A group 15 element forms \( d\pi - d\pi \) bond with transition metals. It also forms a hydride, which is the strongest base among the hydrides of other group members that form \( d\pi - d\pi \) bonds. The atomic number of the element is …….
A group 15 element forms \( d\pi - d\pi \) bond with transition metals. It also forms a hydride, which is the strongest base among the hydrides of other group members that form \( d\pi - d\pi \) bonds. The atomic number of the element is …….
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Solution and Explanation
The element must be from group 15 and form \( d\pi - d\pi \) bonds.
Possible candidates: N, P, As. - Phosphorus (\( P \)) forms \( d\pi - d\pi \) bonds.
- \( PH_3 \) is a strong base compared to \( AsH_3 \).
Atomic number of Phosphorus = 15.
Top Questions on Hydrides
- Consider the following statements:
Statement-I: H$_2$Se is more acidic than H$_2$Te.
Statement-II: H$_2$Se has higher bond dissociation enthalpy than H$_2$Te.
In light of the above statements, choose the correct option: - Identify the hydride which is not correctly matched with the example given in brackets.
- Which of the following reactions give $\mathrm{H_2}$ as one of the products? (Reactions are not balanced.)
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- Statement-I : Among \(15^{th}\) group hydrides reducing character decreases from
\(NH_3\) to \(BiH_3\).
Statement-II : \(E_2O_3\) and \(E_2O_5\) are always basic. [Where \(E\) is group \(15\) element]
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