BrF Reaction Scheme Analysis

The question describes a reaction scheme starting with BrF undergoing disproportionation, followed by the self-ionization of one of the products, X. We need to identify X, and the ions Y and Z formed from its self-ionization, and then evaluate the given statements about their properties.

BrF Disproportionation Reaction

BrF undergoes disproportionation, meaning the bromine atom is simultaneously oxidized and reduced. The reaction given is:

\(\text{BrF} \stackrel{\text { disproportionation }}{\longrightarrow} \text{Br}_2\text{(g)} + \text{X}\)

In BrF, the oxidation state of Br is +1 (assuming F is -1). Br\(_2\) has Br in the 0 oxidation state, representing a reduction of Br. Therefore, in product X, Br must be in a higher oxidation state, representing oxidation.

The product X is stated to have a T-shaped geometry. Common molecules with T-shaped geometry are AB\(_3\)E\(_2\) type (like ClF\(_3\)). Let's consider possible oxidation products of BrF involving fluorine. Br can have oxidation states +3, +5, +7 with fluorine. Possible products are BrF\(_3\), BrF\(_5\), BrF\(_7\).

• BrF\(_3\): Br is in +3 oxidation state. For BrF\(_3\), the central Br atom has 7 valence electrons. It forms 3 single bonds with F atoms, using 3 electrons. This leaves \(7 - 3 = 4\) electrons, forming 2 lone pairs. So, the central Br has 3 bonding pairs and 2 lone pairs. According to VSEPR theory, 5 electron domains (3 bonding + 2 lone pairs) correspond to a trigonal bipyramidal electron geometry. With 2 lone pairs occupying equatorial positions, the molecular geometry is T-shaped. This matches the description of X.
• BrF\(_5\): Br is in +5 oxidation state. Central Br has 7 valence electrons. It forms 5 bonds with F atoms, using 5 electrons. This leaves \(7 - 5 = 2\) electrons, forming 1 lone pair. 6 electron domains (5 bonding + 1 lone pair) correspond to an octahedral electron geometry. With 1 lone pair, the molecular geometry is square pyramidal. This is not T-shaped.
• BrF\(_7\): Br is in +7 oxidation state. Central Br has 7 valence electrons, forming 7 bonds with F atoms, using all 7 electrons. 0 lone pairs. 7 electron domains correspond to a pentagonal bipyramidal geometry. This is not T-shaped.

Therefore, X is BrF\(_3\).

Self-Ionization of X (BrF\(_3\))

BrF\(_3\) undergoes self-ionization. This process typically involves the transfer of fluoride ions (F\(\minus\)). BrF\(_3\) can act as both a fluoride ion acceptor (Lewis acid) and a fluoride ion donor (Lewis base). The self-ionization equilibrium is:

\(\text{2BrF}_3 \rightleftharpoons \text{BrF}_2\(\plus\) + \text{BrF}_4\(\minus\)

Y is the cation and Z is the anion. So, Y is BrF\(_2\)\(\plus\) and Z is BrF\(_4\)\(\minus\).

Evaluating the Statements

Now let's analyze each statement based on X = BrF\(_3\), Y = BrF\(_2\)\(\plus\), and Z = BrF\(_4\)\(\minus\).

Statement A: X, Y, and Z have the same number of lone pairs of electrons.

We interpret this as the number of lone pairs on the central bromine atom.

• X (BrF\(_3\)): Central Br has 7 valence electrons. It forms 3 single bonds. Remaining electrons \(7 - 3 = 4\), which is 2 lone pairs.
• Y (BrF\(_2\)\(\plus\)): Central Br has 7 valence electrons, but the ion has a +1 charge, so we consider \(7 - 1 = 6\) electrons. It forms 2 single bonds. Remaining electrons \(6 - 2 = 4\), which is 2 lone pairs.
• Z (BrF\(_4\)\(\minus\)): Central Br has 7 valence electrons, but the ion has a -1 charge, so we consider \(7 + 1 = 8\) electrons. It forms 4 single bonds. Remaining electrons \(8 - 4 = 4\), which is 2 lone pairs.

All three species (on the central atom) have 2 lone pairs. Therefore, statement A is TRUE.

Statement B: Y has a bent shape.

Y is BrF\(_2\)\(\plus\). The central Br has 2 bonding pairs (to F) and 2 lone pairs. Total electron domains = 2 + 2 = 4. According to VSEPR theory, 4 electron domains lead to a tetrahedral electron geometry. With 2 bonding pairs and 2 lone pairs, the molecular geometry is bent. Therefore, statement B is TRUE.

Statement C: Z is sp\(\textsuperscript{3}\) hybridized and has a tetrahedral shape.

Z is BrF\(_4\)\(\minus\). The central Br has 4 bonding pairs (to F) and 2 lone pairs. Total electron domains = 4 + 2 = 6. According to VSEPR theory, 6 electron domains lead to an octahedral electron geometry. Hybridization for 6 electron domains is sp\(\textsuperscript{3}\)d\(\textsuperscript{2}\). With 4 bonding pairs and 2 lone pairs, the molecular geometry is square planar. Therefore, Z is sp\(\textsuperscript{3}\)d\(\textsuperscript{2}\) hybridized and has a square planar shape. Statement C is FALSE.

Statement D: X is used as a non-aqueous solvent.

X is BrF\(_3\). Interhalogen compounds like BrF\(_3\) and ClF\(_3\) are known to be used as non-aqueous ionizing solvents, similar to water, due to their ability to undergo autoionization and dissolve many substances, especially covalent fluorides. BrF\(_3\) is indeed used as a non-aqueous solvent. Therefore, statement D is TRUE.

Conclusion on Statements

Based on our analysis:

• Statement A is TRUE.
• Statement B is TRUE.
• Statement C is FALSE.
• Statement D is TRUE.

The correct statements are A, B, and D.