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Test: Bond Parameters and VSEPR Theory Hybridisation (May 22) - NEET MCQ


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10 Questions MCQ Test Daily Test for NEET Preparation - Test: Bond Parameters and VSEPR Theory Hybridisation (May 22)

Test: Bond Parameters and VSEPR Theory Hybridisation (May 22) for NEET 2024 is part of Daily Test for NEET Preparation preparation. The Test: Bond Parameters and VSEPR Theory Hybridisation (May 22) questions and answers have been prepared according to the NEET exam syllabus.The Test: Bond Parameters and VSEPR Theory Hybridisation (May 22) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Bond Parameters and VSEPR Theory Hybridisation (May 22) below.
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Test: Bond Parameters and VSEPR Theory Hybridisation (May 22) - Question 1
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The decreasing order of the repulsive interactions between various electron pairs is:

Detailed Solution for Test: Bond Parameters and VSEPR Theory Hybridisation (May 22) - Question 1

The bond pair of electrons is shared by two atoms whereas the lone pair of electrons is only under the influence of central atom.

So the electron cloud of lone pair occupies more space as compared to the bond pair.

This causes greater repulsion between the lone pairs as compared to lone pair-bond pair and bond pair-bond pair interactions.

Test: Bond Parameters and VSEPR Theory Hybridisation (May 22) - Question 2
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The s-orbital does not show preference to any direction because

Detailed Solution for Test: Bond Parameters and VSEPR Theory Hybridisation (May 22) - Question 2

The s-orbital is spherically symmetric in shape so it does not show preference to any direction. It is the same from all the directions.

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Test: Bond Parameters and VSEPR Theory Hybridisation (May 22) - Question 3
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 Among the following species octahedral shape is found in:

Detailed Solution for Test: Bond Parameters and VSEPR Theory Hybridisation (May 22) - Question 3

SF6 has an octahedral geometry.

Test: Bond Parameters and VSEPR Theory Hybridisation (May 22) - Question 4
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A pi-bond is formed by the overlap of:
Detailed Solution for Test: Bond Parameters and VSEPR Theory Hybridisation (May 22) - Question 4

pi-bond is always formed by sidewise overlapping of p – p orbitals in sidewise manner.

Test: Bond Parameters and VSEPR Theory Hybridisation (May 22) - Question 5
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The statement that is true about H2O molecule is:
Detailed Solution for Test: Bond Parameters and VSEPR Theory Hybridisation (May 22) - Question 5

Central atom: O

At. no. = 8,

No. of valence electrons in oxygen is 6

Lewis dot structure of O atom

No. of electrons required for completing the octet = 8 - 6 =2

Atoms attached to the central atom: two H atoms

At. no. of H = 1

H atoms need one electron each for getting fully-filled K shells.

Representation of Lewis dot structures of H2O, which satisfies these conditions:

 

No. of valence shell electron pairs on the central atom O, after forming two covalent bonds with two H atoms = 4.

Geometry of 4 pairs, as predicted by VSEPR theory = Tetrahedral.

No. of lone pairs = Two.

Type, according to VSEPR theory:

So, geometry of pairs and geometry of atoms are different.

The geometry of atoms is derived from the geometry of pairs. At two of the four corners of the tetrahedron, there are no atoms.

Diagrams showing the geometry of 2 H atoms and two lone pairs:

 

Geometry of atoms in H2O is V-shaped. i.e. H2O molecule has a bent geometry

Test: Bond Parameters and VSEPR Theory Hybridisation (May 22) - Question 6
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The species having pyramidal shape is

[IIT JEE 2010]
Detailed Solution for Test: Bond Parameters and VSEPR Theory Hybridisation (May 22) - Question 6


(b) BrF3 - T-shaped (sp3d2)

Test: Bond Parameters and VSEPR Theory Hybridisation (May 22) - Question 7
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ECI3 (where, E = B, P, As, Bi) of these elements are known.
Bond angles  are in the following order
Detailed Solution for Test: Bond Parameters and VSEPR Theory Hybridisation (May 22) - Question 7


Has no lone pair thus, bond angle is 120°. B-atom is sp2-hybridised.

Central atom E is sp3-hybridised. Hence, angle < 120°. As we go down the group, (Ip-bp) repulsion decreases. Hence, angle 
(Cl—E—Cl) PCI3 > AsCI3 > BiCI3.
Thus, order is BCI3 > PCI3 > AsCI3 > BiCI3.

Test: Bond Parameters and VSEPR Theory Hybridisation (May 22) - Question 8
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Which of the following molecule/species has the minimum number of lone pairs?
Detailed Solution for Test: Bond Parameters and VSEPR Theory Hybridisation (May 22) - Question 8

Test: Bond Parameters and VSEPR Theory Hybridisation (May 22) - Question 9
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Which of the following set of molecules have the same shape but different hybridisation?
Detailed Solution for Test: Bond Parameters and VSEPR Theory Hybridisation (May 22) - Question 9

Test: Bond Parameters and VSEPR Theory Hybridisation (May 22) - Question 10
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Direction (Q. Nos. 21-22) This section contains  a paragraph, wach describing theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given options (a),(b),(c),(d).


Following reaction, CIF3 + AsF5 → (CIF2+ ) (AsF-6)

Q. 

Select the correct statements.
Detailed Solution for Test: Bond Parameters and VSEPR Theory Hybridisation (May 22) - Question 10

 sp3d -hybridised Cl-atom - (T-shaped s tructure)
  sp3 -hybridised. Cl-atom (Bent)
 sp3d -hybridised As-atom (Trigonal bipyramidai).
 sp3d2 -hybridised As-atom (Octahedral).

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