Question

\(BeO\) reacts with \(HF\) in presence of ammonia to give \([A]\) which on thermal decomposition produces \([B]\) and ammonium fluoride. Oxidation state of Be in \([A]\) is _______

Answer 1

Correct Answer: 2

Beryllium oxide (\(BeO\)) reacts with hydrofluoric acid (\(HF\)) in the presence of ammonia (\(NH_3\)) to form compound \([A]\). The chemical reaction can be represented as follows: \(BeO + 2HF + NH_3 → BeF_2(NH_3)_2 + H_2O\). Here, the complex \([A]\) is \(BeF_2(NH_3)_2\).

We need to determine the oxidation state of beryllium (Be) in this compound. In the complex \(BeF_2(NH_3)_2\):

• Fluorine (F) typically has an oxidation state of -1.
• Ammonia (NH₃) is a neutral ligand, contributing 0 to the oxidation state.

Calculate the oxidation state of beryllium (Be):\(x + 2(-1) + 2(0) = 0\), where \(x\) is the oxidation state of Be.

Simplifying gives: \(x - 2 = 0\).

Thus, \(x = +2\).

The oxidation state of beryllium in \([A]\) is +2.

This result falls within the specified range of 2,2.