Problem 170
Question
In which of the following arrangements, the sequence is not strictly according to the property written against it? \(\quad\) (a) \(\mathrm{HF}<\mathrm{HCl}<\mathrm{HBr}<\mathrm{HI}\) : increasing acid strength (b) \(\mathrm{NH}_{3}<\mathrm{PH}_{3}<\mathrm{AsH}_{3}<\mathrm{SbH}_{3}:\) increasing basic strength (c) \(\mathrm{B}<\mathrm{C}<\mathrm{O}<\mathrm{N}\) : increasing first ionization enthalpy (d) \(\mathrm{CO}_{2}<\mathrm{SiO}_{2}<\mathrm{SnO}_{2}<\mathrm{PbO}_{2}\) : increasing oxidising power
Step-by-Step Solution
Verified Answer
The incorrect sequence is (b) for increasing basic strength and (c) for ionization enthalpy.
1Step 1: Analyze Option (a)
The sequence provides the acid strength of hydrogen halides: HF, HCl, HBr, HI. As we move down the group in the periodic table, the acid strength increases because the bond strength decreases, making it easier for the hydrogen ion to dissociate. Hence, this sequence is correct.
2Step 2: Analyze Option (b)
The sequence given is for the basic strength of group 15 hydrides: NH3, PH3, AsH3, SbH3. As we move down the group, the ability to donate lone pairs decreases, thus decreasing basic strength rather than increasing it. Therefore, this sequence is incorrect.
3Step 3: Analyze Option (c)
The sequence relates to the first ionization enthalpy of elements: B, C, O, N. According to periodic trends, ionization enthalpy generally increases across a period from left to right. However, nitrogen has a higher ionization enthalpy compared to oxygen due to its half-filled stable configuration. Therefore, this sequence is not entirely correct.
4Step 4: Analyze Option (d)
The sequence is related to the oxidizing power of oxides: CO2, SiO2, SnO2, PbO2. The oxidizing power increases down the group as heavier metals have a lesser ability to hold oxygen. This sequence is correct.
Key Concepts
Acid StrengthBasic StrengthIonization EnthalpyOxidizing Power
Acid Strength
Acid strength refers to an acid's ability to donate a proton (H⁺ ion) to a base. Essentially, the stronger an acid, the easier it donates this proton. In the periodic table, acid strength among hydrogen halides (HF, HCl, HBr, HI) increases as you move down the group. Why does this happen? As the atoms increase in size down the group, the bond between hydrogen and the halide (like fluoride or chloride) becomes weaker. A weaker bond means it’s easier for the hydrogen to break away, enhancing its acidic nature. This is why HI is a stronger acid than HF.
The general trend:
The general trend:
- Bond strength decreases down the group
- Ease of proton donation increases
- Acid strength increases
Basic Strength
Basic strength is all about a molecule's ability to accept protons. In other words, a stronger base readily accepts hydrogen ions. However, when examining basic strength in group 15 hydrides (NH₃, PH₃, AsH₃, SbH₃), the opposite trend to acid strength is observed. As you move down this group, the basic strength decreases.
This happens because of the change in size and electronegativity of the central atom. Larger atoms are less able to hold onto lone pairs of electrons tightly, making them less effective at accepting protons. Therefore, NH₃ is a stronger base compared to SbH₃.
This happens because of the change in size and electronegativity of the central atom. Larger atoms are less able to hold onto lone pairs of electrons tightly, making them less effective at accepting protons. Therefore, NH₃ is a stronger base compared to SbH₃.
- Lone pair availability decreases down the group
- Basic strength decreases
- NH₃ is the strongest base among group 15 hydrides
Ionization Enthalpy
Ionization enthalpy is the energy required to remove an electron from an atom. Generally, moving left to right across a period in the periodic table, ionization enthalpy tends to increase. However, there are exceptions.
Considering elements like B, C, O, N, nitrogen's higher ionization enthalpy compared to oxygen is due to its stable half-filled p orbital. This stability means more energy is required to remove an electron from nitrogen, making the sequence not strictly correct according to the trend. Important considerations include:
Considering elements like B, C, O, N, nitrogen's higher ionization enthalpy compared to oxygen is due to its stable half-filled p orbital. This stability means more energy is required to remove an electron from nitrogen, making the sequence not strictly correct according to the trend. Important considerations include:
- Stability of half-filled and fully filled orbitals
- General trend of increase across a period
- Specific exceptions like nitrogen
Oxidizing Power
Oxidizing power defines a substance’s ability to gain electrons and thereby oxidize other substances. It's an important property in assessing reactivity. Looking at oxides like CO₂, SiO₂, SnO₂, PbO₂, the oxidizing power generally increases down the group.
This phenomenon is explained by the decreasing ability of heavier metals to hold onto oxygen tightly as you go down the group in the periodic table. Basically, they become better oxidizers because they can readily accept electrons.
This phenomenon is explained by the decreasing ability of heavier metals to hold onto oxygen tightly as you go down the group in the periodic table. Basically, they become better oxidizers because they can readily accept electrons.
- Oxidizing power increases down the group
- Heavier metals hold onto oxygen less tightly
- Role in facilitating redox reactions
Other exercises in this chapter
Problem 168
The decreasing values of bond angles from \(\mathrm{NH}_{3}\) \(\left(106^{\circ}\right)\) to \(\mathrm{SbH}_{3}\left(101^{\circ}\right)\) down group 15 of the
View solution Problem 169
The stability of dihalides of \(\mathrm{Si}\), Ge, \(\mathrm{Sn}\) and \(\mathrm{Pb}\) increases steadily in the sequence \(\quad\) [2007] (a) \(\mathrm{SiX}_{2
View solution Problem 171
Which one of the following orders presents the correct sequence of the increasing basic nature of the given oxides? (a) \(\mathrm{Na}_{2} \mathrm{O}
View solution Problem 172
Which of the following statement is wrong? (a) Single \(\mathrm{N}-\mathrm{N}\) bond is weaker than the single \(\mathrm{P}-\mathrm{P}\) bond (b) Nitrogen canno
View solution