Chemistry Test 130

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Chemistry Test 130

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Weekly Quiz Competition

Question 1
4 / -1

The uncertainity found from the uncertainity principle (Δx.Δp = h / 4π) is

A
The minimum value

B
The maximum value

C
The exact value

D
Only an approximate value

Solution

Δx.Δp ≧ h / 4π

So by the equation given in the question, the uncertainity value will be the least.
Question 2
4 / -1

The number of orbitals in any shell (n) is equal to

A
2n²

B
n²

C
2n

D
n

Solution

Maximum number of electrons can be accommodated in a shell of principal quantum number n is 2n². Since each orbital can accommodate only 2 electrons, maximum number of orbitals =2n²/2= n²
Question 3
4 / -1

The maximum number of 4f electrons having spin quantum number (-1/2) is

A
4

B
5

C
7

D
14

Solution

Total no. of electrons in F- subshell = 14 (when we include both spin +1/2 & - 1/2)

But maximum no. of electrons = 7 if only one spin either - 1/2 or +1/2.
Question 4
4 / -1

The number of unpaired electrons in Cu⁺ (Z= 29) is

A
1

B
2

C
0

D
3

Solution

Answer : The electronic configuration of Cu is,

1s²2s²2p⁶3s²3p⁶4s¹3d¹⁰

The reason for abnormal electronic configuration of Cu is because of the half-filled and fully-filled configuration are the more stable configuration. In this configuration, the d-subshell is fully-filled and s-subshell is half-filled which makes the Cu more stable.

Now, for getting a positive charge, it has to lose 1 electron which will be removed from the outermost 4s orbital and hence all the remaining electrons will be paired leading to 0 unpaired electrons.
Question 5
4 / -1

The electrons present in K-shell of the atom will differ in

A
Principal quantum number

B
Spin quantum number

C
Magnetic quantum number

D
Azimuthal quantum number

Solution

Two electrons in K-shell will differ in spin quantum number.

They will have same values for the principal quantum number, azimuthal quantum number, and magnetic quantum number.

For first electron, n = 1,l = 0,m = 0,s = +1 / 2

For second electron, n = 1,l = 0,m = 0,s = −1 / 2
Question 6
4 / -1

Directions For Questions

Consider the following statements. The d-orbitals have

I. four lobes and two nodes

II. four lobes and one node

III. same sign in the opposite lobes

IV. opposite sign in the opposite lobes

...view full instructions

Which of the above statements are correct ?

A
I alone

B
I and III

C
II and III

D
I and IV

Solution

Total no of angular nodes = l (azimuthal quantum no), so l=2 in d orbital, therefore number of nodes is 2 and it is a fact that opposite lobes have same signs and adjacent ones have opposite signs. there are 4 lobes in d orbital.
Question 7
4 / -1

Which of the following substances has the lowest boiling point?

A
CHBr₃

B
CHCl₃

C
CHF₃

D
CHI₃

Solution

Boiling Point is directly proportional to VanderWaals forces of attraction. Down the group, this forces increases. so force of attraction in CHF₃ is weaker than CHL₃, that's why boiling point of CHF₃ is lowest.
Question 8
4 / -1

Which of the following is solid with highest melting point ?

A
CO₂

B
H₂O

C
SiO₂

D
He

Solution

SiO₂ is a covalent 'network' solid. So the energy required to break bonds will be higher. Therefore the melting point will be higher.
Question 9
4 / -1

Out of the two compounds shown below, the vapour pressure of B at a particular temperature is expected to be

A
Higher than that of A

B
Can be higher or lower depending upon the size

C
Lower than that of A

D
Same as that of A

Solution

Due to intramolecular H-bonding in o-nitrophenol it exists as a monomer while due to intermolecular H-bonding in p-nitrophenol it exists as an associate molecule. As a result, vapour pressure of a o-nitrophenol is higher than that of a p-nitrophenol.
Question 10
4 / -1

In which of the following molecules, the central atom does not have sp³ hybridisation?

A
CH₄

B
SF₄

C
BF₄^-

D
NH₄⁺

Solution

When the number of hybrid orbitals, H is 4, the hybridization is sp³ hybridization

H = 1 / 2[V + M - C + A]

where, V = number of monovalent atoms

C = total positive charge

A = negative charge

a) For CH₄

H = 1 / 2[4 + 4 - 0 + 0] = 4

sp³

b) For SF₄,

H = 1 / 2 [6 + 4 - 0 + 0] = 5

sp³d

c) For BF^-₄,

H = 1 / 2 [3 + 4 - 0 + 1] = 4, thus sp³

d) For NH⁺₄

H = 1 / 2[5 + 4 - 1 + 0] = 4 thus sp³

Thus, only in SF₄ the central atom does not have sp³ hybridisation.

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Chemistry Test 130

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Chemistry Test 130

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Question 1

The minimum value

The maximum value

The exact value

Only an approximate value

Solution

Question 2

2n2

n2

2n

n

Solution

Question 3

4

5

7

14

Solution

Question 4

1

2

0

3

Solution

Question 5

Principal quantum number

Spin quantum number

Magnetic quantum number

Azimuthal quantum number

Solution

Question 6

I alone

I and III

II and III

I and IV

Solution

Question 7

CHBr3

CHCl3

CHF3

CHI3

Solution

Question 8

CO2

H2O

SiO2

He

Solution

Question 9

Higher than that of A

Can be higher or lower depending upon the size

Lower than that of A

Same as that of A

Solution

Question 10

CH4

SF4

BF4-

NH4+

Solution

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2n²

n²

CHBr₃

CHCl₃

CHF₃

CHI₃

CO₂

H₂O

SiO₂

CH₄

SF₄

BF₄^-

NH₄⁺