Question

The geometry of \[N{{(C{{H}_{3}})}_{3}}\] and \[N{{(Si{{H}_{3}})}_{3}}\] respectively, is
A. linear, planar
B. pyramidal, pyramidal
C. planar, pyramidal
D. pyramidal, planar

Answer 1

Hint: Hybridization can be defined as the mixing of two atomic orbitals which have the same energy levels and by this mixing they provided a degenerated new type of orbitals and the mixing is based on quantum mechanics.

Complete answer:
Hybridization concept is mainly used to predict the geometry of the compound, hence to find out the geometry first we have to go through the hybridization of that compound and on the basis of mixing hybridization can be classified in 6 categories which can be shown as: \[sp,s{{p}^{2}},s{{p}^{3}},s{{p}^{3}}d,s{{p}^{3}}{{d}^{2}},s{{p}^{3}}{{d}^{3}}\].
In the given question \[N{{(C{{H}_{3}})}_{3}}\] known by the name trimethyl amine in which nitrogen atom represented by \[N\] is \[s{{p}^{3}}\] hybridized and there is no pi-bonding in this compound because carbon atom present has no low-lying d-orbitals and as a result it gives the shape known as pyramidal geometry. On the other hand the compound \[N{{(Si{{H}_{3}})}_{3}}\] known by the name trisilyl amine in which \[N\] atom is \[s{{p}^{2}}\] hybridized and in this compound \[p\pi -d\pi \] bonding is present because \[Si\] has vacant d-orbitals. In this case filled \[2p\] orbitals of \[N\] overlap with the vacant \[3d\] orbitals of \[Si\] to form \[p\pi -d\pi \] and as a result it produces planar geometry.
Hence we can say that geometry of \[N{{(C{{H}_{3}})}_{3}}\] is pyramidal and of \[N{{(Si{{H}_{3}})}_{3}}\] is planar.

Therefore, option D is the correct answer.

Note:
Hybridization can also be defined as the redistribution of the energy of orbitals of individual atoms to give orbitals of equivalent energy happens when two atomic orbitals combine to form hybrid orbital in a molecule and the new orbitals formed in this process are known as hybrid orbitals.