Question

The planar shape is of \[N{\left( {Si{H_3}} \right)_3}\]explained by the:
This question has multiple correct options.
A. Type of hybrid orbitals of nitrogen.
B. Additional $d\pi - p\pi $ overlap along the N-Si bond.
C. Higher electronegativity of nitrogen.
D. Higher electronegativity of silicon.

Answer 1

Hint: Let us discuss some of the properties of silicon, it is not quite electronegative its electronegative value is 1.8 and it is larger in size which makes it easier to attack. Silicon has low energy d orbitals which may be used to form an intermediate compound, thus lowering the activation energy of the process.

Complete step by step answer:
Now let us know the compound, \[N{\left( {Si{H_3}} \right)_3}\]( trisilylamine) the shape of the molecule should be pyramidal since it contains 3 \[\left( {Si{H_3}} \right)\] lone pair of electrons that are present on Nitrogen and vacant d-orbitals in \[\left( {Si{H_3}} \right)\]. Nitrogen donates a lone pair of electrons to one of the $SiH_3$ which leads to the formation of double bond, the nitrogen atom has a sp3 hybridization, 3 of the hybrid orbitals participate in the bond formation.
In trisilylamine three \[s{p^2}\] orbital’s are used for sigma bonding, giving a plane triangular structure. The lone pair of electrons occupy a p orbital at the right angles to the plane triangle. This overlaps over the empty d orbitals on each of three silicon atoms, and results in π bonding, more accurately described as pπ-dπ bonding, because it is from p orbital to an empty d orbital. This shortens the bond lengths N-Si, since the nitrogen no longer has a lone pair of electrons, the molecule has no donor properties.
This type formation of double bond we call it as back bonding between the 2p orbital of nitrogen atom and the empty 3d orbital of Si that is $d\pi - p\pi $ (N-Si). Due to the formation of double bond (back bond) the shape of the molecule changes from pyramidal to trigonal planar. Now the electrons are not available for donation for making bonds with other atoms, as Nitrogen has donated the electron within the molecule and hence it is basic in nature.
Hence we can conclude from the above definition that option A and B are correct.
Higher electronegativity of nitrogen and silicon are not responsible for planarity of the molecule

Hence the option C & D are incorrect.

Note:
\[N{\left( {Si{H_3}} \right)_3}\] tends to form a planar triangular geometrical structure, whereas forms a structure with pyramidal geometry. This is because there is the presence of back bonding in between the 2p orbital of the Nitrogen atom and the empty 3d orbital of Si. The shape of the molecule depends on the lone pair and bonded pair of electrons, if a bonded pair of electrons are present then the shape of the molecule is a regular geometry but if lone pair of electrons are present then the shape of the molecule is an irregular geometry.