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For elements in the third row of the periodic table and beyond, the octet rule is often not obeyed. What factors are usually cited to explain this fact?

Short Answer

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Elements in the third row of the periodic table and beyond do not strictly obey the octet rule due to factors such as the presence of d-orbital electrons, which can accommodate up to 10 electrons, larger atomic size, lower electronegativity, and multiple oxidation states. These factors allow the central atom to expand its octet and bond with more than four electron pairs, deviating from the octet rule.

Step by step solution

01

Understanding the Octet Rule

The octet rule is a chemical rule of thumb stating that atoms tend to combine with other atoms in such a way that each atom has eight valence electrons in its outermost shell, giving it the same electron configuration as a noble gas. This rule mainly applies to elements in the first and second periods of the periodic table.
02

Locating Elements in the Third Row of the Periodic Table and Beyond

Elements in the third row of the periodic table and beyond are those with atomic numbers greater than or equal to 19, starting from potassium (K) onwards. These elements have electron configurations that include s, p, and d orbitals, which are not present in elements from the first and second periods.
03

Understanding the Presence of d-Orbital Electrons in These Elements

In the elements beyond the second period, d-orbital electrons are introduced. Unlike s and p orbitals, which can hold a maximum of 2 and 6 electrons respectively, d orbitals can accommodate up to 10 electrons, thus allowing for more than eight electrons around the central atom.
04

Discussing the Factors Contributing to the Deviation from the Octet Rule

The factors usually cited to explain why the octet rule is not obeyed in elements in the third row of the periodic table and beyond are: 1. The presence of d orbital electrons: As mentioned earlier, the d orbitals can accommodate up to 10 electrons, allowing the central atom to expand its octet and bond with more than four electron pairs. 2. Larger atomic size: As we move from top to bottom in the periodic table, the atomic size of elements increases, which can accommodate more electrons in their outermost shell. This leads to the possibility of more than eight electrons around the central atom. 3. Lower electronegativity: Elements beyond the second period often have lower electronegativity values, which means they tend to share or lose electrons more easily, leading to a deviation from the octet rule. 4. Multiple oxidation states: Since these elements have d-orbital electrons, they can exhibit various oxidation states. The capacity to accommodate more than eight electrons in their outer shells allows them to form compounds in multiple oxidation states, thereby disregarding the octet rule. In conclusion, elements in the third row of the periodic table and beyond do not strictly obey the octet rule due to factors like the presence of d-orbital electrons, larger atomic size, lower electronegativity, and multiple oxidation states.

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Most popular questions from this chapter

(a) What is the octet rule? (b) How many electrons must a sulfur atom gain to achieve an octet in its valence shell? (c) If an atom has the electron configuration \(1 s^{2} 2 s^{2} 2 p^{3}\), how many electrons must it gain to achieve an octet?

Which of the following bonds are polar: (a) \(\mathrm{B}-\mathrm{F}\), (b) \(\mathrm{Cl}-\mathrm{Cl}\), (c) \(\mathrm{Se}-\mathrm{O}\), (d) \(\mathrm{H}-\mathrm{I}\) ? Which is the more electronegative atom in each polar bond?

(a) Use bond enthalpies to estimate the enthalpy change for the reaction of hydrogen with ethene: $$\mathrm{H}_{2}(g)+\mathrm{C}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)$$ (b) Calculate the standard enthalpy change for this reaction, using heats of formation. Why does this value differ from that calculated in (a)?

Average bond enthalpies are generally defined for gasphase molecules. Many substances are liquids in their standard state. \(\mathrm{OB}\) (Section 5.7) By using appropriate thermochemical data from Appendix C, calculate average bond enthalpies in the liquid state for the following bonds, and compare these values to the gas-phase values given in Table 8.4: (a) \(\mathrm{Br}-\mathrm{Br}\), from \(\mathrm{Br}_{2}(l) ;\) (b) \(\mathrm{C}-\mathrm{Cl}\), from \(\mathrm{CCl}_{4}(l) ;\) (c) \(\mathrm{O}-\mathrm{O}\), from \(\mathrm{H}_{2} \mathrm{O}_{2}(l)\) (assume that the \(\mathrm{O}-\mathrm{H}\) bond enthalpy is the same as in the gas phase). (d) What can you conclude about the process of breaking bonds in the liquid as compared to the gas phase? Explain the difference in the \(\Delta H\) values between the two phases.

(a) What is meant by the term electronegativity? (b) On the Pauling scale what is the range of electronegativity values for the elements? (c) Which element has the greatest electronegativity? (d) Which element has the smallest electronegativity?

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