Select correct statement about hydrides of group 15 elements: (1) The central atom in the hydride is sp \(^{2}\) hybriized (2) \(\mathrm{NH}_{3}\) readily form \(\mathrm{NH}_{4}^{+}\) salts with \(\mathrm{H}^{+} ; \mathrm{PH}_{4}^{+}\) saltsare formed with \(\mathrm{H}^{+}\) under anhydrous condition. (3) The tetrahedron is distorted due to repulsion be tween the lone pair of electrons and the bond pairs. (4) The bond energy of the \(\mathrm{M}-\mathrm{H}\) bond decreases from \(\mathrm{NH}_{3}\) to \(\mathrm{BiH}_{3}\) because of increase in the size of the element. (a) \(1,2,3\) (b) \(2,3,4\) (c) 2,4 (d) \(1,3,4\).

Understanding the geometry of molecules is pivotal for students studying chemistry. The concept of hybridization is introduced to explain the shapes and formations of different molecular structures. In the case of Group 15 hydrides, the molecules exhibit 'sp3' hybridization. This means that one s-orbital and three p-orbitals of the central atom combine to form four sp3 hybrid orbitals.

Each sp3 hybrid orbital holds one electron pair, which may be a bonding pair or a lone pair. In ammonia (NH3), for instance, there are three bonding pairs with hydrogen atoms and one lone pair. The shape is a trigonal pyramidal, which is the characteristic shape of sp3 hybridized molecules. This arrangement allows the molecules to minimize electron-pair repulsion, following the VSEPR (Valence Shell Electron Pair Repulsion) theory. Moreover, this concept contradicts Statement 1 in the original exercise, which inaccurately suggests 'sp2' hybridization for such molecules.

The versatility of Group 15 hydrides is also seen in their ability to form additional bonds with hydrogen ions (protons). Particularly, ammonia (NH3), upon accepting a proton (H+), forms the ammonium ion (NH4+). This reaction is fundamental to much of biological chemistry, as it is a key factor in nitrogen metabolism.

Upon accepting a proton, the lone pair electrons of the nitrogen atom in ammonia are donated to the hydrogen ion, creating a new bond. This results in a positively charged ammonium ion with a tetrahedral structure. The formation of this ion is a classic example of a Lewis acid-base reaction. Comparatively, PH3 can also form PH4+ ions, albeit under specific conditions such as the presence of strong acids and lack of water, as stated in Statement 2 of the exercise.

Understanding lone pair and bond pair repulsion is crucial in predicting the shapes of molecules. Lone pair electrons are not shared with another atom and are located closer to the nucleus. As a result, they occupy more space than bonding pairs and repel these bond pairs, causing a distortion in the molecule's shape.

This repulsion is taken into account by the VSEPR model, which helps predict geometries of molecules. In group 15 hydrides, the presence of a lone pair on the central atom results in a trigonal pyramidal geometry instead of a perfect tetrahedral one, affirming Statement 3. This distortion is significant because it affects molecular polarity and, as a consequence, the physical and chemical properties of the substance.

Bond energy is a measure of the strength of a bond between two atoms. In Group 15 hydrides, there is a noticeable trend where bond energy decreases as one moves down the group from NH3 to BiH3. This trend can be attributed to the increasing size of the central atoms as we progress down the group.

With larger atoms, the bond lengths to hydrogen also increase, resulting in weaker M-H bonds due to less efficient orbital overlap. This leads to a decrease in bond energies, which is supported by Statement 4 in the given exercise. Understanding this trend is important as it influences various chemical reactions and processes, including the stability and reactivity of the hydrides in different chemical environments.