On the basis of VSEPR theory, discuss the geometry of the following covalent molecules: (i) BeF2 (ii) BF3 (iii) CH4.
(i) BeF2 : The electronic configuration of Be (Z = 4) is1s22s2. The central atom Be has 2 valence electrons. These two electrons are shared mutually with the electrons of two fluorine atoms to form two Be – F bonds.
To have a minimum repulsion in the electron pairs around the central atom, the geometry of the molecules is regular as well as linear. Bond angle in such cases is 180°.
(iii) BF3: The electronic configuration of B (Z = 5) is 1s2 2s2 2p1x. The central atom B has 3 valence electrons. These three valence electrons are shared mutually with the electrons of the fluorine atoms to form three B - F bonds.
Thus, B atom is surrounded by three bond pairs. These repel each other and go as far apart as possible so that there are no further repulsions. This is so if the three electron pairs are placed at 120° with respect to each other i.e. the most favourable arrangement is the triangular planar geometry.
(iii) CH4 The electronic configuration of C(Z =6) is The central atom has 4 valence electrons. These four valence electrons are shared mutually with the electrons of four hydrogen atoms to form four C - H bonds as shown.
Thus carbon atom is surrounded by four bond pairs. In order to minimise the inter-electron pair repulsions i.e. having a state of minimum enthalpy and hence maximum stability, these bond pairs should be as far apart from one another as possible. This is so if the four electron pairs are placed at an angle of 109.5° with respect to each other i.e. the most favourable arrangement is tetrahedral geometry. In other words, carbon is present in the centre of a regular tetrahedron and four bond pairs are directed to the four corners with bond angle-critical to 109.5°.
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