Question
In which molecule does the central atom have an incomplete octet of electrons?
A. \(\rm H_2Se\)
B. \(\rm PH_3\)
C. \(\rm OF_2\)
D. \(\rm BF_3\)
▶️Answer/Explanation
Markscheme: D
An incomplete octet occurs when a central atom does not have eight electrons in its valence shell. Let’s examine the options:
A. \(\rm H_2Se\) – Selenium (Se) is in Group 16 and can accommodate more than 8 electrons. This molecule has a complete octet around selenium.
B. \(\rm PH_3\) – Phosphorus (P) is in Group 15, so it can have fewer than 8 electrons in its valence shell. This molecule has an incomplete octet around phosphorus.
C. \(\rm OF_2\) – Oxygen (O) is in Group 16 and generally follows the octet rule. This molecule has a complete octet around oxygen.
D. \(\rm BF_3\) – Boron (B) is in Group 13 and often has fewer than 8 electrons in its valence shell. This molecule has an incomplete octet around boron.
Therefore, the molecule in which the central atom has an incomplete octet of electrons is D. \(\rm BF_3\).
Question
What are the formal charges on the atoms in this molecular ion?
▶️Answer/Explanation
Markscheme: B
The formal charge (\(FC\)) on an atom in a Lewis structure is calculated using the formula:
\[FC = \text{Number of valence electrons in the free atom} – \text{Number of lone-pair electrons} – \frac{1}{2} \times \text{Number of bonding electrons}\]
The sum of formal charges in a neutral molecule should be zero, while in an ion, it should be equal to the ion’s charge.
\(\begin{aligned} & \begin{array}{l}\text { } \\ \text { Formal Charge }=\begin{array}{c}\text { Valence } \\ \text { Electrons }\end{array}-\begin{array}{c}\text { NonBonding } \\ \text { Val Electrons }\end{array}-\frac{\text { Bonding Electrons }}{2}\end{array} \\ &\\& \mathrm{~S}=6\quad-\quad4\quad-\quad4 / 2=0 \\ & \text { C }=4\quad-\quad 4\quad-\quad0 / 2=0 \\ & \text { N }=5\quad-\quad4 \quad-\quad 4 / 2=-1 \\ & \end{aligned}\)
Question
Which types of hybridization are present in glycine?
▶️Answer/Explanation
Markscheme: A
The formula for determining the hybridization of an atom in a molecule is given by:
\[ \text{Hybridization} = \frac{1}{2} \left( \text{Number of valence electrons on the atom} + \text{Number of monovalent atoms} – \text{Charge on the ion or molecule} \right) \]
This formula is a simplified approach to understanding the hybridization state based on the number of bonding and non-bonding electron pairs around the central atom.
Here’s a more detailed breakdown:
1. Count the valence electrons:
- Add the number of valence electrons on the central atom.
- Add the number of electrons contributed by monovalent atoms attached to the central atom.
2. Adjust for charge:
- Subtract the charge on the ion or molecule.
3. Divide by 2:
- Divide the result by 2.
The final result is an indication of the hybridization state of the central atom. Common hybridization states include \(sp\), \(sp^2\), \(sp^3\), \(sp^3d\), \(sp^3d^2\), etc.
Question
Which compound has an element with an incomplete octet of electrons?
A. \(BF_3\)
B. \(CF_4\)
C. \(OF_2\)
D. \(ClF_3\)
▶️Answer/Explanation
Markscheme: A
An incomplete octet of electrons refers to a situation where an element has fewer than eight electrons in its valence shell.
Let’s analyze each option:
A. \(BF_3\) (Boron trifluoride): Boron has only six electrons in its valence shell, so it has an incomplete octet.
B. \(CF_4\) (Carbon tetrafluoride): Carbon has a complete octet (four bonds with four fluorine atoms).
C. \(OF_2\) (Oxygen difluoride): Oxygen has a complete octet (two bonds with two fluorine atoms).
D. \(ClF_3\) (Chlorine trifluoride): Chlorine has a complete octet (three bonds with three fluorine atoms).
So, the compound with an element with an incomplete octet is A. \(BF_3\).
Question
What is the correct sequence if the compounds are arranged in order of increasing boiling point?
A. \(CH_3OCH_3\) < \(CH_3CH_2OH\) < \(CH_3CHO\)
B. \(CH_3OCH_3\) < \(CH_3CHO\) < \(CH_3CH_2OH\)
C. \(CH_3CHO\) < \(CH_3CH_2OH\) < \(CH_3OCH_3\)
D. \(CH_3CHO\) < \(CH_3OCH_3\) < \(CH_3CH_2OH\)
▶️Answer/Explanation
Markscheme: B
To determine the order of increasing boiling points, we need to consider the intermolecular forces in each compound.
1. \(CH_3OCH_3\) (dimethyl ether): It has London dispersion forces but lacks hydrogen bonding.
2. \(CH_3CH_2OH\) (ethanol): It has London dispersion forces and can also form hydrogen bonds due to the presence of the -OH group.
3. \(CH_3CHO\) (acetaldehyde): It has London dispersion forces and can form hydrogen bonds due to the carbonyl group.
The general trend is that molecules with stronger intermolecular forces tend to have higher boiling points.
So, the correct sequence of increasing boiling points is:
\[CH_3OCH_3 < CH_3CHO < CH_3CH_2OH.\]
Therefore, the correct answer is B. \(CH_3OCH_3\) < \(CH_3CHO\) < \(CH_3CH_2OH\).