
Which of the following can exhibit only +3 oxidation state?
A. Th
B. Ac
C. Pa
D. U
Answer
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Hint: The oxidation state also known as oxidation number which describes the degree of oxidation i.e. loss of electrons of an atom in a chemical compound. Conceptually the oxidation state may be positive, negative or zero.
Complete Step by step solution: Oxidation states are typically represented by integers which may be positive, zero, or negative. In some cases, the average oxidation state of an element is a fraction. The highest known oxidation state is reported to be +9 in the tetrox iridium (IX) cation \[Ir{{O}_{4}}^{+}\]. It is predicted that even a +10 oxidation state may be achievable by platinum in the tetrox platinum(X) cation \[Pt{{O}_{4}}^{2+}\]. The lowest oxidation state is −5, as for boron. Now let discuss the oxidation states of all the options given in question:
Th having atomic number 90 showing electronic configuration $[Rn]6{{d}^{2}}7{{s}^{2}}$ showing 2 oxidation states of +3 and +4. Ac; atomic number 89 showing electronic configuration $[Rn]6{{d}^{2}}7{{s}^{1}}$ it shows stable oxidation state of +3 only. Pa; atomic number 91; E.C of Pa is $[Rn]5{{f}^{2}}6{{d}^{1}}7{{s}^{2}}$ showing oxidation states of +3, +4 and +5. U mainly used in nuclear reactions having atomic number 92 and showing electronic configuration as $[Rn]5{{f}^{3}}6{{d}^{1}}7{{s}^{2}}$ showing 4 oxidation states +3, +4, +5 and +6. Thus, we can say that Ac is only compound which shows only +3 oxidation state.
Hence option B is the correct answer.
Note: The increase in oxidation state of an atom, through a chemical reaction, is known as an oxidation; a decrease in oxidation state is known as a reduction. Such reactions involve the formal transfer of electrons: a net gain in electrons being a reduction, and a net loss of electrons being an oxidation. For pure elements, the oxidation state is zero.
Complete Step by step solution: Oxidation states are typically represented by integers which may be positive, zero, or negative. In some cases, the average oxidation state of an element is a fraction. The highest known oxidation state is reported to be +9 in the tetrox iridium (IX) cation \[Ir{{O}_{4}}^{+}\]. It is predicted that even a +10 oxidation state may be achievable by platinum in the tetrox platinum(X) cation \[Pt{{O}_{4}}^{2+}\]. The lowest oxidation state is −5, as for boron. Now let discuss the oxidation states of all the options given in question:
Th having atomic number 90 showing electronic configuration $[Rn]6{{d}^{2}}7{{s}^{2}}$ showing 2 oxidation states of +3 and +4. Ac; atomic number 89 showing electronic configuration $[Rn]6{{d}^{2}}7{{s}^{1}}$ it shows stable oxidation state of +3 only. Pa; atomic number 91; E.C of Pa is $[Rn]5{{f}^{2}}6{{d}^{1}}7{{s}^{2}}$ showing oxidation states of +3, +4 and +5. U mainly used in nuclear reactions having atomic number 92 and showing electronic configuration as $[Rn]5{{f}^{3}}6{{d}^{1}}7{{s}^{2}}$ showing 4 oxidation states +3, +4, +5 and +6. Thus, we can say that Ac is only compound which shows only +3 oxidation state.
Hence option B is the correct answer.
Note: The increase in oxidation state of an atom, through a chemical reaction, is known as an oxidation; a decrease in oxidation state is known as a reduction. Such reactions involve the formal transfer of electrons: a net gain in electrons being a reduction, and a net loss of electrons being an oxidation. For pure elements, the oxidation state is zero.
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