Test: Chemical Bonding and Shapes of Compounds - Chemistry MCQ

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

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.

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.

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⁺.

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.

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.

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

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 SiF₆^2-
Both SF₆ and SiF₆^2- have an octahedral geometry around the central atom. In SF₆, there are six fluorine atoms bonded to the central sulfur atom, and in SiF₆^2-, 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: SiF₆^2- and SeF₆^2-
SiF₆^2- has an octahedral geometry around the central silicon atom, while SeF₆^2- 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: XeO₆^4- and TeF₆^2-
XeO₆^4- has an octahedral geometry around the central xenon atom, while TeF₆^2- 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 SiF₆^2- are isostructural.

• 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.

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). 

Answer A is correct because the C₃^4- 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 C₃^4- 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.

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).

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 %

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.

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.

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.

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.

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.

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 p_x 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 p_x orbital of A and the s orbital of B will result in a sigma bond along the x-axis.