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Test: Hybrididation and VSEPR theory - JEE MCQ


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20 Questions MCQ Test - Test: Hybrididation and VSEPR theory

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Test: Hybrididation and VSEPR theory - Question 1

In compounds of type ECl3, where E = B,P, As or Bi, the angles Cl − E − Cl for different E are in the order.

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(Trigonal planar geometry); Bond angle

In these, order of bond angle:
BCl> PCl> AsCl> BiCl3

Test: Hybrididation and VSEPR theory - Question 2

For which of the following sets of geometry, both axial and equatorial positions are present?

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Test: Hybrididation and VSEPR theory - Question 3

Which of the following statements is true?

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SF6 is octahedral and F − S − F bond angle is 90

Test: Hybrididation and VSEPR theory - Question 4
Which one of the following is a correct pair with respect to molecular formula of xenon compound and hybridization state of xenon in it?
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Xe in I excited state
undergoes hybridisation, whereas undergoes hybridisation.
Test: Hybrididation and VSEPR theory - Question 5
Chlorine atom, in its third excited state, reacts with fluorine to form a compound . The formula and shape of are :
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(3rd excited state)

Chlorine atom, in its third excited state, reacts with fluorine to form . Its shape is pentagonal bipyramidal.
Test: Hybrididation and VSEPR theory - Question 6

The bond angle between two hybrid orbitals is Calculate the percentage of -character of hybrid orbital.

Detailed Solution for Test: Hybrididation and VSEPR theory - Question 6

s -character ∝ bond angle

For 25%s character (as in sp3 hybrid orbital), bond angle is 109.5, for 33.3%s character (as in sp2 hybrid orbital), bond angle is 120∘ and for 50%s character (as in sp hybrid orbital), bond angle is 180. Similarly, when the bond angle decreases below 109.5∘, the s -character will decrease accordingly. Decrease in angle = 120 − 109.5 = 10.5

Decrease in s -character = 33.3 − 25 = 8.3
Actual decrease in bond angle = 109.5 − 105 = 4.5
Expected decrease in s -character


Thus, the s -character should decrease by about 3.56%
i.e.,s -character = 25 − 3.56 = 21.44%

Test: Hybrididation and VSEPR theory - Question 7

Main axis of a diatomic molecule is Z. AO's and overlap to form which of the following orbitals?

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and orbitals do not have proper orientation to overlap and hence no bond is formed.

Test: Hybrididation and VSEPR theory - Question 8
Observe the following statements:
i. According to VSEPR theory, and are shown and type molecules respectively.
ii. has "See-saw" shape.
iii. and have same shape.
The statements which are not correct are
Detailed Solution for Test: Hybrididation and VSEPR theory - Question 8
Statements (i) and (iii) are not correct. Their correct explanation are as follows:
(i) According to VSEPR theory has 3 bond pairs and 2 lone pairs thus it is type molecule whereas has bond pairs and one lone pair so it is type molecule.
(iii) has a linear shape because it has 2 bond pairs only whereas has 2 bond pairs with one lone pair thus it has a bent shape.
Test: Hybrididation and VSEPR theory - Question 9

The compound MX4 is tetrahedral. The number of ∠XMX formed in the compound are

Detailed Solution for Test: Hybrididation and VSEPR theory - Question 9


In the tetrahedral model the top X atom is labeled as A, the bottom three are labeled as B, C and D. The atom C is above the plane and the atom D is inside the plane.
Thus the bond angles in MX4 are written as
∠AMB
∠AMC
∠AMD
∠BMC
∠BMD
∠CMD

Test: Hybrididation and VSEPR theory - Question 10

In a regular octahedral molecule, MX6 the number of X−M−X bonds at 180 is

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Thus here bond angles between


Test: Hybrididation and VSEPR theory - Question 11

Which of the following molecular orbitals has two nodal planes?

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It has two nodal planes. It is

Test: Hybrididation and VSEPR theory - Question 12

In the reaction , the change in hybridisation is from

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Test: Hybrididation and VSEPR theory - Question 13

Which of the following represents the given mode of hybridisation sp2 − sp2 − sp − sp from left to right ?

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Test: Hybrididation and VSEPR theory - Question 14
Which of the following statement is incorrect?
(i) being an ionic compound is a good conductor of electricity in the solid state.
(ii) In canonical structures there is no difference in the arrangement of atoms.
(iii) Hybrid orbitals form stronger bonds than pure orbitals.
(iv) VSEPR theory can explain the square planar geometry of .
Detailed Solution for Test: Hybrididation and VSEPR theory - Question 14
When sodium chloride is in solid state, it doesn't have free electrons due to the ionic/electrovalent bond between sodium and chorine but when dissolves in water, the bond is broken and sodium and chlorine separate forming ions like and . Therefore, having free ions, it is able to conduct electricity. So, the option (1) is incorrect.
Test: Hybrididation and VSEPR theory - Question 15

What are the shapes of ethyne and methane?

Detailed Solution for Test: Hybrididation and VSEPR theory - Question 15

Ethyne
Number of -bonds at -atom
Number of lone pair of electrons at -atom
∴ Total
∴ Hybridisation is .
is linear
Methane
Number of bonds at -atom
Number of-lone pair of electrons at C-atom
∴ Total
∴ Hybridisation is .
is tetrahedral.

Test: Hybrididation and VSEPR theory - Question 16
Which of the following is correct?
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does not obey octet rule as in it S-atom has 12 electrons in its valence shell.

Test: Hybrididation and VSEPR theory - Question 17
In which one of the following pairs the two species have identical shape, but differ in hybridisation?
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Central in undergoes hybridisation giving a linear shape with three lone pairs at equatorial positions. Beryllium chloride has linear structure with hybridisation of Be atom.
Test: Hybrididation and VSEPR theory - Question 18
Which one of the following molecules will form a linear polymeric structure due to H-bonding?
Detailed Solution for Test: Hybrididation and VSEPR theory - Question 18
Only forms a linear polymeric structure. There is no hydrogen bonding in form zigzag structure. has the structure in which each molecule is linked to four other molecules.
Test: Hybrididation and VSEPR theory - Question 19
On hybridization of one and one p-orbital we get
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Bond angle between two -hybrid orbitals is .
Test: Hybrididation and VSEPR theory - Question 20
Hybridization of each carbon in is
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