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The increasing order of the first ionization enthalpies of the element \(\mathrm{B}, \mathrm{P}, \mathrm{S}\) and \(\mathrm{F}\) (lowest first) is: (a) \(\mathrm{F}<\mathrm{S}<\mathrm{P}<\mathrm{B}\) (b) \(\mathrm{P}<\mathrm{S}<\mathrm{B}<\mathrm{F}\) (c) \(\mathrm{B}<\mathrm{P}<\mathrm{S}<\mathrm{F}\) (d) \(\mathrm{B}<\mathrm{S} \leq \mathrm{P}<\mathrm{F}\)

Short Answer

Expert verified
The correct order is (c) \(\mathrm{B}<\mathrm{P}<\mathrm{S}<\mathrm{F}\).

Step by step solution

01

Understanding Ionization Enthalpy

Ionization enthalpy is the energy required to remove an electron from an isolated gaseous atom in its ground state. Generally, ionization enthalpy increases across a period from left to right due to increasing nuclear charge and decreases down a group due to increased atomic size and electron shielding.
02

Identify Periods for Elements

Elements given are Boron (B), Phosphorus (P), Sulfur (S), and Fluorine (F). Boron (B) is in the 2nd period, Phosphorus (P) and Sulfur (S) are in the 3rd period, and Fluorine (F) is in the 2nd period. Typically, ionization energy will increase across these elements within a period.
03

Analyze General Ionization Trends

Across a period, ionization enthalpy increases. Hence, Fluorine (F) should have the highest ionization enthalpy since it is at the end of the 2nd period. Boron (B) will likely have the lowest among these due to its position in the periodic table and its lower group. Between Sulfur (S) and Phosphorus (P), since S comes after P in the periodic table, S will have a higher ionization enthalpy.
04

Order the Elements by Ionization Enthalpy

Following the trend, the order from lowest to highest ionization enthalpy is Boron (B), Phosphorus (P), Sulfur (S), Fluorine (F). We map this order to the given options.
05

Check Options Against Calculated Order

By comparing our calculated order (B < P < S < F) with the options provided, we see that option (c) matches our order. Therefore, the increasing order of the first ionization enthalpies is correctly represented by option (c).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Periodic Trends
In the world of chemistry, understanding periodic trends is vital because it helps predict the chemical behavior of elements. The periodic table is arranged in such a way that elements display recurring or "periodic" trends in their physical and chemical properties. One notable periodic trend is ionization enthalpy. As you move from left to right across a period, the nuclear charge increases. This means the nucleus holds onto the outer electrons more tightly, resulting in a higher ionization energy. Conversely, as you move down a group, ionization energy tends to decrease. The increased number of electron shells means the outermost electrons are further from the nucleus, and the inner shell electrons shield them from the nucleus’ pull, making them easier to remove.
Ionization Energy
Ionization energy, also known as ionization enthalpy, is the minimum amount of energy required to remove an electron from a gaseous atom in its ground state. This concept helps explain why elements are more or less likely to form chemical bonds. Important factors affecting ionization energy include:
  • Atomic size: Smaller atoms have higher ionization energies because their outer electrons are closer to the nucleus and more tightly held.
  • Nuclear charge: More protons mean a stronger attraction between the nucleus and electrons, leading to higher ionization energy.
  • Electron shielding: Inner electrons can shield outer electrons from the full pull of the nucleus. More shielding results in lower ionization energy.
Recognizing these factors helps in predicting an element's reactivity and its electrolyte behavior.
Periodic Table
The periodic table is a powerful tool that arranges all known elements in a systematic way based on their atomic numbers, electron configurations, and recurring chemical properties. It enables scientists and students alike to predict the types, strengths, and characteristics of bonds that elements will form.
  • Groups (columns): Elements in the same group typically exhibit similar chemical behavior due to their identical valence electron configurations.
  • Periods (rows): Moving across a period, the properties of elements like ionization energy, electronegativity, and atomic radius gradually change due to increasing nuclear charge.
With its comprehensive data, the periodic table also emphasizes trends like increasing ionization energy across periods and decreasing down groups. This well-structured chart not only aids in learning but also in practical applications like predicting outcomes in chemical reactions.

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Most popular questions from this chapter

The transition metal with least atomic number is: (a) \(\mathrm{Os}\) (b) \(\mathrm{Zr}\) (c) \(\mathrm{Pt}\) (d) \(\mathrm{Ru}\)

The correct statement among the following is: (a) The first ionization potential of \(\mathrm{Al}\) is less than the first ionization potential of \(\mathrm{Mg}\) (b) The second ionization potential of \(\mathrm{Mg}\) is greater than the second ionization potential of Na (c) The first ionization potential of \(\mathrm{Na}\) is less than the first ioniation potential of \(\mathrm{Mg}\) (d) The third ionization potential of \(\mathrm{Mg}\) is greater than the third ionization potential of \(\mathrm{Al}\)

Let IP stand for ionization potential. The IP, and \(\mathrm{IP}_{2}\) of \(\mathrm{Mg}\) are 178 and \(348 \mathrm{kcal} \mathrm{mol}^{-1}\). The energy required for the following reaction is: \(\mathrm{Mg} \rightarrow \mathrm{Mg}^{2+}+2 \mathrm{e}^{-}\) (a) \(+178\) kcal (b) \(+526 \mathrm{kcal}\) (c) \(-170\) kcal (d) \(-526\) kcal

The formation of the oxide ion \(\mathrm{O}^{2-}(\mathrm{g})\) require first an exothermic and then an endothermic step as shown below: \(\mathrm{O}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{O}^{-}(\mathrm{g}) ; \Delta \mathrm{H}^{0}=-142 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\mathrm{O}^{-}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{O}^{2-}(\mathrm{g}) ; \Delta \mathrm{H}^{0}=844 \mathrm{~kJ} \mathrm{~mol}^{-1}\) This is because, (a) Oxygen is more electronegative (b) Oxygen has high electron affinity (c) \(\mathrm{O}^{-}\) ion has comparatively larger size than oxygen atom (d) \(\mathrm{O}\) - ion will tend to resist the addition of another electron

Let IP stand for ionization potential. Then element for which the value of \(\mathrm{IP}_{1}, \mathrm{IP}_{2}, \mathrm{IP}_{3}, \mathrm{IP}_{4}\) are \(50,100,250\) and \(2051 \mathrm{eV}\) respectively, is (a) \(\mathrm{Al}\) (b) \(\mathrm{Na}\) (c) \(\mathrm{Mg}\) (d) Si

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