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Test: Chemical Bonding and Shapes of Compounds - Chemistry MCQ


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20 Questions MCQ Test Inorganic Chemistry - Test: Chemical Bonding and Shapes of Compounds

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Test: Chemical Bonding and Shapes of Compounds - Question 1

In the formation of N2+ from N2, the electron is removed from

Detailed Solution for Test: Chemical Bonding and Shapes of Compounds - Question 1

Constructing the molecular orbital diagram for N2 and N2+  as shows electron is removed from σ orbital.

Test: Chemical Bonding and Shapes of Compounds - Question 2

During the formation of a molecular orbital from atomic orbitals, probability of electron density is

Detailed Solution for Test: Chemical Bonding and Shapes of Compounds - Question 2

In a molecular orbital, a nodal plane is a region where the probability of finding an electron is exactly zero. This occurs because the wave functions of the combining atomic orbitals have opposite signs, and their overlap results in a cancellation effect. As a result, there is no electron density in the nodal plane, which is why the correct answer is option C: zero in the nodal plane.

Options A and B are incorrect because the electron density is neither minimum nor maximum in the nodal plane; it is precisely zero. Option D is also incorrect because the electron density is not zero on the surface of the lobe. In fact, the lobes represent regions of high electron density, where the probability of finding an electron is greatest.

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Test: Chemical Bonding and Shapes of Compounds - Question 3

In the cyanide ion the negative charge is on

Detailed Solution for Test: Chemical Bonding and Shapes of Compounds - Question 3

The cyanide ion is a linear molecule with the chemical formula CN⁻. It consists of a carbon atom (C) triple-bonded to a nitrogen atom (N).

In terms of the distribution of electrons, the carbon atom contributes 4 valence electrons, and the nitrogen atom contributes 5 valence electrons. Additionally, there is an extra electron due to the negative charge of the ion, making a total of 10 valence electrons.

When forming the triple bond between the carbon and nitrogen atoms, each atom contributes 3 electrons, and they share these 6 electrons in the bond. The remaining 4 electrons are distributed as two lone pairs on the nitrogen atom.

Since the negative charge is associated with the extra electron in the ion, and this electron is part of the lone pairs on the nitrogen atom, the negative charge resides on the nitrogen atom in the cyanide ion.

Test: Chemical Bonding and Shapes of Compounds - Question 4

Amongst LiCl , RDCl , BeCl2 and MgCl2 the compounds with the greatest and the least ionic character, respectively are:

Detailed Solution for Test: Chemical Bonding and Shapes of Compounds - Question 4

Fajan's rule explains that the covalent nature of a molecule increases when there is a smaller cation and a larger anion. RbCl is more ionic in nature, while BeCl2 exhibits a higher degree of covalent character due to the strong polarizing power of Be2+.

Test: Chemical Bonding and Shapes of Compounds - Question 5

Which of the set of isomers of C6H4Cl2 is having equal dipole moment with C6H5Cl and C6H6 respectively

Detailed Solution for Test: Chemical Bonding and Shapes of Compounds - Question 5

Test: Chemical Bonding and Shapes of Compounds - Question 6

In ICl4Θ, the shape is square planar The number of bond pair-lone pair repulsion at 90° are:

Detailed Solution for Test: Chemical Bonding and Shapes of Compounds - Question 6

In ICl4Θ (Iodine tetrachloride ion), the central atom is Iodine (I), which is surrounded by four Chlorine (Cl) atoms and has two lone pairs of electrons. The hybridization of Iodine in ICl4Θ is sp3d2, which leads to an octahedral arrangement of electron pairs. However, due to the presence of two lone pairs, the molecular geometry is square planar.
In a square planar geometry, there are four Cl atoms around the Iodine atom, and each Cl atom has a bond pair of electrons shared with the Iodine atom. There are also two lone pairs of electrons on the central Iodine atom. To determine the number of bond pair-lone pair repulsions at 90°, we need to consider the interactions between each bond pair and lone pair.
There are four bond pairs (ICl) and two lone pairs in the molecule. Each bond pair will experience repulsion from the two lone pairs, resulting in 4 x 2 = 8 bond pair-lone pair repulsions at 90°. This is because each bond pair is located 90° away from each lone pair in the square planar geometry.
Therefore, the correct answer is 8 (option B). There are 8 bond pair-lone pair repulsions at 90° in the ICl4Θ molecule.

Test: Chemical Bonding and Shapes of Compounds - Question 7

The structure of XeF6 is :

Detailed Solution for Test: Chemical Bonding and Shapes of Compounds - Question 7

The structure of XeF6 (Xenon hexafluoride) is distorted octahedral (option C) because of its molecular geometry and the number of electron pairs surrounding the central atom.
XeF6 is formed when xenon (Xe) reacts with fluorine (F) to form a compound with six covalent bonds between the central xenon atom and six fluorine atoms. Xenon has 8 valence electrons, and it uses 6 of these electrons to form bonds with the 6 fluorine atoms. The remaining 2 valence electrons form a lone pair.
In the case of XeF6, the central xenon atom is surrounded by six bonding pairs of electrons and one lone pair of electrons. According to VSEPR (Valence Shell Electron Pair Repulsion) theory, the electron pairs will arrange themselves to minimize repulsion, which results in a distorted octahedral molecular geometry.
In a capped octahedral structure, there are six fluorine atoms bonded to the central xenon atom. Five of these fluorine atoms are arranged in a square pyramid around the xenon atom, while the sixth fluorine atom is positioned above the central atom. The lone pair of electrons occupies the position opposite to the sixth fluorine atom, "distorting" the octahedral structure.
This unique molecular geometry of XeF6 distinguishes it from other geometries like pentagonal bipyramidal, octahedral, or square pyramidal.

Test: Chemical Bonding and Shapes of Compounds - Question 8

The shape of [BrF4]- ion is

Detailed Solution for Test: Chemical Bonding and Shapes of Compounds - Question 8

The shape of the [BrF4]- ion is square planar. To understand why, we need to look at the electron configuration of the bromine atom and the number of bonding electrons involved in the ion.

Bromine has 7 valence electrons, and each of the four fluorine atoms contributes one electron for bonding. Additionally, there is one extra electron due to the negative charge of the ion, resulting in a total of 12 valence electrons involved in bonding.

According to the VSEPR (Valence Shell Electron Pair Repulsion) theory, the molecular shape is determined by the repulsion of electron pairs around the central atom. In the case of [BrF4]-, the central bromine atom is surrounded by 4 bonding pairs of electrons and 1 lone pair.

The number of electron pairs around the central atom determines the electron pair geometry. In this case, with 5 electron pairs, the electron pair geometry is trigonal bipyramidal. However, the molecular shape is determined by the arrangement of the bonded atoms around the central atom, not including the lone pairs.

In the case of [BrF4]-, the lone pair of electrons occupies one of the equatorial positions in the trigonal bipyramidal geometry, causing the four fluorine atoms to take up the remaining positions. These fluorine atoms form a square planar arrangement around the central bromine atom.

Therefore, the shape of the [BrF4]- ion is square plan

Test: Chemical Bonding and Shapes of Compounds - Question 9

Which of the following pairs is isostructural?

Detailed Solution for Test: Chemical Bonding and Shapes of Compounds - Question 9

Isostructural compounds have the same arrangement of atoms and the same shape. In other words, they have the same geometry around the central atom.
A: SF4 and SiF4
SF4 has a see-saw shape due to the presence of one lone pair on the sulfur atom, while SiF4 has a tetrahedral shape with no lone pairs on the silicon atom. Therefore, they are not isostructural.
B: SF6 and SiF62-
Both SF6 and SiF62- have an octahedral geometry around the central atom. In SF6, there are six fluorine atoms bonded to the central sulfur atom, and in SiF62-, there are six fluorine atoms bonded to the central silicon atom. Since both compounds have the same arrangement of atoms and the same shape, they are isostructural.
C: SiF62- and SeF62-
SiF62- has an octahedral geometry around the central silicon atom, while SeF62- has an octahedral geometry around the central selenium atom. However, the central atoms are different (Si and Se), so the compounds are not isostructural.
D: XeO64- and TeF62-
XeO64- has an octahedral geometry around the central xenon atom, while TeF62- has an octahedral geometry around the central tellurium atom. However, the central atoms are different (Xe and Te) and the surrounding atoms are also different (O and F), so the compounds are not isostructural.

Therefore, the correct answer is B: SF6 and SiF62- are isostructural.

Test: Chemical Bonding and Shapes of Compounds - Question 10

Which of the following statement is true?

Detailed Solution for Test: Chemical Bonding and Shapes of Compounds - Question 10
  • Covalent bonds are directional:

    • True. Covalent bonds are directional because they form specific angles and orientations based on the overlap of atomic orbitals. This directionality affects the shape and geometry of the molecule.
  • A polar bond is not formed between two atoms which have the same electronegativity value:

    • True. A polar bond occurs when there is a difference in electronegativity between the two atoms involved, causing an uneven distribution of electron density. If the electronegativity values are the same, the bond is nonpolar.
  • The presence of polar bonds in a polyatomic molecule suggests that it has zero dipole moment:

    • False. The presence of polar bonds in a polyatomic molecule does not necessarily mean it has zero dipole moment. The overall dipole moment of the molecule depends on the vector sum of the individual bond dipoles and the molecular geometry. If the dipoles do not cancel out due to the shape of the molecule, it will have a net dipole moment.
Test: Chemical Bonding and Shapes of Compounds - Question 11

A simplified application of MO theory to the hypothetical ‘molecule’ OF would give its bond order as :

Detailed Solution for Test: Chemical Bonding and Shapes of Compounds - Question 11

In MO theory, bond order is used to indicate the stability of a bond. A higher bond order corresponds to a stronger, more stable bond. Bond order can be calculated as the difference between the number of electrons in bonding and antibonding orbitals, divided by 2.
For the hypothetical molecule OF, we can consider the atomic orbitals of oxygen (O) and fluorine (F). Oxygen has the electron configuration 1s2 2s2 2p4, while fluorine has the configuration 1s2 2s2 2p5. The valence electrons, which are the electrons in the outermost shell, are involved in bonding. Oxygen has 6 valence electrons, and fluorine has 7 valence electrons.
When these two atoms come together to form a bond, their atomic orbitals will combine to form molecular orbitals. In this case, the 2p orbitals of oxygen and fluorine will overlap. The 2p orbitals can combine to form two bonding orbitals (σ and π) and two antibonding orbitals (σ* and π*).
The 7 valence electrons from fluorine and the 6 valence electrons from oxygen will fill these molecular orbitals, starting with the lowest energy bonding orbitals and following Hund's rule and the Pauli Exclusion Principle. There will be a total of 13 electrons to distribute:
σ : 2 electrons
π : 4 electrons (2 electrons in each of the two degenerate π orbitals)
σ* : 2 electrons
π* : 5 electrons (3 electrons in one π* orbital, 2 electrons in the other)
Now we can calculate the bond order:
Bond order = (number of electrons in bonding orbitals - number of electrons in antibonding orbitals) / 2
Bond order = (6 - 7) / 2
Bond order = -1 / 2
Bond order = 1.5
Thus, the bond order of the hypothetical molecule OF is 1.5 (option B). 

Test: Chemical Bonding and Shapes of Compounds - Question 12

Mg2C3 reacts with water forming propyne, C34- has:

Detailed Solution for Test: Chemical Bonding and Shapes of Compounds - Question 12

Answer A is correct because the C34- ion in Mg2C3 is a linear acetylide ion with a triple bond between the two carbon atoms. In a triple bond, there are one sigma bond and two pi bonds. A sigma bond is a single covalent bond formed by the direct overlap of bonding orbitals, whereas pi bonds are formed by the sideways overlap of p orbitals. So, in the C34- ion, there are two sigma bonds (one between the two carbon atoms and one between a carbon and a metal atom) and two pi bonds (both in the triple bond between the two carbon atoms). This is in accordance with the chapter on Chemical Bonding and Shapes of Compounds in Inorganic Chemistry, which discusses the different types of covalent bonds and their formation in various compounds.

Test: Chemical Bonding and Shapes of Compounds - Question 13

Compound with maximum ionic character is formed fiom

Detailed Solution for Test: Chemical Bonding and Shapes of Compounds - Question 13

The degree of ionic character in a compound is determined by the difference in electronegativity between the cation and anion. As this difference increases, the compound exhibits a higher degree of ionic character. Additionally, when the cation is larger and the anion is smaller, the electronegativity difference becomes greater, resulting in a higher level of ionic character. Therefore, among the given compounds, CsF is the most ionic due to the significant difference in electronegativity between cesium (Cs) and fluorine (F).

Test: Chemical Bonding and Shapes of Compounds - Question 14

The experimental value of the dipole moment of HCI is 1.03 D. The length of the H-CI bond is 1.275 A. The percentage of ionic character in HCl is

Detailed Solution for Test: Chemical Bonding and Shapes of Compounds - Question 14

The percent ionic character
= Observed dipole moment / Calculated dipole moment (assuming 100% ionic bond) x 100
So, dipole moment (assuming 100% ionic bond) = e × d= 4.8 × 10-10  ×1.275 × 10-8
= 6.12×10-18
So, percent ionic character = (1.03/6.12) × 100 ≈ 17 %

Test: Chemical Bonding and Shapes of Compounds - Question 15

If a molecule MX3 has zero dipole moment, the sigma bonding orbitals used by M (atm. no. < 21) are

Detailed Solution for Test: Chemical Bonding and Shapes of Compounds - Question 15

A molecule with zero dipole moment implies that it has a symmetrical distribution of electron density and charge. MX3 indicates that the central atom M is bonded to three X atoms.
For a molecule to have a zero dipole moment, the three X atoms should be arranged symmetrically around the central atom M. This arrangement is possible in a trigonal planar geometry.
In a trigonal planar geometry, the bond angles between the X atoms are 120 degrees. This geometry can be achieved when the central atom M uses sp2 hybrid orbitals for bonding.
In sp2 hybridization, one s orbital and two p orbitals of the central atom mix to form three sp2 hybrid orbitals that are oriented at 120 degrees to each other. These hybrid orbitals can then form sigma bonds with the X atoms, resulting in a trigonal planar geometry and a molecule with zero dipole moment.

Test: Chemical Bonding and Shapes of Compounds - Question 16

In which of the following molecule are all the bonds not equal ?

Detailed Solution for Test: Chemical Bonding and Shapes of Compounds - Question 16

In CIF3, all the bonds are not equal due to the presence of a lone pair of electrons on the central chlorine atom. This molecule has a T-shaped molecular geometry, and the presence of the lone pair leads to differences in the bond angles and bond lengths.
The central chlorine atom in this molecule is bonded to three fluorine atoms and has one lone pair of electrons. The VSEPR theory predicts a T-shaped geometry for this molecule. The presence of the lone pair on the chlorine atom causes repulsion, leading to unequal bond angles and bond lengths between the chlorine and fluorine atoms. Therefore, all the bonds in CIF3 are not equal.

Test: Chemical Bonding and Shapes of Compounds - Question 17

The cationic part of solid Cl2O6 is having the “_______ ” shape

Detailed Solution for Test: Chemical Bonding and Shapes of Compounds - Question 17

As there are seven electrons in the valence shell of chlorine and one mono-positive charge, Hence chlorine atom in ClO2+ should be sp2 hybridized with a trigonal planar structure.
As the third point of the trigonal planar structure is empty due to the absence of any atom, the structure should be angular in shape.
So, the correct answer of the question is option D, angular.

Test: Chemical Bonding and Shapes of Compounds - Question 18

The molecular shapes of SF4, CF4 and XeF4 are

Detailed Solution for Test: Chemical Bonding and Shapes of Compounds - Question 18

The molecular shapes of SF4, CF4, and XeF4 are different due to the varying number of lone pairs on the central atom.
SF4: Sulfur has 6 valence electrons. In SF4, sulfur forms 4 bonds with 4 fluorine atoms, leaving one lone pair on the central sulfur atom. The molecular shape of SF4 is seesaw (distorted tetrahedral) due to the presence of one lone pair on the sulfur atom.
CF4: Carbon has 4 valence electrons. In CF4, carbon forms 4 bonds with 4 fluorine atoms and has no lone pairs. The molecular shape of CF4 is tetrahedral.
XeF4: Xenon has 8 valence electrons. In XeF4, xenon forms 4 bonds with 4 fluorine atoms, leaving two lone pairs on the central xenon atom. The molecular shape of XeF4 is square planar due to the presence of two lone pairs on the xenon atom.
Thus, the molecular shapes of SF4, CF4, and XeF4 are different with 1, 0, and 2 lone pairs on the central atom, respectively. This illustrates the importance of lone pairs in determining the molecular geometry and shape of compounds in inorganic chemistry.

Test: Chemical Bonding and Shapes of Compounds - Question 19

In the context of carbon, which of the following is arranged in the correct order of electronegativity

Detailed Solution for Test: Chemical Bonding and Shapes of Compounds - Question 19

Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. In general, electronegativity increases with an increasing amount of s character in the hybrid orbitals. This is because s orbitals are more effective at shielding the positively charged nucleus from the negatively charged electrons, resulting in a stronger attraction between the nucleus and the bonding electrons.
In the case of carbon, sp hybrid orbitals have 50% s character, sp2 hybrid orbitals have 33.3% s character, and sp3 hybrid orbitals have 25% s character. Since the s character decreases in the order sp > sp2 > sp3, the electronegativity of carbon atoms in these hybridization states also follows this order.
Therefore, the correct order of electronegativity for carbon is sp > sp2 > sp3. This order plays a significant role in determining the reactivity, stability, and other properties of inorganic compounds containing carbon.

Test: Chemical Bonding and Shapes of Compounds - Question 20

A sigma bond may be formed by the overlap of 2 atomic orbitals of atoms A and B. If the bond is formed along the x-axis, which of the following overlaps is acceptable ?

Detailed Solution for Test: Chemical Bonding and Shapes of Compounds - Question 20

A sigma bond is a type of covalent bond that is formed when two atomic orbitals overlap along the axis connecting the two nuclei. The axis on which the bond is formed is the x-axis, so only orbitals that are oriented along the x-axis can form a sigma bond.
Option D is correct because the px orbital of A is oriented along the x-axis, and the s orbital of B is spherical, so it can overlap with any other orbital. In this case, the overlap of the px orbital of A and the s orbital of B will result in a sigma bond along the x-axis.

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