Question

The bond angle of $H-X-H$is the greatest in the compound:
A. $P{{H}_{3}}$
B. $C{{H}_{4}}$
C. $N{{H}_{3}}$
D. ${{H}_{2}}O$

Answer 1

A bond angle is the angle generated between three atoms across at least two bonds. For four atoms bonded together in a chain, the torsional angle is the angle between the plane generated by the first three atoms and the plane created by the last three atoms.

Complete step by step answer:
->Option 1st: Phosphine is the 2nd row analogue of ammonia. There are 4 regions of electron density around the phosphorus atom, 1 of which is a lone pair. $H-P-H$ bond angles should be $<{{109.5}^{\circ }}$approx. $105-{{107}^{\circ }}$ .
->Option 2nd: In methane, the four hybrid orbitals are located in such a manner so as to decrease the force of repulsion between them. Nonetheless, the four orbitals do repel each other and get placed at the corners of a tetrahedron. $C{{H}_{4}}$has a tetrahedral shape. The $s{{p}^{3}}$hybrid orbitals have a bond angle of${{109.5}^{\circ }}$.
->Option 3rd: The ​ $N{{H}_{3}}$sub-atomic structure (sub-atomic shape) is three-sided pyramidal. When there is one particle in the center, and three others at the corners and all the three atoms are indistinguishable, the sub-atomic calculation accomplishes the state of three-sided pyramidal. Smelling salts have this structure as the Nitrogen has 5 valence electrons and bonds with 3 Hydrogen atoms to finish the octet. The $N{{H}_{3}}$bond point are ${{107}^{\circ }}$(degrees ) in light of the fact that the hydrogen particles are repulsed by the solitary pair of electrons on the Nitrogen molecule.
->Option 4th: In ${{H}_{2}}O$
Sigma bond is 2
Lone pair is 2
Thus hybridization is. $s{{p}^{3}}$ Since it has 2 lone pair so, both the lone pair will repel each other and the bond angle reduces to ${{104.5}^{\circ }}$

Molecule	Hybridization	Bond Angle

1. $P{{H}_{3}}$| $s{{p}^{3}}$| ${{98}^{\circ }}$
2. $C{{H}_{4}}$| $s{{p}^{3}}$| ${{109}^{\circ }}28'$
3. $N{{H}_{3}}$| $s{{p}^{3}}$| ${{107}^{\circ }}$
4. ${{H}_{2}}O$| $s{{p}^{3}}$| ${{104.5}^{\circ }}$

**Hence, the correct option is $C{{H}_{4}}$

Note: **
The bond angle can be separated between linear, three-sided planar, tetrahedral, three-sided bipyramidal, and octahedral. The Ideal bond angles are the angles that show the greatest angles where it would limit repulsion, along these lines confirming the VSEPR hypothesis.