Chemistry - CHEMICAL BONDING Question with Solution | TestHub

The question asks to select the correct statement(s) for the compound $M(E{H}_3{)}_3$. This compound is an amine derivative where M is a central atom and E is a Group 14 or 15 element. The geometry around the E atom (e.g., N, P, Si, Ge) is crucial.

Let's analyze each option:

A. If $M=Nitrogen$ & $E=Carbon$. Geometry is planar at nitrogen.

This option is incorrect. If $M=Nitrogen$ and $E=Carbon$, the compound would be $N(C{H}_3{)}_3$ (trimethylamine). Nitrogen in trimethylamine has three bond pairs and one lone pair, leading to a trigonal pyramidal geometry around nitrogen, not planar.

B. If $M=Phosphorous$ & $E=Si$. Geometry at phosphorous is pyramidal.

This option is correct. If $M=Phosphorous$ and $E=Si$, the compound is $P(Si{H}_3{)}_3$ (trisilyphosphine). Phosphorous in $P(Si{H}_3{)}_3$ has three bond pairs and one lone pair, resulting in a trigonal pyramidal geometry around phosphorous. The lone pair on P is delocalized into the empty d-orbitals of Si, but the overall geometry remains pyramidal.

C. If $M=Nitrogen$ & $E=Germanium$. Geometry at nitrogen is planar.

This option is correct. If $M=Nitrogen$ and $E=Germanium$, the compound is $N(Ge{H}_3{)}_3$ (trigermylamine). Similar to trisilylamine ($N(Si{H}_3{)}_3$), the nitrogen atom in $N(Ge{H}_3{)}_3$ exhibits $s{p}^2$ hybridization due to the delocalization of its lone pair into the empty d-orbitals of germanium. This leads to a planar geometry around nitrogen.

D. If $M=Nitrogen$ & $E=Si$. Geometry at nitrogen is planar.

This option is correct. If $M=Nitrogen$ and $E=Si$, the compound is $N(Si{H}_3{)}_3$ (trisilylamine). The nitrogen atom in trisilylamine is $s{p}^2$ hybridized, and its lone pair is delocalized into the vacant d-orbitals of silicon. This results in a trigonal planar geometry around the nitrogen atom.

The final answer is $\boxed{B,C,D}$.