Chemical Bonding and Molecular Structure Test

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Chemical Bonding and Molecular Structure Test - 2

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Weekly Quiz Competition

Question 1
4 / -1

Which of the following statements is true about hybridisation?

A
The hybridised orbitals have different energies for each orbital

B
The number of hybrid orbitals is equal to the number of atomic orbitals that are hybridised

C
Hybrid orbitals form multiple bonds

D
The orbitals with different energies undergo hybridisation

Solution

The number of orbitals which hybridise remains same after the hybridisation also, e.g.,
Question 2
4 / -1

Which of the following molecules is formed by sp² hybrid orbitals?

A
CH₄

B
CO₂

C
BF₃

D
BeF₂

Solution

Formation of BF₃
Excited state of B :
Question 3
4 / -1

Atomic orbitals of carbon in carbon dioxide are

A
sp²- hybridised

B
sp³d - hybridised

C
Sp - hybridised

D
sp³- hybridised

Solution

In the formation of CO₂ molecule, hybridization of orbitals of carbon occurs only to a limited extent involving only one s and one p orbital to give rise SP hybridization.
Question 4
4 / -1

Which type of hybridisation is shown by carbon atoms from left to right in the given compound;
CH₂ = CH — C ≡ N?

A
sp², sp² sp

B
sp², sp, sp

C
sp, sp², sp³

D
sp³, sp², sp

Solution
Question 5
4 / -1

On hybridisation of one s and three p-orbitals, we get

A
Four orbitals with tetrahedral orientation

B
Three orbitals with trigonal orientation

C
Two orbitals with linear orientation

D
Two orbitals with perpendicular orientation

Solution

Four sp³ hybrid orbitals are formed when one sand three p-orbitals hybridise.
Question 6
4 / -1

The molecules like BrF₅ and XeOF₄ are square pyramidal in shape.What is the type of hybridisation shown in these molecules?

A
dsp³

B
dsp²

C
sp³d

D
sp³d²

Solution

BrF₅ and XeOF₄ have sp³d²hybridisation
hence their shape is square pyramidal.
Question 7
4 / -1

Which of the following shows dsp² hybridisation and a square planar geometry?

A
SF₆

B
BeF₅

C
PCl₅

D
[Ni(CN)₄]^2-

Solution
Question 8
4 / -1

Which of the following pairs are isostructural?

A
SO^2-₄ and BF^-₄

B
NH₃ and NH⁺₄

C
CO₃^2- and CO₂

D
CH₄ and BF₃

Solution

SO^2-₄ and BF^-₄ have sp³ hybridisation and are tetrahedral in shape.
Question 9
4 / -1

Hybridisation state of Xe in XeF₂, XeF₄ and XeF₆ respectively are

A
sp², sp³d, sp³d²

B
sp³d, sp³d², sp³d³

C
sp³d², sp³d, sp³d³

D
sp², sp³, sp³d

Solution

Total no. of valence electrons = 22
22/8 = 2(Q) + 6(R), 6/2 = 3(Q)
X = 2+3 = 5
Hybridisation is sp³d.
Hybridisation is sp³d².
Total no. of electrons in outermost shells = 8 + 28 = 36
36/8 = 4(Q) + 4(R), 4/2 = 2(Q) + 0(R)
X = 4+2 + 0 = 6
hybridisation is sp³d².
Total no. of valence electrons = 8 + 42 = 50
50/8 = 6(Q) + 2(R), 2/2 = 1(Q), X = 6+1 = 7
hybridisation is sp³d³.
Question 10
4 / -1

What is the hybrid state of carbon in ethyne, graphite and diamond?

A
sp²,sp,sp³

B
sp, sp², sp³

C
sp³,sp²,sp

D
sp², sp³, sp

Solution

In ethyne, H−C≡C−H, C is sp hybridised.
In graphite, one C atom is attached to 3 other C atoms which are sp² hybridised.
In diamond, one C atom is attached to 4 other C atoms which are sp³ hybridised.
Question 11
4 / -1

Given below is the bond angle in various types of hybridisation. Mark the bond angle which is not correctly matched.

A
dsp²−90^∘

B
sp³d²−90^∘

C
sp³d−90^∘

D
sp³−109.5^∘

Solution

In sp³d, the bond angle is 120^∘ and 90^∘.
Question 12
4 / -1

Order of size of sp, sp²and sp³ orbitals is

A
sp³< sp² < sp

B
sp < sp² < sp³

C
sp² < sp < sp³

D
sp² < sp³ < sp

Solution

The percentage of s-character in sp³,sp² and sp is 25%, 33% and 50% respectively. Order of size of orbitals is sp23.
Question 13
4 / -1

Match the column I with column II and mark the appropriate choice.

A
(A)→(i),(B)→(iii),(C)→(ii),(D)→(iv)

B
(A)→(iii),(B)→(i),(C)→(iv),(D)→(ii)

C
(A)→(ii),(B)→(iii),(C)→(i),(D)→(iv)

D
(A)→(iv),(B)→(i),(C)→(iii),(D)→(ii)

Solution

In C₂H₂,C undergoes, sp hybridisation.
In SF₆, Sundergoes sp³d² hybridisation.
In SF₆,S undergoes sp³d² hybridisation.
In IF₇,I undergoes sp³d³ hybridisation
Question 14
4 / -1

In formation of ethene, the bond formation between s and p-orbitals takes place in the following manner.

A
sp² hybridised orbitals form sigma bond while the unhybridised (p_x or p_y) overlaps sidewiseto form π-bond

B
sp² hybridised orbitals form π-bond while the unhybridised (p_z) overlaps to form σ-bond

C
sp² hybridised orbitals overlap with s-orbitals of H atoms while unhybridised orbitals form C—C bond

D
sp² hybridised orbitals form sigma bonds with H atoms while unhybridised orbitals form π-bonds between C atoms

Solution
Question 15
4 / -1

The ground state electronic configuration of S is 3s²3p⁴. How does it form the compound SF₆?

A
Due to octahedral shape ofSatoms

B
Due to presence of vacant 3d-orbitals which provide 6 unpaired electrons in excited state

C
Due to sp³ hybridisation of S atom which provides 6 electrons to 6 F atoms

D
Due to presence of 3 sigma and 3 pi bonds between S and F

Solution

S can go into excited state with 6 unpaired electrons due to presence of vacant 3d-orbitals, which are overlapped by six F electrons.