Question

(a) the electron - configuration of carbon may be written as: 1s²2s²2pₓ¹2pᵧ¹
(i) explain each of the symbols in 2pₓ¹.
(ii) considering the configuration of carbon stated, explain why carbon forms four equivalent bonds.
(b) (i) what is inductive effect?
(ii) explain why chloroethanoic acid is a stronger acid than ethanoic acid.
(c) (i) classify the following species into nucleophiles and electrophiles: h₂o, nh₃, br⁻, h⁺, alcl₃, oh⁻
(ii) explain why the reactions of alkenes are described as electrophilic addition reactions.
(d) distinguish between the terms homolytic fission and heterolytic fission.
end of essay test

Explanation:

(a)(i)

The symbol $2p$ indicates the second - energy level p - orbital. The superscript 1 means there is 1 electron in that particular p - orbital. The sub - script $x$ or $y$ (in this case $2p_x^1$) is used to distinguish between the three p - orbitals ($p_x$, $p_y$, $p_z$) in a given energy level.

(a)(ii)

Carbon has an electron configuration of $1s^22s^22p^2$. In order to achieve a stable octet, carbon can promote one of its $2s$ electrons to the empty $2p$ orbital, giving it four unpaired electrons ($2s^12p^3$). These four unpaired electrons can then form four covalent bonds with other atoms to achieve a stable electron configuration.

(b)(i)

The inductive effect is the polarization of a $\sigma$ - bond due to the electronegativity difference between atoms in the bond. A more electronegative atom pulls the bonding electrons towards itself, creating a partial positive charge on the adjacent atom and a partial negative charge on itself.

(b)(ii)

Chloroethanoic acid ($ClCH_2COOH$) is a stronger acid than ethanoic acid ($CH_3COOH$) because the chlorine atom is highly electronegative. The inductive effect of the chlorine atom withdraws electron density from the $-COOH$ group. This stabilizes the carboxylate ion ($ClCH_2COO^-$) formed when the acid dissociates, making it easier for chloroethanoic acid to lose a proton ($H^+$), thus it is a stronger acid.

(c)(i)

Nucleophiles: $H_2O$, $NH_3$, $Br^-$, $OH^-$; these species have lone pairs of electrons or a negative charge and can donate electrons to an electron - deficient species.
Electrophiles: $H^+$, $AlCl_3$; $H^+$ has a positive charge and can accept an electron pair, and $AlCl_3$ has an incomplete octet and can accept an electron pair to complete its octet.

(c)(ii)

Alkenes have a double bond ($C = C$) which is a region of high electron density. Electrophiles are attracted to this electron - rich double bond. In electrophilic addition reactions, the electrophile attacks the double bond, breaking one of the $\pi$ - bonds and forming a new bond with the electrophile. Then, a second species (nucleophile) adds to the remaining positively charged carbon atom, resulting in the addition of two groups across the double bond.

(d)

Homolytic fission is the breaking of a covalent bond in such a way that each atom retains one of the bonding electrons, resulting in the formation of free radicals. For example, $A:B
ightarrow A\cdot + B\cdot$.
Heterolytic fission is the breaking of a covalent bond in such a way that one atom takes both of the bonding electrons, resulting in the formation of ions. For example, $A:B
ightarrow A^+ + :B^-$

Answer:

(a)(i) See above explanation for symbols in $2p_x^1$.
(a)(ii) Carbon forms four bonds to achieve octet by promoting an electron.
(b)(i) Inductive effect is $\sigma$ - bond polarization due to electronegativity difference.
(b)(ii) Chloroethanoic acid is stronger due to chlorine's inductive effect.
(c)(i) Nucleophiles: $H_2O$, $NH_3$, $Br^-$, $OH^-$; Electrophiles: $H^+$, $AlCl_3$.
(c)(ii) Alkenes have electron - rich double bonds, attacked by electrophiles.
(d) Homolytic fission forms free radicals, heterolytic fission forms ions.