Among the following statements which is INCORRECT : (1) In the preparation of compounds of \(X e\), Bartlett had taken \(\mathrm{O}_{2} \mathrm{PtF}_{6}\) as a base compound because both \(\mathrm{O}_{2}\) and Xe have almost same ionisation enthalpy. (2) Nitrogen does not show allotropy. (3) A brown ring is formed in the ring test for \(\mathrm{NO}_{3}^{-}\)ion. It is due to the formation of \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{NO})\right]^{2 *}\) (4) On heating with concentrated \(\mathrm{NaOH}\) solution in an inert atmosphere of \(\mathrm{CO}_{2}\), red phosphorus gives \(\mathrm{PH}_{3}\) gas.

Short Answer

Expert verified
Statement 2 is incorrect.

Step by step solution

01

Analyze Statement 1

Evaluate if it is true that Bartlett used \(\text{O}_2 \text{PtF}_6\) as a base compound for preparing compounds of \(Xe\) due to similar ionization enthalpies between \(O_2\) and \(Xe\). This statement is correct because Bartlett demonstrated that \(Xe\) could form stable compounds by comparing it to the behavior of \(O_2\) in the context of ionization energies.
02

Analyze Statement 2

Check the validity of the statement 'Nitrogen does not show allotropy.' This statement is false because nitrogen does exist in different forms (allotropes), such as diatomic nitrogen (N2) and ionic forms.
03

Analyze Statement 3

Determine if the statement regarding the brown ring test for \(NO_3^-\) ion, indicating the formation of \([Fe(H_2O)_5(NO)]^{2+}\), is correct. This statement is correct because the brown ring formed in this test is due to the formation of the complex \([Fe(H_2O)_5(NO)]^{2+}\).
04

Analyze Statement 4

Judge whether the statement about red phosphorus reacting with concentrated \(NaOH\) in an inert \(CO_2\) atmosphere to produce \(\text{PH}_3\) gas is accurate. This statement is correct because red phosphorus does produce phosphine gas (\(\text{PH}_3\)) under these conditions.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ionization Enthalpy
Ionization enthalpy is the energy required to remove an electron from an atom or ion in its gaseous state. It plays a crucial role in understanding chemical reactivity and the formation of compounds. Higher ionization enthalpy means an atom holds onto its electrons more tightly, making it less likely to form positive ions.
In the context of xenon (Xe) and oxygen (O2), the similarity in their ionization enthalpies allowed chemist Neil Bartlett to hypothesize that xenon could form stable compounds much like oxygen does. He used \(\text{{O}}_2 \text{{PtF}}_6\) to demonstrate this because both elements have closely matched ionization energies. This groundbreaking discovery led to the synthesis of xenon compounds, changing the landscape of chemical bonding theories.
Nitrogen Allotropes
Allotropy refers to the existence of an element in more than one form, each having different physical or chemical properties. Nitrogen (N) exhibits allotropy primarily in the form of diatomic nitrogen (N2) and ionic nitrogen compounds. Diatomic nitrogen (N2) is the most common and stable form, making up about 78% of Earth's atmosphere.
Despite the claim in the exercise, nitrogen does show allotropy. Besides N2, nitrogen can exist in other forms under certain conditions. These allotropes are crucial for various chemical reactions and industrial processes.
Brown Ring Test
The brown ring test is a classic analytical chemistry technique used to detect nitrate ions (NO3-) in a solution. When nitric acid (HNO3) and iron(II) sulfate (FeSO4) are combined, and concentrated sulfuric acid (H2SO4) is carefully added, a brown ring forms at the interface.
The brown ring is due to the formation of the complex ion \(\text{{[Fe(H}}_2\text{{O}})_5(\text{{NO}})]^{2+}\). This test is highly sensitive and provides a visual confirmation of the presence of nitrate ions. It’s a reliable method commonly used in chemistry laboratories.
Phosphorus Reactions
Phosphorus, particularly red phosphorus, is known for its various chemical reactions. One notable reaction is when red phosphorus is heated with concentrated sodium hydroxide (NaOH) in an inert atmosphere of carbon dioxide (CO2). This reaction produces phosphine gas (PH3).
This gas is highly toxic and flammable. The reaction is written as:
\[ \text{4P + 3NaOH + 3H}_2\text{O} \rightarrow \text{PH}_3 + \text{3NaH}_2\text{PO}_2 \]
This reaction is significant in the study of phosphorus chemistry and is often used to demonstrate the properties and reactivity of phosphorus compounds in educational settings.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) are converted into monoanions \(\mathrm{N}_{2}^{-}\)and \(\mathrm{O}_{2}^{-}\)respectively. Which of the following statements is wrong? (1) In \(\mathrm{N}_{2}^{-}\), N-N bond weakens (2) In \(\mathrm{O}_{2}^{-}\), O-O bond order increases (3) In \(\mathrm{O}_{2}^{-}, \mathrm{O}-\mathrm{O}\) bond order decreases (4) \(\mathrm{N}_{2}^{-}\)becomes paramagnetic

Which equilibrium can be described as an acid base reaction using the Lewis acid - base definition but not using the Bronsted- Lowry definition? (1) \(2 \mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightleftharpoons 2 \mathrm{NH}_{4}^{+}+\mathrm{SO}_{4}^{2-}\) (2) \(\mathrm{NH}_{3}+\mathrm{CH}_{3} \mathrm{COOH} ?\mathrm{NH}_{4}{ }^{*}+\mathrm{CH}_{3} \mathrm{COO}^{-}\) (3) \(\mathrm{H}_{2} \mathrm{O}+\mathrm{CH}_{3} \mathrm{COOH} \quad \mathrm{H}_{3} \mathrm{O}++\mathrm{CH}_{3} \mathrm{COO}^{-}\) (4) \(\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{2+}+4 \mathrm{NH}_{3}=\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) \(+4 \mathrm{H}_{2} \mathrm{O}\)

A piston can slide without friction inside a horizontal cylindrical vessel, which contains an ideal monoatomic gas. The piston and the cylinder both are made of a perfect heat insulating material. Initially, the piston is in equilibrium and divides the cylindrical into two part \(\mathrm{A}\) and \(\mathrm{B}\), which are not necessarily equal. The temperatures of the gases in both the parts are equal. Now, the piston is held in its initial position and the gas in part A is supplied some amount of heat, then the piston is released. What will the piston do in its subsequent motion. (1) It will execute oscillatory motion of definite amplitude. (2) It will stop exactly at middle of the cylinder after several oscillations of decreasing amplitude. (3) It will stop exactly where it has started after several oscillations of decreasing amplitude. (4) Certainly, it will execute oscillations of decreasing amplitude but where it will finally stop depends on the amount of heat supplied.

STATEMENT - 1 : If the roots of equation \(a x^{4}+b x^{3}+c x^{2}+d x+e=0\) are in H.P then the roots of the equation \(e x^{4}+d x^{3}+c x^{2}+b x+a=0\) are in A.P. STATEMENT- 2 : The equation having reciprocals of the roots of the equation \(f(x)=0\), as roots is \(f\left(\frac{1}{x}\right)=0\) (1) Statement \(-1\) is True, Statement \(-2\) is True ; Statement \(-2\) is a correct explanation for Statement \(-1\) (2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement- 1 (3) Statement \(-1\) is True, Statement \(-2\) is False (4) Statement \(-1\) is False, Statement \(-2\) is True

For a particle moving along \(x\)-axis, the acceleration a of the particle in terms of its \(x\)-coordinate \(x\) is given by \(a=-9 x\), where \(x\) is in meters and a is in \(\mathrm{m} / \mathrm{s}^{2}\). Take acceleration, velocity and displacement in positive \(x\)-direction as positive. The initial velocity of particle at \(x=0\) is \(u=+6 \mathrm{~m} / \mathrm{s}\). The velocity of particle at \(x=2 \mathrm{~m}\) will be : \((1)+6 \sqrt{2} \mathrm{~m} / \mathrm{s}\) (2) \(-6 \sqrt{2} \mathrm{~m} / \mathrm{s}\) (3) \(72 \mathrm{~m} / \mathrm{s}\) (4) 0

See all solutions

Recommended explanations on English Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free