes. The repulsion between the fluorine atom is greater than that between the chlorine. Thus PF₃Cl₂ is polar.
Question 9
4 / -1

The BCI₃ is a planar molecule whereas NCI₃ is pyramidal, because

A
B-Cl bond is more polar than N-Cl bond

B
N-Cl bond is more covalent than B-Cl bond

C
Nitrogen atom is smaller than boron atoms

D
BCI₃ has no lone pair but NCI₃has a lone pairofelectrons

Solution

One lone pair of electrons is available on N atom, occupies a corner in the tetrahedral arrangement. Therefore, NCI₃ appears pyramidal in shape.
Question 10
4 / -1

Bond order of (B— H) x and y bonds in diborane are respectively

A
1, 1

B
0, 0.5

C
0.5, 0.5

D
1, 0.5

Solution