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Current Question (ID: 8658)

Question:

$\text{Ionic hydrides are formed by:}$

Options:

1. $\text{Transition metals.}$
2. $\text{Elements of very high electropositivity.}$
3. $\text{Elements of very low electropositivity.}$
4. $\text{Metalloids.}$

Solution:

$\text{Hint: Compounds form between hydrogen and the most active metals, especially with the alkali and alkaline-earth metals of group one and two elements.}$ $\text{Explanation:}$ $\text{I group (alkali metals) and II group (alkaline earth metals except for Be, Mg) element forms ionic hydrides.}$ $\text{The I group elements and II group elements are electropositive in nature.}$ $\text{So they form ionic hydrides. The ionic hydride generates } \text{H}^-\text{.}$ $\text{Transition metals form metallic hydrides or Interstitial hydrides. For example, } \text{TiH}_2$ $\text{Elements of very low electropositivity form covalent hydride.}$ $\text{For example, some covalent hydrides are } \text{NH}_3\text{, } \text{H}_2\text{O}\text{, } \text{H}_2\text{O}_2 \text{ and } \text{HF}\text{.}$

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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
}