Question

2. Choose the correct code of characteristics for the given order of hybrid orbitals of same atom, $Sp<sp^2<sp^3$ (i) Electronegativity (ii) Bond angle between same hybrid orbitals (iii) Size (iv) Energy level

• A. ii, iii and iv
• B. iii, iv
• C. ii and iv
• D. i, ii, iii and iv

Answer 1

Answer: (B)

b) iii, iv

Answer 2

The order $sp < sp^2 < sp^3$ corresponds to a decrease in s-character (50% > 33.3% > 25%).

(i) Electronegativity: Electronegativity increases with increasing s-character. Therefore, as s-character decreases, electronegativity decreases. The order of electronegativity is $sp > sp^2 > sp^3$. Thus, the statement "Electronegativity increases" (as implied by $sp < sp^2 < sp^3$) is false.

(ii) Bond angle: The bond angle decreases as s-character decreases (180° for sp, 120° for $sp^2$, 109.5° for $sp^3$). Thus, the statement "Bond angle increases" (as implied by $sp < sp^2 < sp^3$) is false.

(iii) Size: The size of hybrid orbitals increases as s-character decreases because s-orbitals are closer to the nucleus. Thus, the order of size is $sp < sp^2 < sp^3$. This statement is true.

(iv) Energy level: The energy of hybrid orbitals increases with increasing s-character. Therefore, as s-character decreases, the energy level decreases. Thus, the order of energy is $sp < sp^2 < sp^3$. This statement is true.

Therefore, only (iii) and (iv) correctly follow the given inequality.