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Chapter 6: Problem 74

Among the following statements which is INCORRECT: (1) In the preparation of compounds of \(\mathrm{Xe}\), 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 4 is incorrect.

Step by step solution

01

Analyze Statement 1

The statement mentions that \(\mathrm{O}_{2} \mathrm{PtF}_{6}\) was used by Bartlett as a base compound due to the similar ionisation enthalpy of \(\mathrm{O}_{2}\) and \(\mathrm{Xe}\). This is a correct statement based on known chemical principles.
02

Analyze Statement 2

The statement claims that nitrogen does not show allotropy. This is a correct statement as nitrogen exists only as \(\mathrm{N}_{2}\) under normal conditions and does not have different allotropic forms.
03

Analyze Statement 3

This statement discusses the brown ring test for \(\mathrm{NO}_{3}^{-}\) ion and attributes it to the formation of \([\mathrm{Fe}(\mathrm{H}_{2}\mathrm{O})_{5}(\mathrm{NO})]^{2+}\). This is a correct statement, as the brown ring formed in the test is indeed due to the mentioned complex.
04

Analyze Statement 4

The statement suggests that red phosphorus gives \(\mathrm{PH}_{3}\) gas when heated with concentrated \(\mathrm{NaOH}\) solution in an inert atmosphere of \(\mathrm{CO}_{2}\). This is an incorrect statement. The correct reaction conditions require heating white phosphorus with concentrated \(\mathrm{NaOH}\) in an inert atmosphere to produce \(\mathrm{PH}_{3}\). Red phosphorus does not give \(\mathrm{PH}_{3}\) under these conditions.

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Key Concepts

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

Ionisation Enthalpy
Ionisation enthalpy is the energy required to remove an electron from an isolated gaseous atom or ion. It indicates an element's tendency to lose electrons.

Higher ionisation enthalpy means it is harder to remove an electron. Factors affecting ionisation enthalpy:
  • Atomic size: Larger atoms have lower ionisation enthalpy.
  • Nuclear charge: Greater charge increases ionisation enthalpy.
  • Electron shielding: More shielding lowers ionisation enthalpy.
  • Sublevel arrangement: Electron removal from a stable sublevel requires more energy.
In this exercise, \(\text{O}_2\) and \(\text{Xe}\) were compared. Their similar ionisation enthalpies justified using \[ \text{O}_2 \text{PtF}_6 \] for preparing Xe compounds.
Allotropy
Allotropy is the existence of an element in different forms in the same physical state. These forms, called allotropes, possess distinct physical and chemical properties. Notably:
  • Carbon has allotropes like diamond, graphite, and graphene.
  • Oxygen's allotropes include \( \text{O}_2 \) and ozone (\( \text{O}_3 \)).
The statement claims that nitrogen does not exhibit allotropy, which is correct. Nitrogen is diatomic (\( \text{N}_2 \)) and does not exist in different allotropes under normal conditions.
Brown Ring Test
The brown ring test detects nitrates (\( \text{NO}_3^- \)) in a solution. Adding iron(II) sulfate followed by concentrated sulfuric acid forms a brown ring. The ring results from the reaction:

\[ \text{NO}_3^- + 3 \text{FeSO}_4 + 4 \text{H}_2\text{SO}_4 \to [\text{Fe}(\text{H}_2\text{O})_5(\text{NO})]^{2+} + 2 \text{H}_2O + 3 \text{FeSO}_4 \]

The complex \[ \text{Fe}(\text{H}_2\text{O})_5(\text{NO})]^{2+}\] forms and appears as a brown ring, verifying the presence of nitrate ions.
Red Phosphorus and Phosphine Gas
Phosphorus has several allotropes, mainly white and red. White phosphorus reacts with concentrated \(\text{NaOH}\) in an inert atmosphere, producing phosphine gas (\(\text{PH}_3 \)).

The reaction:
\[ 4 \text{P} + 3 \text{KOH} + 3 \text{H}_2O \to \text{PH}_3 + 3 \text{KH}\text{2}\text{PO}_2 \].

However, red phosphorus does not produce phosphine gas under these conditions. This difference is essential in understanding the chemical behavior of phosphorus allotropes and their reaction mechanisms.

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Most popular questions from this chapter

Which of the following ions does not have S-S linkage? (1) \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) (2) \(\mathrm{S}_{2} \mathrm{O}_{6}^{2-}\) (3) \(\mathrm{S}_{2} \mathrm{O}_{5}^{2-}\) (4) \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\)

A source emit sound waves of frequency \(1000 \mathrm{~Hz}\). The source moves to the right with a speed of \(32 \mathrm{~m} / \mathrm{s}\) relative to ground. On the right a reflecting surface moves towards left with a speed of \(64 \mathrm{~m} / \mathrm{s}\) relative to ground. The speed of sound in air is \(332 \mathrm{~m} / \mathrm{s}\), then which of the following options is incorrect : (1) wavelength of sound in ahead of source is \(0.3 \mathrm{~m}\) (2) number of waves arriving per second which meets the reflected surface is 1320 (3) speed of reflected wave is \(268 \mathrm{~m} / \mathrm{s}\) (4) wavelength of reflected waves is nearly \(0.2 \mathrm{~m}\)

Unequal bond lengths are present in : (1) \(\mathrm{BF}_{3}\) (2) \(\mathrm{CO}_{3}{ }^{2-}\) (present in \(\left.\mathrm{CaCO}_{3}\right)\) (3) \(\mathrm{HNO}_{3}\) (4) \(\mathrm{SO}_{4}^{2-}\)

Which one is incorrect statement among the following? (1) \(\mathrm{PH}_{5}, \mathrm{SCl}_{6}\) and \(\mathrm{FCl}_{3}\) do not exist. (2) \(p \pi-d \pi\) bond is present in \(\mathrm{SO}_{2}\) molecule. (3) \(12 \mathrm{P}-\mathrm{O}\) bonds are present in \(\mathrm{P}_{4} \mathrm{O}_{6}\) molecule. (4) Bond angle in \(\mathrm{SiH}_{4}\) less than that in \(\mathrm{CH}_{4}\).

The set of values of \(p\) for which both the roots of the quadratic equation, \(4 x^{2}-20 p x+\left(25 p^{2}+15 p-66\right)=0\) are less than 2 will be (1) \((4 / 5,2)\) (2) \((2, \infty)\) (3) \((-1,4 / 5)\) \((4)(-\infty,-1)\)
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