Chemical bonding 3 Flashcards
Cu+ and Na+ are of same size but CuCl is insoluble while NaCl is soluble in water. Explain.
In NaCl, the hydration energy is sufficient to overcome the lattice energy of ionic NaCl. Hence NaCl is soluble in water. The high hydration energy of NaCl is because of relatively high ionic character of NaCl with the result that water dipoles interact strongly with Na+ and Cl- ions.
On the other hand, CuCl has sufficiently covalent character because of greater polarization of Cu+ ion (refer Q.4). As a result of covalent character, its hydration energy is low and is not sufficient to overcome the lattice energy of CuCl. Consequently, it is insoluble in water.
Account for the linear shape of I 3-ion.
The central I atom has the ground state electronic configuration as 5s2 5p5. It involves sp3d hybridization involving one 5s, three 5p and one vacant 5d orbital. The half filled sp3d hybrid orbital forms a covalent bond with one I atom and an empty sp3d hybrid orbital accepts an electron pair from I-ion to form a coordinate bond. Thus, the geometry of the molecule is trigonal bipyramidal in which three positions are occupied by lone pairs.
XeF2, molecule is linear molecule but it is sp3d hybridized. Why?
Due to sp3d hybridisation, it has trigonal bipyramidal geometry in which three positions are occupied by lone pairs of electrons. The lone pairs are present in equatorial positions and their resultant effect is zero. As a result, the bonds are formed by axial Xe-F bonds only which are at an angle of 180° to each other.
BeF2, molecule is linear while SF, is angular though both are triatomic.
In BeF2, Be is surrounded only by 2 electron bond pairs and therefore, the molecule is linear. On the other hand, in SF2, the central atom S, is surrounded by 2 bond pairs and 2 lone pairs. Since it is surrounded by 4 electron pairs, the geometry is expected to be tetrahedral with two positions occupied by lone pairs.
Why does PCI5, exist as [PC1]+ [PCl ̧] ̄ in the crystalline state?
PCI5 has unsymmetrical structure because of different axial and equatorial bonds. Therefore, it is unstable and it changes to tetrahedral [PCl]+ and octahedral [PCl]- ions, which have stable geometries.
Explain the geometry of H2O molecule based on VSEPR theory.
The O atom in H2O has 6 V.E. 2 of them are shared with 2 hydrogen atoms. There are 4 electron pairs around oxygen atom, 2 bond pairs and 2 lone pairs. The arrangement of this molecule is tetrahedral but due to the presence of 2 lone pairs on oxygen, it causes distortion in the geometry. The lone pair exerts greater repulsion on bond pair, so the bond angle decreases from 109.5 to 104.5 the shape changes to bent shape.
Explain the geometry of ClF3 molecule based on VSEPR theory.
In chlorine trifluoride, central atom is chlorine and valence electrons on central atom is 7. Also, contribution of three fluorine atoms is 1 electron each. Therefore, there are total 10 electrons or five electron pairs. The high lp-lp repulsion than lp-bp and bp-bp causes distortion in the geometry.
this results in a trigonal bipyramidal geometry for the shape-determining five electron pairs. Three of these are bond pairs and two are lone pairs. These keep as far apart as possible, minimising repulsion between each of the negatively charged clouds by adopting a trigonal bipyramidal arrangement. The two lone pairs occupy equatorial position at an angle of 120° to each other, this gives the lowest energy arrangement of electron pairs in the molecule. Because repulsion involving lone pairs are stronger than bond pairs. Thus, F-Cl-F angle is a little less than 180°. Therefore, molecule has a T-shape geometry.
Describe the geometry of XeF4
Xenon (Xe) is in Group 8 (or 18), so it has 8 valence electrons. Fluorine (F) has 7 valence electrons each.
XeF4 has 6 fluorine atoms bonded to xenon. Each fluorine atom forms a single bond with xenon, utilizing one electron pair. Therefore, there are 6 electron pairs around xenon.
In XeF4, there are no lone pairs on xenon, only bonding pairs.
In this case, the 6 bonding pairs arrange themselves in an octahedral electron pair geometry, with 90° angles between them.
In the case of XeF4, the lone pair on each fluorine atom occupies more space than the bonding pairs, causing the fluorine atoms to push away from each other. As a result, the molecular geometry of XeF4 is distorted from the ideal octahedral geometry, resulting in a square planar molecular geometry.
Describe the change in hybridisation (if any) of the Al atom in the following reaction.
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AlCl3 + Cl- → AlCl 4-
In AlCl3, Thus, it has three unpaired electrons. Chlorine needs one electron to complete its octet. Three Cl atoms hybridise with Al to undergo sp2 hybridisation to give a trigonal planar arrangement in AlCl3. But in the formation of AlCl 4 the empty 3pz orbital will also be involved in the hybridisation and thus the hybridization changes from sp2 to sp3. As a result, the shape gets changed to tetrahedral.
Is there any changes in the hybridisation of B and N atoms as a result of the following reaction. BF3 + NH3 → F3B - NH3 ?
B atom in BF 3 has 3 bond pairs of electrons and 0 lone pairs of electrons. It is sp 2 hybridised. B atom in
F3B←NH 3
has 4 bond pairs of electrons and 0 lone pairs of electrons. It is sp 3 hybridised.
Hence, the formation of molecular complex F 3B←NH3
results in a change in hybridization of boron from sp 2to sp 3
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