Answer the question on basis of passage given below:

P block elements are placed in groups 13 to 18 of the periodic table. Their valence shell electronic configuration is \(ns^2\, \ np^{1-6}\). Group 16 of of the p-block elements are known as the group of chalcogens having \(ns^2\, \ np^4\) as their general electronic configuration. They exhibit number of oxidation states but the stability of \(-2\) oxidation state decreases down the group. They are sometimes also known as group of chalcogens as the name is derived from the Greek word for brass and points to the association of sulphur and its congeners with copper. Their ionization enthalpy decreases down the group. Oxygen shows anomalous behaviour due to its small size and high electronegativity.

What will be the products formed in the following equation?

\(U (s) + 3ClF_3 (l) \longrightarrow _{------} + _{------}\)

\(UF_6(g)\) and \(3ClF (g)\)

The correct answer is option 2. \(UF_6(g)\) and \(3ClF (g)\).

Let us break down the reaction between uranium (U) and chlorine trifluoride (ClF₃) in detail.

Uranium is a transition metal that can form several compounds with different oxidation states. Common fluorides include uranium tetrafluoride (\(UF_4\)) and uranium hexafluoride (\(UF_6\)). The oxidation states of uranium in these compounds are +4 and +6, respectively. Chlorine trifluoride is a strong fluorinating agent where chlorine is in the +3 oxidation state. It reacts readily with various elements and compounds, particularly those that can form fluorides.

Reaction Analysis

Uranium (U) can react with chlorine trifluoride (ClF₃) to form uranium fluorides. The fluorine atoms from ClF₃ will be incorporated into the uranium fluoride compounds.

Possible Products:

Uranium Hexafluoride (\(UF_6\)): Uranium in the +6 oxidation state reacts with fluorine to form uranium hexafluoride. This is the most stable and common fluoride of uranium.

Chlorine Fluoride (\(ClF\)): Chlorine trifluoride (ClF₃) can be reduced to chlorine fluoride (ClF) in the reaction. For each ClF₃ molecule, one chlorine fluoride (ClF) molecule is typically produced.

Balancing the Reaction: To balance the reaction, you need to ensure that both the number of atoms of each element and the charge are balanced on both sides of the equation.

Detailed Balancing Steps

Determine the oxidation states: Uranium (U) starts with an oxidation state of 0. Chlorine in ClF₃ has an oxidation state of +3.

General reaction:

\(U + \text{ClF}_3 \rightarrow UF_m + \text{ClF} \)

The number of fluorine atoms required: Uranium hexafluoride (\(UF_6\)) requires 6 fluorine atoms. Each ClF₃ molecule provides 3 fluorine atoms. Therefore, 3 ClF₃ molecules will provide the 6 fluorine atoms needed for 1 UF_6 molecule.

Balancing the reaction: The formation of 1 UF_6 from 3 ClF₃ will consume all the fluorine atoms and leave behind chlorine fluoride (ClF) as the by-product.

The balanced reaction is:

\(U + 3 \text{ClF}_3 \rightarrow UF_6 + 3 \text{ClF} \)

The correct products formed in the reaction between uranium and chlorine trifluoride are: 2. \(UF_6(g)\) and \(3ClF (g)\)

This reaction results in the formation of uranium hexafluoride and chlorine fluoride, reflecting the stoichiometric balance of the reactants and products.