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BrF3 is a liquid which considerably unde...

`BrF_3` is a liquid which considerably undergoes self ionization to form cationic and anionic species.Based on VSEPR theory, number of 90 degree F-Br-F bond angles is ……. In anionic species.
`2BrF_3 hArr [BrF_2]^+ + [BrF_4]^(-)`

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To solve the problem regarding the number of 90-degree F-Br-F bond angles in the anionic species \( \text{BrF}_4^- \), we can follow these steps: ### Step 1: Understand the self-ionization of \( \text{BrF}_3 \) The self-ionization of \( \text{BrF}_3 \) can be represented as: \[ 2 \text{BrF}_3 \rightleftharpoons [\text{BrF}_2]^+ + [\text{BrF}_4]^- \] This indicates that \( \text{BrF}_3 \) can lose a fluorine atom to form a cationic species \( \text{BrF}_2^+ \) and an anionic species \( \text{BrF}_4^- \). ### Step 2: Determine the electron configuration of bromine in \( \text{BrF}_4^- \) Bromine (Br) has 7 valence electrons. In \( \text{BrF}_4^- \): - 4 electrons are used to form bonds with 4 fluorine atoms. - The remaining 3 electrons will include 2 lone pairs and 1 additional electron due to the negative charge. ### Step 3: Identify the molecular geometry using VSEPR theory According to VSEPR theory, the arrangement of electron pairs around the central atom (Br) determines the molecular geometry. - We have 4 bond pairs (from the 4 Br-F bonds) and 2 lone pairs. - The total number of electron pairs is 6 (4 bond pairs + 2 lone pairs). ### Step 4: Determine the hybridization and shape The hybridization for 6 electron pairs is \( \text{sp}^3\text{d}^2 \), which corresponds to an octahedral arrangement. However, with 2 lone pairs, the molecular shape becomes square planar. ### Step 5: Calculate the bond angles In a square planar geometry: - The bond angles between the fluorine atoms around the bromine atom are 90 degrees. ### Step 6: Count the number of 90-degree angles In \( \text{BrF}_4^- \): - There are 4 F-Br-F bond angles, and all of them are 90 degrees. ### Final Answer Thus, the number of 90-degree F-Br-F bond angles in the anionic species \( \text{BrF}_4^- \) is **4**. ---

To solve the problem regarding the number of 90-degree F-Br-F bond angles in the anionic species \( \text{BrF}_4^- \), we can follow these steps: ### Step 1: Understand the self-ionization of \( \text{BrF}_3 \) The self-ionization of \( \text{BrF}_3 \) can be represented as: \[ 2 \text{BrF}_3 \rightleftharpoons [\text{BrF}_2]^+ + [\text{BrF}_4]^- \] This indicates that \( \text{BrF}_3 \) can lose a fluorine atom to form a cationic species \( \text{BrF}_2^+ \) and an anionic species \( \text{BrF}_4^- \). ...
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in an ionic bond the cation tends to polarise the electron cloud of the anion by pulling electron density towards itself This causes development of covalent character in the ionic bond because the electron density gets localised in between the nuclei The tendency of the cation to bring about the polarisation of the anion is expressed as its polasising power The ability of ion to undergo polarisation is called its polarisability The polarising power of a cation or an anion is decided on the basis of F ajans' rules as follows . (i) The smaller the cation, the higher is its polarising power (ii) Cations with pseudo-noble gas configuration (ns^(2)np^(6)nd^(10)) have relatively high polarising power than those with noble gas configuration (ns^(2) np^(6)) (iii) The larger the size of the anion, the higher is its polarisability Choose the correct order of polarisability for the following I^(Θ),Br^(Θ),CI^(Θ)F^(Θ) .

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