Among the following species, identity the isostructural pairs NF3, $NO_3^-$, BF3, H3O+ , HN3
A.
[NF3, $NO_3^-$] and [BF3, H3O+]
B.
[NF3, HN3] and [$NO_3^-$, BF3]
C.
[NF3, H3O+] and [$NO_3^-$, BF3]
D.
[NF3, H3O+] and [HN3, BF3]
Correct Answer: C
Explanation:
To identify the isostructural pairs among the given species, we first need to understand their molecular geometries and the electron-pair arrangements.
1. NF3: Nitrogen trifluoride has a trigonal pyramidal structure due to the presence of one lone pair of electrons on nitrogen. It follows the AX3E steric number formula (where A is the central atom, X represents bonded atoms, and E represents lone pairs).
2. $NO_3^-$: The nitrate ion has a trigonal planar structure. This can be rationalized using VSEPR theory, where it follows the AX3 steric number formula with no lone pairs on the central nitrogen atom.
3. BF3: Boron trifluoride is also trigonal planar. It conforms to the AX3 steric number formula with no lone pairs on boron.
4. H3O+: The hydronium ion has a trigonal pyramidal structure, similar to NF3, with one lone pair of electrons. It follows the AX3E formula.
5. HN3: Hydronitrenium ion has a linear or slightly bent structure, depending on how we look at the resonance structures. Overall, it does not share the same geometric configuration with any of the molecules listed above.
From the above analysis:
Isostructural Pair 1: NF3 and H3O+ (both have a trigonal pyramidal structure)
Isostructural Pair 2: $NO_3^-$ and BF3 (both have a trigonal planar structure)
The number and type of bonds between two carbon atoms in CaC2 are:
A.
one sigma ($\sigma$) and one pi ($\pi$) bonds
B.
one sigma ($\sigma$) and two pi ($\pi$) bonds
C.
one sigma ($\sigma$) and one and a half pi ($\pi$) bonds
D.
one sigma ($\sigma$) bond
Correct Answer: B
Explanation:
Calcium carbide ($CaC_2$) is an ionic compound consisting of ${Ca^{2+}}$ and carbide ions ${C_2^{2-}}$. In this compound, the two carbon atoms are bonded together in the carbide ion ${C_2^{2-}}$. The bonding in the carbide ion involves one sigma ($\sigma$) bond and two pi ($\pi$) bonds, which forms a triple bond between the two carbon atoms.
The reasoning is as follows:
Each carbon atom in the ${C_2^{2-}}$ ion is sp hybridized, having one sigma ($\sigma$) bond from the overlap of sp hybrid orbitals and two pi ($\pi$) bonds from the lateral overlap of p orbitals. Therefore, the carbide ion ${C_2^{2-}}$ has a carbon-carbon triple bond.
Thus, the correct option is:
Option B: one sigma ($\sigma$) and two pi ($\pi$) bonds
To determine the number of paired electrons in an O2 molecule, we need to understand its electronic configuration and molecular orbital configuration. Each oxygen atom has an atomic number of 8, so its electron configuration is:
$1s^2 2s^2 2p^4$
When two oxygen atoms form a diatomic O2 molecule, their atomic orbitals combine to form molecular orbitals. For O2, the relevant molecular orbitals and their filling order are:
Filling these molecular orbitals with the 16 electrons (8 from each oxygen atom) according to the aufbau principle and Hund's rule, we get the following configuration:
Here’s the breakdown of paired and unpaired electrons:
$ \sigma_{1s} $ -> 2 paired electrons
$ \sigma_{1s}^* $ -> 2 paired electrons
$ \sigma_{2s} $ -> 2 paired electrons
$ \sigma_{2s}^* $ -> 2 paired electrons
$ \sigma_{2p_z} $ -> 2 paired electrons
$ \pi_{2p_x} $ -> 2 paired electrons
$ \pi_{2p_y} $ -> 2 paired electrons
The antibonding $ \pi_{2p_x}^* $ and $ \pi_{2p_y}^* $ orbitals each have one unpaired electron, but since the question asks for paired electrons, these do not count.
The dipole moment of KCl is 3.336 $\times$ 10-29 Coulomb meters which indicates that it is a highly polar molecule. The interatomic distance between K+ and Cl- in this molecule is 2.6 $\times$ 10-10. Calculate the dipole moment of KCl molecule if there were opposite charges of one fundamental unit located at each nucleus. Calculate the percentage ionic character of KCl.
Correct Answer: 80.11 %
Explanation:
To calculate the dipole moment assuming one elementary charge of opposite kinds located at each nucleus, and to determine the percentage ionic character of KCl, we will use the concepts of charge and distance in the dipole moment equation, and understand the measure of ionic character compared to a hypothetical purely ionic bond.
Step 1: Calculating theoretical dipole moment
The elementary charge (e) is $ 1.602 \times 10^{-19} $ Coulombs. If KCl were purely ionic, the potassium and chlorine atoms would carry charges of +e and -e respectively. The dipole moment $ \mu $ is calculated by the formula:
$ \mu = q \times d $
where:
- $ q $ is the charge in Coulombs (+e for K+ and -e for Cl-)
- $ d $ is the separation distance between the charges, which is 2.6 $ \times 10^{-10} $ meters.
Thus, using the values:
$ \mu = 1.602 \times 10^{-19} \, \text{Coulombs} \times 2.6 \times 10^{-10} \, \text{meters} = 4.1652 \times 10^{-29} \, \text{Coulomb meters} $
Step 2: Actual dipole moment of KCl
The actual measured dipole moment of KCl is given as 3.336 $ \times $ 10-29 Coulomb meters. This is due to the actual electron distribution in the bond being not purely ionic.
Step 3: Percentage ionic character
The percentage ionic character can be calculated by comparing the actual dipole moment to the theoretical dipole moment for fully ionic charges, using the formula:
This calculation shows that while KCl is primarily ionic, it has less than 100% ionic character, indicating some degree of covalent character in its bonding, where the electron distribution is not completely transferred but shared to a certain extent.
2, 2-dimethylpropane and 2, 2, 3, 3 - tetramethylbutane are symmetrical molecules so their net dipole moment is zero.
(b) and (c) have resultant dipole moment.
To determine which molecule or ion assumes a linear structure, we need to consider the molecular geometry, which is often predicted by the VSEPR (Valence Shell Electron Pair Repulsion) theory.
Let’s analyze each option:
Option A: SnCl2
Tin(II) chloride, SnCl2, has a bent structure rather than a linear one due to the presence of a lone pair of electrons on the tin atom, which repels the bonds between Sn and Cl.
Option B: NCO-
The cyanate ion, NCO-, has a linear structure because the central atom carbon forms a double bond with nitrogen and a triple bond with oxygen, aligning all atoms in a straight line.
Option C: CS2
Carbon disulfide, CS2, has a linear structure. The central carbon atom forms double bonds with each sulfur atom, allowing the molecule to maintain a straight line.
Option D: $NO_2^+$
The nitronium ion, $NO_2^+$, also has a linear structure. It has a central nitrogen atom bonded to two oxygen atoms with no lone pairs on nitrogen, thus the molecule assumes a linear shape.
So, the linear structures are assumed by Options B, C, and D.
The strength of hydrogen bonding is directly proportional to the electronegativity and inversely proportional to the size of the atom bonded to the hydrogen atom.
Electronegativity: A more electronegative atom attracts the electron density in the covalent bond more strongly, leaving the hydrogen atom with a partial positive charge (δ+). This partial positive charge allows for a stronger electrostatic attraction to the lone pair of electrons on another electronegative atom, forming a stronger hydrogen bond.
Size: A smaller atom has a more concentrated electron density, leading to a stronger electrostatic attraction and a stronger hydrogen bond.
Therefore, among the given atoms, the order of increasing hydrogen bond strength is:
Cl < S < N < O < F
With fluorine (F) forming the strongest hydrogen bonds due to its high electronegativity and small size.
Additionally, chlorine (Cl) is relatively large in size and less electronegative compared to the other atoms listed. This makes it less likely to participate in significant hydrogen bonding.
Isostructural compounds are those that have the same shape and similar bond angles. To determine which compound is isostructural with carbon dioxide (CO2), we need to analyze the molecular geometry of each of the options.
1. CO2 has a linear structure with the carbon atom in the center, double-bonded to two oxygen atoms, giving it a bond angle of 180°. The structure can be represented as:
$ \text{O} = \text{C} = \text{O} $
2. HgCl2 also has a linear structure. The mercury atom in the center is single-bonded to two chlorine atoms, giving it a bond angle of 180°. The structure can be represented as:
$ \text{Cl} - \text{Hg} - \text{Cl} $
3. SnCl2 has a bent (or V-shaped) structure due to lone pairs on the tin atom. Hence, it is not isostructural with CO2.
4. C2H2 (acetylene) has a linear structure with a carbon-carbon triple bond and hydrogen atoms at each end, giving it a bond angle of 180°. The structure can be represented as:
5. NO2 has a bent structure due to the presence of a lone electron on the nitrogen atom, which causes repulsion and bends the molecule. Therefore, it is not isostructural with CO2.
Given this analysis, CO2 is isostructural with both HgCl2 and C2H2.
How many sigma bonds and how many pi-bonds are present in a benzene molecule?
Correct Answer: 12 $$\sigma$$, 3 $$\pi$$
Explanation:
Benzene, with the chemical formula $C_6H_6$, is an aromatic hydrocarbon consisting of six carbon atoms arranged in a hexagonal ring, with alternating single and double bonds. Each carbon atom also bonds with one hydrogen atom.
The structure includes three double bonds and three single bonds interconnecting the carbon atoms in the ring. Here's how these bonds are categorized:
Sigma Bonds ($\sigma$-bonds): Sigma bonds are the strongest type of covalent chemical bond. They are formed by head-on overlapping between atomic orbitals. In benzene:
Each of the six carbon atoms forms a sigma bond with one hydrogen atom, giving us total six C-H sigma bonds.
Each carbon-carbon (C-C) single bond in the ring is also a sigma bond. Since there are three C-C single bonds, this adds another three sigma bonds.
For double bonds, each contains one sigma and one pi bond. However, only the sigma component is counted here. Therefore, the three double bonds contribute another three sigma bonds from their C-C interactions.
Total Sigma Bonds: $6 \,(\text{C-H}) + 3 \,(\text{C-C single bonds}) + 3 \,(\text{C-C part of double bonds}) = 12 \,\sigma\text{-bonds}$.
Pi Bonds ($\pi$-bonds): Pi bonds are formed by the side-to-side overlap of atomic orbitals and are generally weaker than sigma bonds. In benzene, the pi bonds are part of the double bonds. Each double bond contributes one pi bond. Thus:
There are three double bonds in benzene, contributing one pi bond each.
Total Pi Bonds: $3 \,\pi\text{-bonds}$.
Therefore, a benzene molecule contains 12 sigma bonds and 3 pi bonds.
The statement "SnCl2 is a non-linear molecule" is TRUE.
To understand why SnCl2 (Tin(II) chloride) is a non-linear molecule, we need to look at its molecular geometry. Tin (Sn) is the central atom and it has a lone pair of electrons along with two bonding pairs of electrons with the chlorine (Cl) atoms. This creates a V-shaped or bent molecular geometry.
According to the Valence Shell Electron Pair Repulsion (VSEPR) theory, the lone pair of electrons on the tin atom will repel the bonding pairs, causing the molecule to take on a shape that minimizes repulsion. Therefore, the electron geometry of SnCl2 is trigonal planar, but because of the lone pair, the molecular shape is bent.
The bond angle in a bent molecule like SnCl2 deviates from the typical 180 degrees and is approximately 120 degrees. This bent shape confirms that SnCl2 is non-linear.
