Carbon is in the lowest oxidation state in (a) \(\mathrm{CH}_{4}\) (b) \(\mathrm{CCl}_{4}\) (c) \(\mathrm{CF}_{4}\) (d) \(\mathrm{CO}_{2}\)

Short Answer

Expert verified
Carbon is in the lowest oxidation state in \(\mathrm{CH}_{4}\).

Step by step solution

01

Understand the concept of oxidation state

The oxidation state is the hypothetical charge that an atom would have if all bonds to atoms of different elements were completely ionic. A lower oxidation state means the atom has more electrons compared to its standard state.
02

Determine the oxidation state of carbon in each compound

Assign oxidation states based on known rules: Hydrogen is typically +1, oxygen is -2, fluorine is always -1, and chlorine is usually -1. Add the oxidation states of all atoms in the molecule to equal the overall charge (which is zero for all molecules listed).
03

Compare the oxidation states for carbon in each molecule

\(\mathrm{CH}_{4}\): Each hydrogen is +1, so carbon must be -4. \(\mathrm{CCl}_{4}\): Each chlorine is -1, so carbon is +4. \(\mathrm{CF}_{4}\): Each fluorine is -1, so carbon is +4. \(\mathrm{CO}_{2}\): Each oxygen is -2, so carbon is +4. The lowest oxidation state for carbon is -4.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Oxidation State Rules
Understanding the rules of oxidation states is crucial for analyzing chemical reactions and bonding. An oxidation state, often referred to as oxidation number, is a theoretical number assigned to an element in a chemical compound that represents its loss or gain of electrons when compared to its elemental state.

Oxidation states help in determining how electrons are distributed in a chemical compound and can indicate whether a compound is likely to be oxidized or reduced in a chemical reaction. Here are some key rules for assigning oxidation states:
  • The oxidation state of a pure element is always zero.
  • For monoatomic ions, the oxidation state is equal to the charge on the ion.
  • Oxygen usually has an oxidation state of -2, except in peroxides where it is -1.
  • Hydrogen has an oxidation state of +1 when combined with nonmetals, and -1 when bonded to metals.
  • Alkali metals (group 1) always have an oxidation state of +1, and alkaline earth metals (group 2) are always +2.
  • Fluorine always has an oxidation state of -1 as it is the most electronegative element.
  • The sum of oxidation states in a neutral compound is zero, while in polyatomic ions it equals the charge of the ion.
These rules form a foundation for systematically assigning an oxidation state to each element within a compound.
Assigning Oxidation Numbers
Assigning oxidation numbers is like solving a puzzle where the oxidation rules are your guide. The goal is to find the hypothetical charges for each atom that make the overall charge of the compound come out right.

Here's a simple approach for assigning oxidation numbers: Start by assigning the elements with known oxidation states based on the rules, such as hydrogen (+1), oxygen (-2), and fluorine (-1). For a molecule like water (H2O), you'd assign hydrogen as +1 and oxygen as -2. Since there are two hydrogens, their combined charge is +2, and oxygen is -2, which equals out to zero – the overall charge for a molecule.

When dealing with molecules where the oxidation states are not immediately obvious, like in transition metal complexes, you can still apply the oxidation state rules in combination with information about the compound's charge and the known oxidation states of other ligands or atoms to deduce the oxidation state of the metal. The process requires careful analysis and sometimes a bit of trial and error to balance out the charges.
Chemical Bonding
Chemical bonding is the force that holds atoms together in molecules and compounds. It encompasses a variety of interactions including ionic, covalent, and metallic bonds, all of which are influenced by the oxidation states of the participating atoms.

In ionic bonding, atoms transfer electrons to achieve noble gas electron configurations, resulting in the formation of positively and negatively charged ions. For example, when sodium (Na) combines with chlorine (Cl) to form sodium chloride (NaCl), Na gives up one electron to achieve an oxidation state of +1, while Cl accepts an electron, resulting in an oxidation state of -1.

Covalent bonding, on the other hand, involves the sharing of electron pairs between atoms, leading to the formation of molecules. The oxidation state in a covalent bond can be determined by assigning shared electrons to the more electronegative atom. For instance, in the molecule of carbon dioxide (CO2), oxygen is more electronegative than carbon, so the electrons in the C=O bonds are considered closer to oxygen, resulting in a +4 oxidation state for carbon and a -2 state for each oxygen.

Metallic bonding occurs between metal atoms, which pool their valence electrons into a 'sea' of delocalized electrons, resulting in metals' characteristic properties like conductivity and malleability. Understanding the concept of oxidation states is less directly applicable to metallic bonding, but it's still essential for grasping how metals can give up electrons to form positive ions in ionic compounds.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What is the correct representation of reaction occurring when \(\mathrm{HCl}\) is heated with \(\mathrm{MnO}_{2} ?\) (a) \(\mathrm{MnO}_{4}^{-}+5 \mathrm{Cl}^{-}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+5 \mathrm{Cl}^{-}+5 \mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{MnO}_{2}+2 \mathrm{Cl}^{-}+4 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+\mathrm{Cl}_{2}+2 \mathrm{H}_{2} \mathrm{O}\) (c) \(2 \mathrm{MnO}_{2}+4 \mathrm{Cl}^{-}+8 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+}+2 \mathrm{Cl}_{2}+4 \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{MnO}_{2}+4 \mathrm{HCl} \rightarrow \mathrm{MnCl}_{4}+\mathrm{Cl}_{2}+\mathrm{H}_{2} \mathrm{O}\)

Given below are few statements regarding electrode potentials. Mark the correct statements. (i) The potential associated with each electrode is known as electrode potential. (ii) A negative \(E^{\circ}\) means that the redox couple is a stronger reducing agent than \(\mathrm{H}^{+} / \mathrm{H}_{2}\) couple. (iii) A positive \(E^{\circ}\) means that the redox couple is a weaker reducing agent than \(\mathrm{H}^{+} / \mathrm{H}_{2}\) couple.

What are the oxidation states of phosphorus in the following compounds? \(\mathrm{H}_{3} \mathrm{PO}_{2}, \mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}, \mathrm{PH}_{3}, \mathrm{HPO}_{3}\) (a) \(+1,+3,+3,+3,+5\) (b) \(+3,+3,+5,+5,+5\) (c) \(+1,+2,+3,+5,+5\) (d) \(+1,+5,+5,-3,+5\)

Which compound amongst the following has the highest oxidation number of Mn? (a) \(\mathrm{KMnO}_{4}\) (b) \(\mathrm{K}_{2} \mathrm{MnO}_{4}\) (c) \(\mathrm{MnO}_{2}\) (d) \(\mathrm{Mn}_{2} \mathrm{O}_{3}\)

Given \(E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0}=+0.80 \mathrm{~V} ; \quad E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=+0.34 \mathrm{~V}\) \(E_{\mathrm{Ft}^{3}+i \mathrm{Fe}^{2}}=+0.76 \mathrm{~V} ; E^{\circ} \mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}=+1.60 \mathrm{~V}\) Which of the following statements is not correct? (a) \(\mathrm{Fe}^{3+}\) does not oxidise \(\mathrm{Ce}^{3+}\). (b) Cu reduces \(\mathrm{Ag}^{+}\)to \(\mathrm{Ag}\). (c) Ag will reduce \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\). (d) \(\mathrm{Fe}^{3+}\) reduces \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free