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Mock Exam IB DP Chemistry HL Paper 1 Set 4

Question

The observed specific optical rotation, [α ], of a compound is +7.00°. What is the specific optical rotation of a racemate of this compound?
A. -7.00°
B. 0.00°
C. +7.00°
D. +14.00°

▶️Answer/Explanation

Markscheme: B

The specific optical rotation (\([ \alpha ]\)) is a measure of the rotation of plane-polarized light caused by a chiral compound. The specific optical rotation for an enantiomerically pure substance is denoted as \([ \alpha ]\) and its enantiomer would have the opposite sign. For a racemate (a mixture of equal amounts of both enantiomers), the specific optical rotation is zero.

Given that the specific optical rotation of the compound is +7.00°, and a racemate is a 1:1 mixture of the two enantiomers, the specific optical rotation of the racemate is 0.00°. Therefore, the correct answer is:

B. 0.00°

Question

Which mechanism does the nitration of benzene proceed by?

A. electrophilic addition
B. electrophilic substitution
C. nucleophilic addition
D. nucleophilic substitution

▶️Answer/Explanation

Markscheme: B

The nitration of benzene proceeds by electrophilic aromatic substitution. Therefore, the correct answer is:

B. Electrophilic substitution

Question

Which trend is correct, going down group 1 ?

A. Melting point increases
B. Reactivity decreases
C. First ionisation energy increases
D. Electronegativity decreases

▶️Answer/Explanation

Solution:

The Group 1 elements, also known as the alkali metals, include lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr). Going down Group 1, several trends can be observed:

A. Melting point: The melting point of Group 1 elements generally decreases as we move down the group. This is because the size of the atoms increases, and the metallic bond between the atoms weakens due to an increase in the distance between the nuclei and valence electrons. As a result, less energy is required to overcome the metallic bond and melt the metal. Therefore, option A is incorrect.

B. Reactivity: The reactivity of Group 1 elements increases as we move down the group. This is because the valence electron is further away from the nucleus, shielded by more electron shells, and experiences a weaker attractive force from the nucleus. As a result, it is easier for the valence electron to be removed and participate in chemical reactions. Therefore, option B is incorrect.

C. First ionisation energy: The first ionisation energy of Group 1 elements generally decreases as we move down the group. This is because the valence electron is further away from the nucleus, shielded by more electron shells, and experiences a weaker attractive force from the nucleus. As a result, less energy is required to remove the valence electron and form a positively charged ion. Therefore, option C is incorrect.

D. Electronegativity: The electronegativity of Group 1 elements generally decreases as we move down the group. This is because the valence electron is further away from the nucleus, shielded by more electron shells, and experiences a weaker attractive force from the nucleus. As a result, the ability of the atom to attract electrons towards itself and form covalent bonds with other atoms decreases. Therefore, option D is correct.

Therefore, the correct trend going down Group 1 is that electronegativity decreases.

Question

What is the overall charge, $x$, of the chromium (III) complex?

$$
\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right]^{\mathrm{x}}
$$

A. $0$
B. $1+$
C. $2-$
D. $3+$

▶️Answer/Explanation

Solution:

To determine the overall charge, $x$, of the chromium (III) complex, we need to consider the charges of each individual ion in the complex.

The chromium ion, $\mathrm{Cr}^{3+}$, has a charge of $3+$.

Each water molecule, $\mathrm{H}_2\mathrm{O}$, is neutral and has a charge of $0$.

Each chloride ion, $\mathrm{Cl}^-$, has a charge of $1-$.

Therefore, the total charge of the complex can be calculated as:

Total charge $=\left(\right.$ Charge of $\left.\mathrm{Cr}^{3+}\right)+\left(\right.$ Charge of $\left.4 \mathrm{H}_2 \mathrm{O}\right)+\left(\right.$ Charge of $\left.2 \mathrm{Cl}^{-}\right)$
$$
\text { Total charge }=(+3)+(0 \times 4)+(-1 \times 2)=+1
$$

Since the total charge of the complex is $+1$, the answer is option B, $1+$.

Question

Which compound contains both ionic and covalent bonds?

A. $\mathrm{MgO}$
B. $\mathrm{CH}_2 \mathrm{Cl}_2$
C. $\mathrm{CH}_3 \mathrm{COOH}$
D. $\mathrm{NaOH}$

▶️Answer/Explanation

Solution:

The compound that contains both ionic and covalent bonds is $\mathrm{NaOH}$, option D.

$\mathrm{NaOH}$ contains both an ionic bond between sodium cation ($\mathrm{Na}^+$) and hydroxide anion ($\mathrm{OH}^-$) and a covalent bond between the oxygen atom and the hydrogen atom within the hydroxide molecule. The ionic bond is formed by the transfer of one electron from sodium to hydroxide while the covalent bond is formed by sharing of electrons between the oxygen and hydrogen atoms.

On the other hand, options A and C contain purely covalent bonds while option B contains purely covalent bonds between carbon and hydrogen atoms and polar covalent bonds between carbon and chlorine atoms.

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