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For the atom \(^{119} \mathrm{Sn}\), indicate the number of (a) protons in the nucleus; (b) neutrons in the nucleus; (c) \(4 d\) electrons; (d) 3s electrons; (e) 5 \(p\) electrons; (f) electrons in the valence shell.

Short Answer

Expert verified
The atom \(^{119} Sn\) has (a) 50 protons, (b) 69 neutrons, (c) 10 4d electrons, (d) 2 3s electrons, (e) 2 5p electrons, and (f) 4 electrons in the valence shell.

Step by step solution

01

Protons from Atomic Number

Search for Sn on the periodic table, it has atomic number 50. Hence, it has 50 protons.
02

Neutrons from Mass Number

Subtract atomic number from mass number, i.e., \(neutrons = 119 - 50 = 69\).
03

Electrons in the 4d Orbital

According to the electron distribution in atomic orbitals, 4d orbital fills after the 5s orbital. Tin corresponds to the electron configuration [Kr]4d105s25p2. Since the 4d orbital is filled before the 5s and 5p orbitals, it's full with 10 electrons.
04

Electrons in the 3s Orbital

The 3s orbital fills before the 3p and 4s orbitals. It always holds only 2 electrons when full. Therefore, there are 2 electrons in the 3s orbital of Tin.
05

Electrons in the 5p Orbital

In configuration [Kr]4d105s25p2, it's evident that 5p orbital has 2 electrons.
06

Electrons in the Valence Shell

Valence shell is the outermost shell, which is the 5th shell for Tin. This shell includes 5s and 5p orbitals, hence holds \(2+2=4\) electrons.

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Key Concepts

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

Atomic Structure
Atomic structure is the foundation of understanding how atoms behave and interact. An atom consists of a nucleus, packed with protons and neutrons, surrounded by clouds of electrons. This setup is often depicted as electrons orbiting the nucleus, but in reality, electrons are found in regions called orbitals. Every atom has a nucleus at the center, which contains protons and neutrons, while electrons circle around in space defined by orbitals.
  • Protons are positively charged particles.
  • Neutrons have no charge.
  • Electrons are negatively charged and occupy various energy levels.
Understanding atomic structure is essential for determining an atom's reactivity, electronegativity, and bonding behavior. The arrangement of electrons is particularly crucial, as it defines how elements will bond and form compounds.
Neutrons
Neutrons, along with protons, form the nucleus of an atom. These subatomic particles play a pivotal role in the stability of the atomic nucleus. While they are electrically neutral, neutrons contribute significantly to the atomic mass. To find the number of neutrons in an atom, subtract the atomic number from the mass number. For example, in Tin ( {119} ext{Sn}), there are 69 neutrons (119 mass number - 50 protons = 69 neutrons).
Neutrons are vital for:
  • Stability: They offset the repulsive forces between positively charged protons within the atomic nucleus.
  • Isotopic identity: Variations in neutron number between atoms of the same element produce different isotopes with distinct properties.
Understanding neutrons helps in recognizing isotopes and predicting nuclear reactions.
Valence Electrons
Valence electrons are the outermost electrons of an atom and determine the chemical properties and reactivity of the element. In Tin ( {119} ext{Sn}), these are the electrons found in the 5s and 5p orbitals.
Tin has four valence electrons in total, with two in the 5s orbital and two more in the 5p orbital. Valence electrons are crucial for:
  • Bond Formation: They are the electrons involved when atoms form chemical bonds.
  • Reactivity: The number and distribution of valence electrons dictate how and with what elements an atom will react.
  • Electronegativity: The readiness with which an atom attracts and holds onto electrons in a bond.
Understanding valence electrons is key to mastering the concept of chemical bonding and reaction mechanisms.
Periodic Table
The periodic table is a systematic arrangement of elements based on their atomic number, electron configurations, and recurring chemical properties. It is a powerful tool for predicting and understanding the behavior of elements. Elements are organized into groups and periods, with groups being the vertical columns and periods being the horizontal rows. Each position on the periodic table reflects an element’s electron configuration and its place in terms of reactivity and properties.
The periodic table provides insight into:
  • Trends in electronegativity and atomic size.
  • Electron configurations, indicating why particular elements display behavior such as forming certain types of ions or bonds.
  • The classification of elements as metals, nonmetals, or metalloids.
Utilizing the periodic table helps in predicting the behavior of elements and their compounds in various chemical reactions.

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Most popular questions from this chapter

The following melting points are in degrees Celsius. Show that melting point is a periodic property of these elements: \(\mathrm{Al}, 660 ; \mathrm{Ar},-189 ; \mathrm{Be}, 1278 ; \mathrm{B}, 2300 ; \mathrm{C}\) \(3350 ; \mathrm{Cl},-101 ; \mathrm{F},-220 ; \mathrm{Li}, 179 ; \mathrm{Mg}, 651 ; \mathrm{Ne},-249 ; \mathrm{N}\) \(-210 ; \mathrm{O},-218 ; \mathrm{P}, 590 ; \mathrm{Si}, 1410 ; \mathrm{Na}, 98 ; \mathrm{S}, 119.\)

A method for estimating electron affinities is to extrapolate \(Z_{\text {eff }}\) values for atoms and ions that contain the same number of electrons as the negative ion of interest. Use the data in the table on the next page to answer the questions that follow. $$\begin{array}{lll} \hline \begin{array}{l} \text { Atom or lon: } \\ \text { I(kJmol }^{-1} \text {) } \end{array} & \begin{array}{l} \text { Atom or lon: } \\ \text { I(kJmol }^{-1} \text {) } \end{array} & \begin{array}{l} \text { Atom or lon: } \\ \text { I(kJmol }^{-1} \text {) } \end{array} \\ \hline \text { Ne: 2080 } & \text { F: 1681 } & \text { O: } 1314 \\ \text { Na }^{+}: 4565 & \text { Ne }^{+}: 3963 & \text { F }^{+}: 3375 \\ \text { Mg }^{2+} \text { : 7732 } & \text { Na }^{2+}: 6912 & \text { Ne }^{2+}: 6276 \\ \text { A1 }^{\text {3 }^{+}: 11,577} & \text { Mg }^{3+}: 10,548 & \text { Na }^{3+}: 9540 \\ \hline \end{array}$$ (a) Estimate the electron affinity of \(F\), and compare it with the experimental value. (b) Estimate the electron affinities of \(\mathrm{O}\) and \(\mathrm{N}\) (c) Examine your results in terms of penetration and screening.

Refer only to the periodic table on the inside front cover, and arrange the following ionization energies in order of increasing value: \(I_{1}\) for \(\mathrm{F} ; I_{2}\) for \(\mathrm{Ba}\) \(I_{3}\) for \(\mathrm{Sc} ; I_{2}\) for \(\mathrm{Na} ; I_{3}\) for \(\mathrm{Mg}\). Explain the basis of any uncertainties.

Arrange the following in expected order of increasing radius: \(\mathrm{Br}, \mathrm{Li}^{+}, \mathrm{Se}, \mathrm{I}^{-} .\) Explain your answer.

Two elements, \(A\) and \(B\), have the electron configurations shown. $$ \mathrm{A}=[\mathrm{Kr}] 4 s^{2} \quad \mathrm{B}=[\mathrm{Ar}] 3 d^{10} 4 s^{2} 4 p^{5} $$ (a) Which element is a metal? (b) Which element has the greater ionization energy? (c) Which element has the larger atomic radius? (d) Which element has the greater electron affinity?

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