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Test: VSEPR Theory & shapes of molecules - Chemistry MCQ


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10 Questions MCQ Test - Test: VSEPR Theory & shapes of molecules

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Test: VSEPR Theory & shapes of molecules - Question 1

Square planar complex results from ___________ hybridization.

Detailed Solution for Test: VSEPR Theory & shapes of molecules - Question 1

Concept-

​Crystal field theory:

  • A metal ion in a complex is surrounded by coordinating ligands anions or negative ends.
  • An electric field is produced at the metal ion by the surrounding ligands.
  • This electrostatic approach towards the interaction between metals and ligands is known as crystal field theory.
  • CFT theory considers the ligands as point charges.
  • There is no overlap between the ligand orbitals and metal ion orbitals.
  • Ligands donate lone pair of electrons to the metal atoms and form coordinate complexes.

The geometry of the complex formed:

  • Placement of the donor atoms at different stereochemical positions will perturb the metal ions to a different extent.
  • This will result in the formation of different stereochemistries or structures.
  • The stereochemistry also depends on the coordination number and the type of orbitals involved.


Explanation:
The formation of square planar complexes involves one s, two p, and dx2-y2 orbitals.

  • Four pure orbitals mix and results in the formation of four dsp2 orbitals.
  • The four orbitals are distributed along four corners of a square plane as shown above.
  • Thus, the complexes formed via dsp2 hybridization are also square planar in shape.

Hence, the square planar complex results from  dsp2 hybridization.

Test: VSEPR Theory & shapes of molecules - Question 2

Which has a minimum bond angle?

Detailed Solution for Test: VSEPR Theory & shapes of molecules - Question 2

H2O:

  • In water, the center atom oxygen is sp3 hybridized.
  • Out of the four bond pairs, two are bond pairs and the other two are lone pairs.
  • Due to the presence of lone electrons, there is repulsion between bond pairs and lone pairs.
  • The repulsion between l.p is > b.p and thus pushes the two bonds to come closer decreasing the bond angle from ideal 109.8°.

H2S:

  • In water, the center atom oxygen is sp3 hybridized.
  • Out of the three bond pairs, two are bond pairs and the other two are lone pairs.
  • Due to the presence of lone electrons, there is repulsion between bond pairs and lone pairs.
  • The repulsion between l.p is > b.p and thus pushes the two bonds to come closer decreasing the bond angle from ideal 109.8°.
  • As sulfur is a larger molecule than oxygen, its electron density is more dispersed. It is more easily distorted by the repulsion from the lone pair of electrons and the bond pairs come closer to each other.
  • This decreases the bond angle more than in water.
  • The bond angle in H2S is 92.10.

NH3:

  • In ammonia, the center atom oxygen is sp3 hybridized.
  • Out of the three bond pairs, three are bond pairs and the other is lone pair.
  • Due to the presence of lone electrons, there is repulsion between bond pairs and lone pairs.
  • The repulsion between l.p > b.p and thus pushes the two bonds to come closer decreasing the bond angle from ideal 109.8° to 107°.

CH4:

  • CH4 molecule has sp3 hybridization.
  • Each of these 4 sp3 hybrid orbitals overlaps with one 1s atomic orbital of hydrogen.
  • The arrangement of the atoms gives tetrahedral geometry.
  • There is no lone pair and the bond angle is ideal 109.5 °.


Hence, the minimum bond angle is present in H2S.

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Test: VSEPR Theory & shapes of molecules - Question 3

Which type of bond is present in NaCl molecule

Detailed Solution for Test: VSEPR Theory & shapes of molecules - Question 3

Ionic Bond:

  • Low positive charge and large size of cation and a small charge on anion and small size of anion favor the formation of ionic compounds.
  • Ionic bond formation occurs via electrostatic attraction between two ions.
  • One ion is positively charged and the other is negatively charged like in NaCl.


Hence, we can conclude that Ionic bond is present in NaCl molecule.

Test: VSEPR Theory & shapes of molecules - Question 4

Which of the following has sp3d hybridization?

Detailed Solution for Test: VSEPR Theory & shapes of molecules - Question 4

SF4:

  • Contribution of electrons of sulfur toward the bonding = 6
  • Contribution from Fluorine = 4
  • The total number of electron pairs is 5, which corresponds to sp3d hybridization.
  • However, out of these 5 pairs, 4 are bond pairs and one is lone pair.
  • The geometry is trigonal bipyramidal but the shape is a see-saw.


Hence, SF4 has sp3d hybridization.

Test: VSEPR Theory & shapes of molecules - Question 5

Pick out the isoelectronic structure from the Following:

I.CH3+, II.H3O+, III. NH3, IV. CH3-

Detailed Solution for Test: VSEPR Theory & shapes of molecules - Question 5

The species and their structures and number of electrons are given below:

  • From the table above we see that CH3-, NH3, H3O+ all have 10 electrons in them.
  • Hence, they are isoelectronic species.
Test: VSEPR Theory & shapes of molecules - Question 6

The atomic number of magnesium is 12 and mass number is 24. Which one of the following is the correct representation of the ion formed from it and its valency?

Detailed Solution for Test: VSEPR Theory & shapes of molecules - Question 6
  • Atomic Number of Magnesium (Mg) is - 12
  • Atomic Configuration of Mg = 2, 8, 2
  • Its valance electrons is less than 4.
  • So, its Valancy is 2 i.e electrons in the last cell.
  • Therefore the stable configuration of Mg will be when it loses 2 electrons i.e. Mg 2+
Test: VSEPR Theory & shapes of molecules - Question 7

The mode of hybridization of carbon in CO is

Detailed Solution for Test: VSEPR Theory & shapes of molecules - Question 7
  • In CO, the structure is:
    : C ≡ O :
  • There exist three bonds between carbon and oxygen out of which one is sigma and two are pi.
  • So, two pi bond corresponds to 'sp' hybridization.
  • The sigma bond between the two atoms is made by the head-on overlap of 2p - 2p orbitals of Carbon and Oxygen.
  • The pi bonds are formed by the lateral overlap between p orbitals of carbon and oxygen.

Hence, the mode of hybridization of carbon in CO is 'sp'.

Test: VSEPR Theory & shapes of molecules - Question 8

What is the hybridisation of the oxygen atom in water?

Detailed Solution for Test: VSEPR Theory & shapes of molecules - Question 8

H2O:

Oxygen is the central molecule.
Oxygen donates six electrons towards hybridization and one electron is donated by each hydrogen atom.
So total number of electrons = 2 + 6 = 8 and number of pairs = 4. Hence, H = 4 and the hybridisation is sp3
Out of four pairs, two are bond pairs and two are lone pairs.
The expected geometry is tetrahedral but the actual shape is bent.

Hence, the hybridization of the oxygen atom in water is sp3.

Test: VSEPR Theory & shapes of molecules - Question 9

The shape a molecule occupies allows to minimize repulsions among them and maximize the space between them.

Detailed Solution for Test: VSEPR Theory & shapes of molecules - Question 9

The correct answer is Option 4, i.e IF7

  • In IF7, the central atom I (Iodine) has seven valance electrons.
  • Each of these valance electrons form bond with F-atom and therefore, the molecule has pentagonal bipyramidal geometry.
  • According to VSEPR theory, the shape of a molecule depends upon the number of valence shell electron pairs (bonded or no bonded) around the central atom.
  • PF5 has square pyramidal geometry.
  • SF6  and TeF6  have octahedral geometry.
Test: VSEPR Theory & shapes of molecules - Question 10

The VSEPR Model is used for predicting the ______________ of the molecules.

Detailed Solution for Test: VSEPR Theory & shapes of molecules - Question 10

The correct answer is Geometrical shapes.

Key Points

  • The VSEPR model used for predicting the geometrical shapes of molecules is based on the assumption that electron pairs repel each other and, therefore, tend to remain as apart as possible.
  • According to this model, molecular geometry is determined by repulsions between lone pairs and lone pairs, lone pairs and bond pairs and bond pairs and bond pairs. 
  • The order of these repulsions being: lone pairs - lone pairs > lone pairs - bond pairs > bond pairs - bond pairs.
  • Sidgwick and Powell in the year 1940, proposed a simple theory based on the repulsive interactions of the electron pairs in the valence shell of the atoms.
  • The VSEPR theory can predict the geometry of a large number of molecules, especially the compounds of p-block elements accurately.
  • It is also quite successful in determining the geometry quite-accurately even when the energy difference between possible structures is very small.

Additional Information

  • Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule.
  • Bond lengths are measured by spectroscopic, X-ray diffraction, and electron diffraction techniques.
  • Bond order is the number of bonds between the two atoms in a molecule.
  • In the Lewis description of the covalent bond, the bond Order is given by the number of bonds between the two atoms in a molecule.
  • With an increase in bond order, bond enthalpy increases and bond length decreases.
  • The concept of resonance says that whenever a single Lewis structure cannot describe a molecule accurately, a number of structures with similar energy, positions of nuclei, bonding, and non-bonding pairs of electrons are taken as the canonical structures of the hybrid which describes the molecule accurately.
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