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

Hint: 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.