Import Question JSON

Current Question (ID: 18794)

Question:

$\text{The correct statement(s) about transition elements is/are:}$

Options:

1. $\text{The lowest oxide of a transition metal is basic, the highest is amphoteric/acidic.}$
2. $\text{A transition metal exhibits the highest oxidation state in oxides and fluorides.}$
3. $\text{The highest oxidation state is exhibited in the oxoanion of a metal.}$
4. $\text{All of the above.}$

Solution:

$\text{HINT: Oxygen and fluorine act as strong oxidizing agents because of their high electronegativities and small sizes.}$ $\text{Explanation:}$ $\text{(i) In the case of a lower oxide of a transition metal, the metal atom has a low oxidation state. This means that some of the valence electrons of the metal atom are not involved in bonding. As a result, it can donate electrons and behave as a base.}$ $\text{On the other hand, in the case of a higher oxide of a transition metal, the metal atom has a high oxidation state.}$ $\text{This means that the valence electrons are involved in bonding and so, they are unavailable. There is also a high effective nuclear charge. As a result, it can accept electrons and behave as an acid.}$ $\text{For example, MnO is basic and Mn}_2\text{O}_7 \text{ is acidic.}$ $\text{(ii) Oxygen and fluorine act as strong oxidizing agents because of their high electronegativities and small sizes. Hence, they bring out the highest oxidation states from the transition metals.}$ $\text{In other words, a transition metal exhibits higher oxidation states in oxides and fluorides. For example, in OsF}_6 \text{ and V}_2\text{O}_5, \text{ the oxidation states of OS and V are +6 and +5 respectively.}$ $\text{(iii) Oxygen is a strong oxidizing agent due to its high electronegativity and small size.}$ $\text{So, oxo-anions of a metal have the highest oxidation state. For example in MnO}_4^-, \text{ the oxidation state of Mn is +7.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}