Question:medium
Consider \(|x|\) is the difference in oxidation states of Mn in highest manganese fluoride and highest manganese oxide. The ions with \(|x|\) number of unpaired electrons from the following are:
A. \(Sc^{3+}\)
B. \(Zn^{2+}\)
C. \(V^{2+}\)
D. \(Fe^{2+}\)
E. \(Co^{2+}\)
Choose the correct answer from the options given below:
Consider \(|x|\) is the difference in oxidation states of Mn in highest manganese fluoride and highest manganese oxide. The ions with \(|x|\) number of unpaired electrons from the following are:
A. \(Sc^{3+}\)
B. \(Zn^{2+}\)
C. \(V^{2+}\)
D. \(Fe^{2+}\)
E. \(Co^{2+}\)
Choose the correct answer from the options given below:
A. \(Sc^{3+}\)
B. \(Zn^{2+}\)
C. \(V^{2+}\)
D. \(Fe^{2+}\)
E. \(Co^{2+}\)
Choose the correct answer from the options given below:
Updated On: Jun 6, 2026
- A and B Only
- C, D and E Only
- C and E Only
- B and E Only
Show Solution
The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
We first need to identify the highest oxidation states of Manganese (Mn) when combined with Fluorine and Oxygen. Due to oxygen's ability to form multiple pi-bonds, it can stabilize a higher oxidation state than fluorine. After finding the difference $|x|$, we determine the electron configurations of the given transition metal ions to count their unpaired electrons.
Step 2: Key Formula or Approach:
Highest Mn Fluoride: $MnF_4$ (Oxidation state = +4).
Highest Mn Oxide: $Mn_2O_7$ (Oxidation state = +7).
Unpaired electrons = number of electrons in singly occupied d-orbitals.
Step 3: Detailed Explanation:
The highest known fluoride of manganese is $MnF_4$, where Mn is in the $+4$ oxidation state.
The highest known oxide of manganese is $Mn_2O_7$, where Mn is in the $+7$ oxidation state.
The difference is $|x| = |4 - 7| = |-3| = 3$.
So we are looking for ions with exactly 3 unpaired electrons.
Let's write the electronic configurations for the given ions:
A. $Sc^{3+}$: Neutral Sc is $[Ar] 3d^1 4s^2$. $Sc^{3+}$ is $[Ar] 3d^0$. (0 unpaired electrons)
B. $Zn^{2+}$: Neutral Zn is $[Ar] 3d^{10} 4s^2$. $Zn^{2+}$ is $[Ar] 3d^{10}$. (0 unpaired electrons)
C. $V^{2+}$: Neutral V is $[Ar] 3d^3 4s^2$. $V^{2+}$ is $[Ar] 3d^3$. (3 unpaired electrons)
D. $Fe^{2+}$: Neutral Fe is $[Ar] 3d^6 4s^2$. $Fe^{2+}$ is $[Ar] 3d^6$. (4 unpaired electrons)
E. $Co^{2+}$: Neutral Co is $[Ar] 3d^7 4s^2$. $Co^{2+}$ is $[Ar] 3d^7$. (3 unpaired electrons)
The ions containing exactly 3 unpaired electrons are $V^{2+}$ and $Co^{2+}$ (C and E).
Step 4: Final Answer:
Options C and E Only are correct.
We first need to identify the highest oxidation states of Manganese (Mn) when combined with Fluorine and Oxygen. Due to oxygen's ability to form multiple pi-bonds, it can stabilize a higher oxidation state than fluorine. After finding the difference $|x|$, we determine the electron configurations of the given transition metal ions to count their unpaired electrons.
Step 2: Key Formula or Approach:
Highest Mn Fluoride: $MnF_4$ (Oxidation state = +4).
Highest Mn Oxide: $Mn_2O_7$ (Oxidation state = +7).
Unpaired electrons = number of electrons in singly occupied d-orbitals.
Step 3: Detailed Explanation:
The highest known fluoride of manganese is $MnF_4$, where Mn is in the $+4$ oxidation state.
The highest known oxide of manganese is $Mn_2O_7$, where Mn is in the $+7$ oxidation state.
The difference is $|x| = |4 - 7| = |-3| = 3$.
So we are looking for ions with exactly 3 unpaired electrons.
Let's write the electronic configurations for the given ions:
A. $Sc^{3+}$: Neutral Sc is $[Ar] 3d^1 4s^2$. $Sc^{3+}$ is $[Ar] 3d^0$. (0 unpaired electrons)
B. $Zn^{2+}$: Neutral Zn is $[Ar] 3d^{10} 4s^2$. $Zn^{2+}$ is $[Ar] 3d^{10}$. (0 unpaired electrons)
C. $V^{2+}$: Neutral V is $[Ar] 3d^3 4s^2$. $V^{2+}$ is $[Ar] 3d^3$. (3 unpaired electrons)
D. $Fe^{2+}$: Neutral Fe is $[Ar] 3d^6 4s^2$. $Fe^{2+}$ is $[Ar] 3d^6$. (4 unpaired electrons)
E. $Co^{2+}$: Neutral Co is $[Ar] 3d^7 4s^2$. $Co^{2+}$ is $[Ar] 3d^7$. (3 unpaired electrons)
The ions containing exactly 3 unpaired electrons are $V^{2+}$ and $Co^{2+}$ (C and E).
Step 4: Final Answer:
Options C and E Only are correct.
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