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Current Question (ID: 9131)

Question:

$\text{An alkali metal that reacts with nitrogen to form nitride is -}$

Options:

1. $\text{Li}$
2. $\text{Na}$
3. $\text{Cs}$
4. $\text{None of the above.}$

Solution:

$\textbf{HINT:} \text{ Li form } \text{Li}_3\text{N} \text{ directly with } \text{N}_2$$ $\textbf{Explanation:}$ $\text{Only Li combines directly with nitrogen to form lithium nitride. The reaction is as follows:}$ $6\text{Li} + \text{N}_2 \rightarrow 2\text{Li}_3\text{N}$ $\text{Li reacts directly with nitrogen to form lithium nitride.}$ $\text{This is because } \text{Li}^+ \text{ is very small in size having high polarising power and so it's size is the most compatible with the } \text{N}^{3-} \text{ ion.}$ $\text{This energy also overcomes the high amount of energy required for the formation of the } \text{N}^{3-} \text{ ion.}$

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
}