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Using only the periodic table as your guide, select the most electronegative atom in each of the following sets: (a) Na, \(\mathrm{Mg}, \mathrm{K}, \mathrm{Ca} ;(\mathbf{b}) \mathrm{P}, \mathrm{S},\) As, \(\mathrm{Se} ;(\mathbf{c}) \mathrm{Be}, \mathrm{B}, \mathrm{C}, \mathrm{Si} ;(\mathbf{d}) \mathrm{Zn}, \mathrm{Ge}, \mathrm{Ga}, \mathrm{Gs}\)

Short Answer

Expert verified
The most electronegative elements in each set are: (a) $\mathrm{Mg}$; (b) $\mathrm{S}$; (c) $\mathrm{C}$; (d) $\mathrm{Ga}$.

Step by step solution

01

Set (a): Na, Mg, K, Ca

To determine the most electronegative atom among Na, Mg, K, and Ca, we need to locate these elements in the periodic table. They are all in the same period (Period 3) and belong to Groups 1 and 2. As electronegativity increases from left to right across a period, Mg (Group 2) is more electronegative than Na (Group 1). Similarly, Ca (Group 2) is more electronegative than K (Group 1). Between Mg and Ca, since electronegativity decreases down a group, Mg is more electronegative than Ca. Therefore, Mg is the most electronegative element in the given set.
02

Set (b): P, S, As, Se

To determine the most electronegative atom among P, S, As, and Se, we need to locate these elements in the periodic table. They belong to Groups 15 and 16. From left to right in the periodic table, electronegativity increases, so S (Group 16) is more electronegative than P (Group 15) and Se (Group 16) is more electronegative than As (Group 15). Between S and Se, since electronegativity decreases down a group, S is more electronegative than Se. Therefore, S is the most electronegative element in the given set.
03

Set (c): Be, B, C, Si

To determine the most electronegative atom among Be, B, C, and Si, we need to locate these elements in the periodic table. They are all in the same period (Period 2) and belong to Groups 13 and 14. As electronegativity increases from left to right across a period, C (Group 14) is more electronegative than B (Group 13) and Si (Group 14). Between C and Si, since electronegativity decreases down a group, C is more electronegative than Si. Therefore, C is the most electronegative element in the given set.
04

Set (d): Zn, Ge, Ga, Gs

To determine the most electronegative atom among Zn, Ge, Ga, and Gs, we need to locate these elements in the periodic table. They belong to Groups 12, 13, and 14. From left to right in the periodic table, electronegativity increases, so Gs (Group 14) is more electronegative than Ge (Group 13) and Ga (Group 13) is more electronegative than Zn (Group 12). Between Gs and Ga, since electronegativity increases from bottom to top in a group, Ga is more electronegative than Gs. Therefore, Ga is the most electronegative element in the given set.

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Key Concepts

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

periodic table
The periodic table is a powerful tool for chemists and chemistry students. It arranges all known chemical elements in an order that exhibits recurring properties. Each element is placed in a specific position based on its atomic number, electron configuration, and chemical properties. This layout helps us predict the behavior of elements in chemical reactions and bonds.

The periodic table is made of rows and columns. The rows are called periods, and the columns are called groups or families. As you move across a period from left to right, the number of protons and electrons increases, which leads to changes in the properties of the elements. On the other hand, moving down a group keeps the electron configuration in the outer shell similar, giving elements similar characteristics.
groups and periods
Groups and periods in the periodic table are crucial in understanding element properties, including electronegativity.

Groups refer to the vertical columns. Elements in the same group have the same number of electrons in their outer shell, often resulting in similar chemical properties. For example, all elements in Group 1 are metals with one electron in their outer shell, making them very reactive.

Periods, on the other hand, are the horizontal rows. As you move across a period from left to right, the atomic number increases, and elements become less metallic and more non-metallic. For example, in Period 2, you start with metals like Lithium and end with the noble gas Neon. This trend is significant when discussing properties like electronegativity because it generally increases across a period but decreases down a group.
electronegativity trends
Electronegativity refers to the ability of an atom to attract and hold onto electrons in a chemical bond. It's a crucial concept for understanding how atoms interact in molecules. In the periodic table, electronegativity follows distinct trends:

- **Across a period**: Electronegativity values generally increase from left to right as you move across a period. This is because atoms with more protons can attract the bonding electron pair more strongly. For example, in Period 2, Carbon is more electronegative than Boron.

- **Down a group**: Electronegativity decreases as you move down a group. This is because the addition of electron shells increases the distance between the nucleus and the bonding electrons, making the attraction weaker. For example, in Group 16, Oxygen is more electronegative than Sulfur.

This knowledge helps predict how different elements will bond and react with each other, which is essential for mastering chemistry.
chemical elements comparison
When comparing chemical elements, particularly for properties like electronegativity, it's important to use the periodic table guidelines. Understanding these comparisons helps provide insights into why elements behave the way they do chemically.

For instance, when looking at a set of elements, you can determine the most electronegative by considering their positions in the periodic table. Magnesium (Mg), for example, is more electronegative than Sodium (Na) because it is further to the right in Period 3. Similarly, Sulfur (S) is more electronegative than Phosphorus (P) because it is found further right in Period 3.
  • Remember: Electronegativity increases across a period.
  • Electronegativity decreases down a group.
By using these patterns, we can efficiently and accurately assess how elements will interact in chemical reactions.

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Most popular questions from this chapter

Ammonia reacts with boron trifluoride to form a stable compound, as we saw in Section 8.7 . (a) Draw the Lewis structure of the ammonia-boron trifluoride reaction product. (b) The B-N bond is obviously more polar than the \(\mathrm{C}-\mathrm{C}\) bond. Draw the charge distribution you expect on the \(\mathrm{B}-\mathrm{N}\) bond within the molecule (using the delta plus and delta minus symbols mentioned in Section 8.4\()\) . ( ) Boron trichloride also reacts with ammonia in a similar way to the trifluoride. Predict whether the \(B-N\) bond in the trichloride reaction product would be more or less polar than the \(B-N\) bond in the trifluoride product, and justify your reasoning.

By referring only to the periodic table, select (a) the most electronegative element in group \(6 \mathrm{A} ;(\mathbf{b})\) the least electronegative element in the group Al, Si, P; (c) the most electronegative element in the group \(\mathrm{Ga}, \mathrm{P}, \mathrm{Cl}, \mathrm{Na} ;(\mathbf{d})\) the element in the group \(\mathrm{K}\) \(\mathrm{C}, \mathrm{Zn}, \mathrm{F}\) that is most likely to form an ionic compound with Ba.

In the following pairs of binary compounds, determine which one is a molecular substance and which one is an ionic substance. Use the appropriate naming convention (for ionic or molecular substances) to assign a name to each compound: (a) \(\mathrm{TiCl}_{4}\) and \(\mathrm{CaF}_{2},(\mathbf{b}) \mathrm{ClF}_{3}\) and \(\mathrm{VF}_{3},(\mathbf{c}) \mathrm{SbCl}_{5}\) and \(\mathrm{AlF}_{3} .\)

Under special conditions, sulfur reacts with anhydrous liquid ammonia to form a binary compound of sulfur and nitrogen. The compound is found to consist of 69.6\(\% \mathrm{S}\) and 30.4\(\% \mathrm{N} .\) Measurements of its molecular mass yield a value of 184.3 \(\mathrm{g} / \mathrm{mol}\) . The compound occasionally detonates on being struck or when heated rapidly. The sulfur and nitrogen atoms of the molecule are joined in a ring. All the bonds in the ring are of the same length. (a) Calculate the empirical and molecular formulas for the substance. (b) Write Lewis structures for the molecule, based on the information you are given. (Hint: You should find a relatively small number of dominant Lewis structures.) (c) Predict the bond distances between the atoms in the ring. (Note: The \(S-S\) distance in the \(S_{8}\) ring is 2.05 A.) ( d.) The enthalpy of formation of the compound is estimated to be 480 \(\mathrm{kJ} / \mathrm{mol}^{-1} . \Delta H_{f}^{9}\) of \(\mathrm{S}(g)\) is 222.8 \(\mathrm{kJ} / \mathrm{mol} .\) Estimate the average bond enthalpy in the compound.

(a) Does the lattice energy of an ionic solid increase or decrease (i) as the charges of the ions increase, (ii) as the sizes of the ions increase? (b) Arrange the following substances not listed in Table 8.1 according to their expected lattice energies, listing them from lowest lattice energy to the highest: MgS, Ki, GaN, LiBr.

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