Chemical Bonding and Molecular Structure Class Notes PDF

# Chemical Bonding and Molecular Structure ## **Date: / /** ## **Page: / /** ### L.O:- To study about chemical bonds and draw Lewis structures. ## Bond ### Ionic 1. **Ionic bond:** Transfer of a valance electron from one element to another. (Mostly Metal to Non-Metal) ### G.N Lewis Structure 1...

# Chemical Bonding and Molecular Structure ## **Date: / /** ## **Page: / /** ### L.O:- To study about chemical bonds and draw Lewis structures. ## Bond ### Ionic 1. **Ionic bond:** Transfer of a valance electron from one element to another. (Mostly Metal to Non-Metal) ### G.N Lewis Structure 1. **Na-11 2,8,1** > $Na \rightarrow Na^+ + e^-$ > $Cl + e^- \rightarrow Cl^-$ 2. **Mg-12 2,8,2** > $Mg \rightarrow Mg^{2+} + 2e^-$ > $O + 2e^- \rightarrow O^{2-}$ 3. **K-19 2,8,8,1** > $K \rightarrow K^+ + e^-$ > $Cl + e^- \rightarrow Cl^-$ 4. **Ca-20 2,8,8,2** > $Ca \rightarrow Ca^{2+} + 2e^-$ > $2Cl + 2e^- \rightarrow 2Cl^-$ 5. **Na-11 2,8,1** > $Na + Na + O \rightarrow (Na)_2O$ > $O-8 2,6$ ### Covalent **Covalent bond:** Sharing of valance electrons between two atoms to become an octet. #### $Cl_2$ - 2,8,7 > $Cl \: : Cl$ > **Single covalent bond** 1. **$H_2$ - 1** > $ H-H$ 2. **$O_2$ - 2,6** > $O=O$ > **Double covalent bond** 3. **$N_2$ - 2,5** > $N \equiv N$ 4. **$CH_4$** > $C-2,4$ > $H-1$ > $H$ > $H-C-H$ > $H$ 5. **$CHCl_3$** > $C-2,4$ > $H-1$ > $Cl-2,8,7$ > > $Cl$ > $Cl-C-Cl$ > $Cl$ > $H$ ## Lewis Structures 1. **Hydronium ions** - $H_3O^+$ * **Step 1:** Calculation of total valence electrons > $H + 3 + 6 - 1 = 8$ * **Step 2:** Draw the skeletal structure > $H$ > $H-O-H$ > $H$ * **Step 3:** Subtract Bond pair electrons from Total valence electrons > $8 - 6 = 2$ > $H$ > $\ddot{O}$ > $|$ > $H$ > $H$ > $lone \: pair$ 2. **Ammonium ions** - $NH_4^+$ * **Step 1:** $5 + 4 - 1 = 8$ * **Step 2:** > $H$ > $H-N-H$ > $H$ * **Step 3:** $8 - 8 = 0$ > $H$ > $H-N-H$ > $H$ 3. **Carbonate ions** - $CO_3^{2-}$ * **Step 1:** $4 + 18 + 2 = 24$ * **Step 2:** > $O$ > $|\:\:\:\:\:\:\:\:\:\:$ > $O-C-O$ * **Step 3:** $24 - 6 - 18$ > $O$ > $||$ > $O-C-O$ ## **Date: / /** ## **Page: / /** 4. **Nitrate ions** - $NO_3^-$ * **Step 1:** $5 + 18 + 1 = 24$ * **Step 2:** > $O$ > $| \:\:\:\:\:\:\:\:\:\:\:$ > $O-N-O$ * **Step 3:** $24 - 6 = 18$ > $O$ > $| \:\:\:\:\:\:\:\:\:\:\:$ > $O-N-O$ 5. **Nitrite ions** - $NO_2^-$ * **Step 1:** $5 + 12 + 1 = 18$ * **Step 2:** $O-N-O$ * **Step 3:** $18 - 4 = 14$ > $O$ > $| \:\:\:\:\:\:\:\:\:\:\:$ > $O-N-O$ 6. **Ozone ion** - $O_3$ * **Step 1:** $6 x 3 = 18$ * **Step 2:** $O-O-O$ * **Step 3:** $18 - 4 = 14$ ## **Date: / /** ## **Page: / /** ### L.O: To calculate formal charge. The individual charge present on each atom is called formal charge. **Formal Charge = V.E. - L.P. - 1/2 (B.P)** 1. **Hydronium ion:** $H_3O^+$ > $H - 1 - 0 - 1 x 2 = 1 - 1 = 0 $ > $H_2 = 0$ > $H_3 = 0$ > $O = 6 - 2 - 1 x 6 / 2 = 6 - 5 = 1$ > $F.C = 1$ 2. **Ammonium ion:** $NH_4^+$ > $N = 5 + 0 - 1 x 8 / 2 = 5 - 4 = 1$ > $H_1 - 1 - 0 - 1 x 2 / 2 = 1 - 1 = 0$ > $H_2 = 0$ > $H_3 = 0$ > $H_4 = 0$ > $F.C = 1$ ## **Date: / /** ## **Page: / /** 3. **Carbonate ion:** $CO_3^{2-}$ > $C = 4 - 0 - 1 x 8/2 = 4 - 4 = 0$ > $O_1 = 6 - 4 - 1 x 4 / 2 = 6 - 6 = 0$ > $O_2 = 6 - 6 - 1 x 2/2 = 0 - 1 = -1$ > $O_3 = 6 - 6 - 1 x 2 / 2 = 0 - 1 = -1$ > $F.C = -2$ 4. **Nitrate ion:** $NO_3^-$ > $N = 5 - 0 - 1 x 8 / 2 = 5 - 4 = 1$ > $O_1 = 6 - 4 - 1 x 4 / 2 = 6 - 6 = 0$ > $O_2 = 6 - 6 - 1 x 2 /2 = 0 - 1 = -1$ > $O_3 = 6 - 6 - 1 x 2 / 2 = 0 - 1 = -1$ > $F.C. = -1$ 5. **Nitrite ion:** $NO_2^-$ > $N = 5 - 2 - 1 x 8 / 3 = 5 - 5 = 0$ > $O_1 = 6 - 4 - 1 x 4 / 2 = 6 - 6 = 0$ > $O_2 = 6 - 6 - 1 x 2 / 2 = 0 - 1 = -1$ ## **Date: / /** ## **Page: / /** 6. **Ozone ion: Os** > $O_1 = 6 - 2 - 1 x 6 / 2 = 1$ > $O_2 = 6 - 6 - 1 x 2 / 2 = -1$ > $O_3 = 0$ > $F.C = -1$ **Hydronium ion** + $H_3O^+$ > $H-O-H$ > **Both e from same atom: Coordinate Bond** **Ammonium ion** + $NH_4$ > $H-N-H$ > **Both e from same atom: Coordinate Bond** **Carbonate ion** $CO_3^{2-}$ > $O$ > $| \:\:\:\:\:\:\:\:\:\:\:$ > $O-C-O$ ## **Date: / /** ## **Page: / /** **Nitrate ion** $NO_3^-$ > $O$ > $| \:\:\:\:\:\:\:\:\:\:\:$ > $O-N-O$ > **Coordinate Bond** **Nitrite ion** $NO_2^-$ > $O-N-O$ **Ozone ion** $O_3$ > $O-O-O$ > **Coordinate Bond** ### NCERT Questions: 4.1 to 4.4 4.1 A chemical bond is a force of attraction between atoms. Bonds form when atoms share or transfer valance electrons. 4.2: Mg, Na, N, Br 4.3: >$ S, Al, Al, H, H$ 4.4 $H_2S$ > **Step 1:** $2 + 6 = 8$ > **Step 2:** $H-S-H$ > **Step 3:** $8 - 4 - 4 \rightarrow H-S-H$ ## **Date: / /** ## **Page: / /** #### $SiCl_4$ > **Step 1:** $4 + 4 x 7 = 4 + 21 = 25$ > **Step 2:** $Cl$ > $Cl-Si-Cl$ > $Cl$ > **Step 3:** $25 - 8 = 17$ > $Cl$ > $Cl-Si-Cl$ > $Cl$ #### $CO_3^{2-}$ > **Step 1:** $4 + 18 + 2 = 24$ > **Step 2:** $O$ > $||$ > $O-C-O$ > **Step 3:** $24 - 6 = 18$ > $O$ > $||$ > $O-C-O$ ## **Date: / /** ## **Page: / /** #### $HCOOH$ > $H - C - O - H$ > $O$ 4.5 **Octet rule** states that every atom tends to have eight, valence in its outermost shell. It successfully explain the formation of chemical bonds and determine stability. It cannot be applied to non-metals after silicon as they can expand their octet. ## Bond Parameters ### L.O: To understand octet rule. **Octet rule** is defined as achievement of stability by having 8 e in the outermost shell. There are limitations as well which do not follow octet rule. 1. **En: NO** $5 + 6 = 11 - 2 = 9$ > * $N = O $ > * $odd e$ 2. **NO** $5 + 12 = 17 - 4 = 13$ > * $O-N-O$ ## **Date: / /** ## **Page: / /** 3. **$BF_3$** $3 + 7(3) \rightarrow 24 - 6 = 18$ > $F-B-F$ > $F$ 4. **$BeF_2$** $ 2 + 7(2) = 16 - 4 = 12$ > $F-Be-F$ > $F$ 5. **$PCl_5$** > $Cl$ > $Cl-P-Cl$ > $Cl$ > $Cl$ >> $5 + 7(5) = 40 - 10 = 30$ ### L.O:- To understand Bond Parameters. **Bond Parameters** | Bond Length | Bond Energy | Bond Angle | Bond Order | |---|---|---|---| | $O-O$ | Break bond between 2 atoms | $F-Be-F$ | $H - H$ | | | | | $N \equiv N$ | | | | | $O = O$ | | | | | Bond order | ## **Date: / /** ## **Page: / /** ### L.O: To understand the phenomena of resonance. When a single Lewis structure cannot describe a molecule accurately there more number of structures are drawn with similar energy. A hybrid structure is obtained which explains the molecule accurately. This phenomena is called resonance. **Ozone:** > $O$ > $O$ > $O$ > * ova single bond > * double bond $O \longleftrightarrow O \longleftrightarrow O$ **Carbonate:** > $O - C - O$ > $O = C = O$ > $O - C - O$ **Nitrate:** > $O-N-O$ > $O-N=O$ > $O-N=O$ ## **Date: / /** ## **Page: / /** ### L.O: To understand Polarity of Bonds. **Polarity of Bonds** mean covalent compounds can still have partial ionic character because of the difference in electronegativity. In case of HF, F being highly electronegative attracts the shared pair of electrons to itself and hence acquires partial negative charge and H acquires partial positive charge which creates dipole between them. > $ H-F$ > **Covalent Bond** > **(Polar Covalent Bond)** > $ H-Cl$ > $H-Cl$ **Q. Why is the dipole movement of $BeF_2$ and $BCl_3$ zero?** **A.** In $BeF_2$, due to the linear structure the two dipoles cancel out each other in the opposite direction making the resultant dipole movement zero. > $F-Be-F$ > $Cl$ > $Cl - B - Cl$ > $Cl$ **Resultant Dipole Movement comes out to be zero.** ## **Date: / /** ## **Page: / /** Draw the arrows for dipole movement for: 1. **$NH_3$** > $N \rightarrow H$ > $H$ > $H$ 2. **$NF_3$** > $ F \leftarrow N \rightarrow F$ > $F$ **Q. Why is the dipole movement of $NH_3$ greater than $NF_3$ ?** The dipole movement of $NH_3$ is greater than $NF_3$ because the movement of $N$ in $NH_3$ is upwards, however the movements in $NF_3$ are in opposite directions which cancels out the movements. Hence dipole movement of $NH_3 > NF_3$. ### L.O:- To draw structure of molecules based on VSEPR theory. **VSEPR Theory:** VSEPR stands for Valence Shell Electron Pair Repulsion. Sidgwick and Powell in 1940 proposed a simpler theory based on repulsion of electron pairs of the valence shell. It helps to determine the shape of the molecule as these pair of e- will occupy such positions so as to have minimum repulsion and maximum distance b/w them. ## **Date: / /** ## **Page: / /** For example: 1. **$BeF_2$** $ 2 + 2(7) = 16 = 2$ > $F-Be-F$ > * 180° > * Linear 2. **$BF_3$** $3 + 3(7) = 24 = 3$ > $F-B-F$ > $F$ > * 120° > * Trigonal Planar 3. **$CH_4$** $4 + 4(1) = 8 = 4$ > $H - C-H$ > $H$ > * 109.5° > * Tetrahedral 4. **$PCl_5$** $5 + 5 (7) = 40 = 5$ > $Cl-P-Cl$ > $Cl$ > $Cl$ > * Trigonal BiPlanar There are 2 bonds in $PCl_5$, Axial and Equatorial. The bond angle between 3 equatorial is 120° and the bond angle between 2 axial is 180° and the bond angle between axial and equatorial is 90°. 5. **$SF_6$** $6 + 6(7) = 48 = 6$ > $F$ > $S-F$ > $F$ > $F$ > $F$ > * 90° > * Octahedral 6. **$H_2O$** $2 + 6 = 2 + 6 = 4$ > $H-O-H$ > * Tetrahedral (Geometry) > * V shape/Bent ## **Date: / /** ## **Page: / /** 7. **$H_2S$** $ 2 x 1 + 6 = 8 = 4$ > $H-S-H$ > * Tetrahedral (Geometry) > * V shape/Bent 8. **$NH_3$** $5 + 3 x 1 = 8 = 4$ > $H-N-H$ > $H$ > * Tetrahedral (Geometry) > * Trigonal Pyramidal 9. **$PH_3$** (phosphine) $5 + 3 x 1 = 8 = 4$ > $H-P-H$ > $H$ > * Tetrahedral (Geometry) > * Trigonal Pyramidal 10. **$ClF_3$** $7 + 3(7) = 28 = 5$ > $F-Cl-F$ > $F$ > * TBP (Geometry) > * T shape 11. **$XeF_4$** $8 + 4 (7) = 8 + 28 = 36 = 6$ > $F$ > $F-Xe - F$ > $F$ > * Octahedral (Geometry) > * Square planar ## **Date: / /** ## **Page: / /** 12. **$BrF_5$** $7 + 5(7) = 42 = 6$ > $F$ > $F-Br-F$ > $F$ > $F$ > $F $ > * Octahedral (Geometry) > * Square Pyramidal 13. **$XeF_2$** $8 + 14 = 22 = 5$ > $F-Xe-F$ > * TBP (Geometry) > * Linear *Note: L.P occupy equatorial position* 14. **$SF_4$** $6 + 28 = 34 = 5$ > $F-S-F$ > $F$ > * TBP (Geometry) > * Seesaw ## **Date: / /** ## **Page: / /** 12/10/2024 **L.O: To understand the formation of chemical bond from VBT (Valence Bond Theory).** **Valence Bond Theory:** Valence Bond Theory helps understand how a chemical bond is formed by overlapping of orbitals of atomic. This was given by Pauling in 1928. There are 2 types of overlapping. **Sigma Bond (σ)** The axial overlapping of atomic orbitals. > * Head-2-Head > * $8 + 8$ **Pi Bond (π)** The parallel overlapping of atomic orbitals. > * Side-2-Side > * $8 + 8$ **Hybridization:** The intermixing of different atomic orbitals to produce hybrid orbitals of similar energy and shape is called as hybridization. *Number of atomic orbitols = Number of hybrid orbitals* There are different types of hybridization. * **sp** * **sp²** * **sp³** * **sp³d** * **sp³d²** ## **Date: / /** ## **Page: / /** **Examples:** 1. **$BeF_2$** (Central atom: Be (2nd period)) > * E.C = $1s^2 2s^2 2p^0$ > *  2s 2p >  $ ↑↓ $ >  $ ↑↓ ↑ $ > *  Ground state > *  2s 2p >  $ ↑ ↑ $ >  $ ↑↓ ↑ $ > * Encited state > * 2 A.O combine to produce 2 H.O. > $ + \infty = (11.0)$ > * $(2\ H.O)$ >     > * $F = 1s^2 2s^2 2p^5$     2. **$B Cl_3$** (Central atom: B (2nd period)) >  * E.C = $ 1s^2 2s^22p^1$ > * 2s 2p >  $ ↑↓ $ >  $ ↑ $ > *  Ground state > *  2s 2p >    $ ↑ $ >    $ ↑ ↑ $ > * Encited state > * 3 A.O 3 H.O > *  $(1s^2 2s^2 2p^1 3s^2 3p^5)$ > * Total = 3 ## **Date: / /** ## **Page: / /** 3. **$CH_4$** (Central atom: C (2nd period)) > (4+4)-4 > $E.C= 1s^2 2s^2 2p^2$ > *  2s 2p >  $ ↑↓ $ >  $ ↑↓ ↑ $ > *  Ground state > *  2s 2p >    $ ↑ $ >    $ ↑ ↑ ↑ $ > * Encited state > *   $sp^3 \rightarrow 4A.O = 4 \: H.O$ > * $H = 1s^1$ > $H$ > $H - C - H$ > $H$ > $4 - bonds$ > $H$ **Q. Find out hybridization in (a) $PCl_5$ (b) $SF_6$** **(a)** $PCl_5$ (Central atom: P (3rd period)) > $E.C = 1s^2 2s^2 2p^6 3s^2 3p^3$ > *  3s 3p 3d >  $ ↑↓ $ >  $ ↑↓ ↑ $ > *  Ground state > *  3s 3p 3d >    $ ↑ $ >    $ ↑ ↑ ↑ $ > * Encited state > *  $sp^3d \rightarrow 5A.O = 5 \: H.O$ ## **Date: / /** ## **Page: / /** **(b)** $SF_6$ (Central atom: S (3rd period)) > $E.C = 1s^2 2s^2 2p^6 3s^2 3p^4$ > *  3s 3p 3d >  $ ↑↓ $ >  $ ↑↓ ↑ ↑ $ > *  Ground state > *  3s 3p 3d >    $ ↑ $ >    $ ↑ ↑ ↑ ↑ ↑ $ > * Encited state > *  $sp^3d^2 \rightarrow 6A.O = 6 \: H.O$