Question

The correct order of bond angle in ${H_2}S,N{H_3},B{F_3}$ and $Si{H_4}$ is:
A. ${H_2}S < N{H_3} < Si{H_4} < B{F_3}$
B. $N{H_3} < {H_2}S < Si{H_4} < B{F_3}$
C. ${H_2}S < Si{H_4} < N{H_3} < B{F_3}$
D. ${H_2}S < N{H_3} < B{F_3} < Si{H_4}$

Answer 1

Hint: We are given to arrange the compounds in ascending order of their bond angles.So after checking their structure on the basis of their hybridization we can know their bond lengths , bond angles.Then we can arrange them in the required order.

Complete step by step answer:
The hybridization , geometry and bond angles of the given molecules are;
In the ${H_2}S$ molecule, two Hydrogen atoms form a bond with the central Sulfur atom. Two single bonds are formed in the molecule. These bonds take up four valence electrons, and hence there are four other valence electrons left. While forming a bond the s orbital of the Hydrogen atom overlaps with p orbital of the Sulfur atom. The lone pair of electrons take up two of the $s{p^3}$ orbitals. The other two orbitals of $s{p^3}$ overlap with 1s orbital of the Hydrogen atom.As S is less electronegative it’s lone pair- lone pair repulsion is less therefore the angle between H and S is \[{92.1^ \circ }\] .

$B{F_3}$ molecule is $s{p^2}$ hybridized as there are three σ bonds and no lone pair. The geometry of the BF 3 molecule is named trigonal planar. The fluorine atoms are positioned at the vertices of an equiangular triangle . The F-B-F angle is 120 and every one four atoms dwell on an equivalent plane.
During the formation of ammonia, one 2s orbital and three 2p orbitals of nitrogen combine to make four hybrid orbitals having equivalent energy which is then considered as an $s{p^3}$ type of hybridization.The molecular geometry of ammonia is trigonal pyramidal or distorted tetrahedral structure. This is mainly due to the presence of a lone non-bonding pair which usually exerts greater repulsion on the bonding orbitals. The bond angle in ammonia is less than the standard, the bond angle is ${107^ \circ }$.
$Si{H_4}$ is $s{p^3}$ hybridized with four σ bonds and zero lone pairs .It has tetrahedral geometry just like $C{H_4}$ . The bond angle is ${109.5^ \circ }$ .
Now the arrangement of the given molecules in increasing order of bond angle is:
So, the correct answer is “Option A”.

Note: When one or more of the groups is a lone pair of electrons (non-bonded electrons), the experimentally-observed geometry around an atom is slightly different than in the case where all groups are bonds. The actual bond angles are similar, but not exactly the same, as those predicted based on the total number of groups (the "parent" geometry).