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The electronic configuration of four different elements is given below. Identify the group IV element among these. (a) \([\mathrm{He}] 2 \mathrm{~s}^{1}\) (b) \([\mathrm{Ne}] 3 \mathrm{~s}^{2}\) (c) \([\mathrm{Ne}] 3 \mathrm{~s}^{2} 3 \mathrm{p}^{2}\) (d) \([\mathrm{Ne}] 3 \mathrm{~s}^{2} 3 \mathrm{p}^{5}\)

Short Answer

Expert verified
The Group IV element is silicon with configuration [ ext{Ne}] 3 ext{s}^{2} 3 ext{p}^{2}.

Step by step solution

01

Review the Periodic Table Structure

Understand that the periodic table is organized in groups (vertical columns) and periods (horizontal rows). Elements in the same group typically have similar properties and react similarly. Group IV elements are in Group 14 of the periodic table.
02

Understand Electron Configuration

Electron configurations describe the distribution of electrons in shells and subshells. For identifying elements, particularly their group, the valence electron shell and its electrons (outermost shell) are crucial.
03

Analyze Each Configuration

(a) [ ext{He}] 2 ext{s}^{1} is for lithium, which is in Group 1. (b) [ ext{Ne}] 3 ext{s}^{2} is for magnesium, which is in Group 2. (c) [ ext{Ne}] 3 ext{s}^{2} 3 ext{p}^{2} is for silicon, which is in Group 14. (d) [ ext{Ne}] 3 ext{s}^{2} 3 ext{p}^{5} is for chlorine, which is in Group 17.
04

Identify the Group IV Element

From the analysis, configuration (c) [ ext{Ne}] 3 ext{s}^{2} 3 ext{p}^{2} corresponds to silicon, which belongs to Group 14 (the same as Group IV element).

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Key Concepts

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

Electron Configuration
Electron configuration is the arrangement of electrons in an atom's electron shells and subshells. Each element has a unique electron configuration that determines its chemical properties and its position in the periodic table.
For instance, the electron configuration provides insight into how an atom will bond with other atoms. It describes the distribution of electrons among the orbitals of an atom.
The configuration not only shows which "blocks" of elements in the periodic table the atom belongs to (s, p, d, or f) but also helps predict group behavior among elements. Understanding electron configuration allows us to identify valence electrons, which are essential for determining how an element will react chemically.
  • This configuration is typically written using the notation of an element's nearest noble gas in brackets, followed by the electron configuration of the outer electrons.
  • For example, \([\mathrm{Ne}] 3 \mathrm{s}^{2} 3 \mathrm{p}^{2}\) tells us that beyond neon's inner stable configuration, the atom has additional electrons distributed in 3s and 3p orbitals.
Group IV Elements
Group IV elements, better known as Group 14, include carbon, silicon, germanium, tin, lead, and flerovium. These elements are notable for their versatility in forming various types of bonds.
They sit in the p-block of the periodic table and possess four valence electrons. This means they can form covalent bonds by sharing these electrons easily, making them vital in forming a wide range of compounds.
Silicon, which was identified in the provided exercise as having the electron configuration \([\mathrm{Ne}] 3 \mathrm{s}^{2} 3 \mathrm{p}^{2}\), is a classic Group 14 element known for its role in the technology sector due to its semiconducting properties.
  • Group 14 elements show a progression of non-metallic to metallic characteristics as you move down the group.
  • While carbon displays significantly non-metallic traits, lead is more metallic.
  • This diversity in property is due to changes in atomic size and electron shielding as we descend the group.
Valence Electrons
Valence electrons are the outermost electrons of an atom and are crucial in determining an element's chemical properties. They participate in chemical bonds and influence how an element will interact with others.
The number of valence electrons is critical as it dictates the bonding capacity of the atom.
In context, Group IV elements have four valence electrons, which can form stable bonds either by sharing all four (covalently) or through other types of interactions. This allows these elements to form a range of molecules and compounds.
  • Atoms strive to achieve a stable electron configuration, often resembling that of noble gases, by gaining, losing, or sharing electrons.
  • This is known as the octet rule, which Group 14 elements can achieve by sharing their four valence electrons (thus forming covalent bonds).
Understanding these valence electrons and their configuration not only helps predict reactivates but also assists in constructing models for new compounds and materials, such as those seen in organic chemistry and semiconductor design.

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Most popular questions from this chapter

Among the elements W, X, Y and Z having atomic numbers \(9,10,11\) and 12 respectively, the correct order of ionization energies is: (a) \(\mathrm{W}>\mathrm{Y}>\mathrm{X}>\mathrm{Z}\) (b) \(\mathrm{X}>\mathrm{W}>\mathrm{Z}>\mathrm{Y}\) (c) \(X>Z>Y>W\) (d) \(\mathrm{Z}>\mathrm{Y}>\mathrm{X}>\mathrm{W}\)

Which of the electronic configuration represents a noble gas? (a) \(1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{6}\) (b) \(1 \mathrm{~s}^{2}, 2 \mathrm{~s}^{2}, 2 \mathrm{p}^{6}, 3 \mathrm{~s}^{2}, 3 \mathrm{p}^{3}\) (c) \(1 \mathrm{~s}^{2}, 2 \mathrm{~s}^{2}, 2 \mathrm{p}^{6}, 3 \mathrm{~s}^{2}, 3 \mathrm{p}^{1}\) (d) \(1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{4}\)

The electron affinity of \(\mathrm{Be}\) is similar to (a) He (b) B (c) \(\mathrm{Li}\) (d) \(\mathrm{Na}\)

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

Correct order of ionization energy among the elements Be, B, C, N, O is (a) \(\mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N}\) (b) \(\mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{N}<\mathrm{O}\) (c) \(\mathrm{Be}<\mathrm{B}<\mathrm{C}<\mathrm{N}<\mathrm{O}\) (d) \(\mathrm{Be}<\mathrm{B}<\mathrm{O}<\mathrm{N}<\mathrm{C}\)

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