Chemistry Test 1

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Chemistry Test 1

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Weekly Quiz Competition

Question 1
4 / -1

What is the percentage of s-character in a hybrid orbital having a bond angle of 103°?

A
100- 80 %

B
50 – 55%

C
22- 25%

D
9 – 12%

Solution

For sp³ orbital, the bond angle is 109.5° and has 25% s-character. Further, the higher the bond angle, the more the s-character. Therefore, for the given bond angle, the s-character will lies between 22 – 25%.
Question 2
4 / -1

Which of the following is the correct order of radii?

A
Co³⁺ < Co²⁺ < Co⁴⁺

B
F^-< O ^2- < N^3-

C
Na < Li < K

D
N < Be < B

Solution

In an isoelectronic species the radii decreases with the increase in atomic number as the effective nuclear charge is inversely proportional to the radii.
Question 3
4 / -1

Which of the following is false about the Kossel- Lewis Approach to Chemical Bond Formation and Octet Rule?

A
Atoms achieve a stable octet on linking by chemical bonds.

B
The outer electrons participate in the chemical bonding.

C
During the bond formation, the hybrid orbitals take part.

D
Ions are formed by electron transfer.

Solution

The concept of hybrid orbitals was given by Pauling while explaining the concept of hybridization. Hence it is not a part of
Question 4
4 / -1

According to the Molecular Orbital Theory, which of the following statements is not correct?

A
He₂⁺ can exist.

B
Be₂ is not stable.

C
In N₂, the order of energies of MO is σ2s<σ∗2s<σ2pz<π2px = π2py<π∗2px= π∗2py<σ∗2pz.

D
In the second period, out of all the homonuclear diatomic molecules, the bond strength of N₂ is found to be the maximum.

Solution

The correct order of energies for the molecular orbital of N₂ is σ1s<σ∗1s<σ2s<σ∗2s< π2px =π2py<σ2pz<π∗2px= π∗2py<σ∗2pz.
Question 5
4 / -1

On the basis of VSEPR theory, which of the following has the same shape as NH₃?

A
NO₂

B
NH₂^-

C
BF₃

D
H₃O⁺

Solution

H₃O⁺has 3 bond pairs and one lone pair exactly like NH_3.
Question 6
4 / -1

Which of the following is the correct order of the size of d-orbitals in the given elements?

A
Si> N > P > C

B
P > Si > N > C

C
Si > P > C > N

D
P > N > Si > C

Solution

The size of the d-orbital is directly proportional to the size of the element. Therefore, (c) is the correct option.
Question 7
4 / -1

Which of the following assumptions is explained with the help of VBT?

A
The difference between the strong and weak ligands.

B
The interpretation of the thermodynamic stabilities of complexes.

C
The spectral properties of the complexes.

D
The donation of ligand to the higher energy orbitals of metal.

Solution

The (a), (b), (c) are the limitations of VBT in coordination chemistry.
Question 8
4 / -1

Which of the following is the most stable coordination compound?

A
[Cu(NH₃)₂(H₂O)₄]²⁺

B
[Cu(H₂O)₄(en)]²⁺

C
[Cu(en)₃]²⁺

D
[Cu(H₂O)₆]²⁺

Solution

Complexes involving multidentate ligands are more stable than those with only unidentate.
Question 9
4 / -1

Which of the following is an ionisation isomer of [Cr(H₂O)₄Cl(NO₂)]Cl?

A
Tetraaquanitrochromium(III) chloride

B
Tetraaquadichlorochromium(III) nitrite

C
Tetraaquadinitritochromium(III) chloride

D
Triaquadichloronitritochromium(III) aqua

Solution

[Cr (H₂O)₄ Cl₂ ]NO₂is the ionisation isomer of [Cr (H₂O)₄ Cl (NO₂)] Cl.

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Chemistry Test 1

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Chemistry Test 1

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Question 1

100- 80 %

50 – 55%

22- 25%

9 – 12%

Solution

Question 2

Co3+ < Co2+ < Co4+

F-< O 2- < N3-

Na < Li < K

N < Be < B

Solution

Question 3

Atoms achieve a stable octet on linking by chemical bonds.

The outer electrons participate in the chemical bonding.

During the bond formation, the hybrid orbitals take part.

Ions are formed by electron transfer.

Solution

Question 4

He2+ can exist.

Be2 is not stable.

In N2, the order of energies of MO is σ2s<σ∗2s<σ2pz<π2px = π2py<π∗2px= π∗2py<σ∗2pz.

In the second period, out of all the homonuclear diatomic molecules, the bond strength of N2 is found to be the maximum.

Solution

Question 5

NO2

NH2-

BF3

H3O+

Solution

Question 6

Si> N > P > C

P > Si > N > C

Si > P > C > N

P > N > Si > C

Solution

Question 7

The difference between the strong and weak ligands.

The interpretation of the thermodynamic stabilities of complexes.

The spectral properties of the complexes.

The donation of ligand to the higher energy orbitals of metal.

Solution

Question 8

[Cu(NH3)2(H2O)4]2+

[Cu(H2O)4(en)]2+

[Cu(en)3]2+

[Cu(H2O)6]2+

Solution

Question 9

Tetraaquanitrochromium(III) chloride

Tetraaquadichlorochromium(III) nitrite

Tetraaquadinitritochromium(III) chloride

Triaquadichloronitritochromium(III) aqua

Solution

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Co³⁺ < Co²⁺ < Co⁴⁺

F^-< O ^2- < N^3-

He₂⁺ can exist.

Be₂ is not stable.

In N₂, the order of energies of MO is σ2s<σ∗2s<σ2pz<π2px = π2py<π∗2px= π∗2py<σ∗2pz.

In the second period, out of all the homonuclear diatomic molecules, the bond strength of N₂ is found to be the maximum.

NO₂

NH₂^-

BF₃

H₃O⁺

[Cu(NH₃)₂(H₂O)₄]²⁺

[Cu(H₂O)₄(en)]²⁺

[Cu(en)₃]²⁺

[Cu(H₂O)₆]²⁺