Number of CORRECT order(s) against the mentioned characteristic among the following. (i) O 2 O (IE) (ii) O 2 − O 2 + (bond length) (iii) H 2 ⊕ = H 2 ⊖ (bond energy) (iv) [ V ( CO ) 6 ]

Question

Number of CORRECT order(s) against the mentioned characteristic among the following.

(i) $O_{2} > O$ (IE)

(ii) $O_{2}^{-} > O_{2}^{+}$ (bond length)

(iii) $H_{2}^{\oplus} = H_{2}^{⊖}$ (bond energy)

(iv) $[V (CO)_{6}]^{-} > Cr (CO)_{6}$ (C-O bond length)

(v) $N_{2}^{+} > N_{2}^{-}$ (Bond order)

(vi) $CO^{+} > CO$ (bond order)

TestHub · Accepted Answer

(i) O₂ > O (IE): Incorrect. Ionization energy (IE) of O is greater than O₂. O is an atom, O₂ is a molecule.

(ii) O₂⁻ > O₂⁺ (bond length): Correct. Bond order (BO) of O₂⁻ is 1.5, BO of O₂⁺ is 2.5. Lower BO means longer bond length.

(iii) H₂⁺ = H₂⁻ (bond energy): Incorrect. H₂⁻ is unstable.

(iv) [V(CO)₆]⁻ > Cr(CO)₆ (C-O bond length): Correct. V is more electron-rich, leading to more back-bonding, weakening C-O bond and increasing its length.

(v) N₂⁺ > N₂⁻ (Bond order): Incorrect. BO of N₂⁺ is 2.5, BO of N₂⁻ is 2.5. They have the same bond order.

(vi) CO⁺ > CO (bond order): Correct. BO of CO is 3, BO of CO⁺ is 3.5. Removal of an electron from an antibonding orbital increases bond order.

Correct orders: (ii), (iv), (vi). Total = 3.

Chemistry - CHEMICAL BONDING Question with Solution | TestHub

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