Phosphorus forms a compound with hydrogen called phosphine, PH₃. This compound can react with a hydrogen ion, H⁺. Which type of interaction occurs between PH₃ and H⁺?
▶️ Answer/Explanation
Ans: A
Phosphine (PH₃) has a lone pair of electrons on the phosphorus atom. When it reacts with H⁺ (which has an empty 1s orbital), the lone pair is donated to form a dative covalent bond (also called a coordinate bond). This results in the formation of PH₄⁺ (phosphonium ion). Since the bond is formed by sharing a lone pair, the correct answer is A.
Phosphorus forms a compound with hydrogen called phosphine, PH₃. This compound can react with a hydrogen ion, H⁺. Which type of interaction occurs between PH₃ and H⁺?
▶️ Answer/Explanation
Ans: A
Phosphine (\(PH_3\)) has a lone pair of electrons on the phosphorus atom. When \(H^+\) (which has an empty 1s orbital) interacts with \(PH_3\), the lone pair is donated to form a dative covalent bond (also called a coordinate bond). This results in the formation of \(PH_4^+\), where the bond is formed by the sharing of electrons provided entirely by \(PH_3\).
In which pairs are both species free radicals?
1 Cl and O
2 \(Cl^–\) and \(O_2^–\)
3 Cl and \(O^–\)
4 Cl⁻ and O₂⁺
▶️ Answer/Explanation
Ans: A
A free radical is a species with an unpaired electron. Analyzing the pairs:
1. Cl and O – Both are neutral atoms with unpaired electrons (radicals).
3. Cl and \(O^–\) – Cl is a radical, and \(O^–\) has an unpaired electron (radical).
4. Cl⁻ and O₂⁺ – Cl⁻ has no unpaired electrons, but O₂⁺ has one unpaired electron (radical).
Thus, pairs 1, 3, and 4 include at least one radical, making option A correct.
In which species is there a lone pair of electrons?
▶️ Answer/Explanation
Ans: C
To find a lone pair, we calculate the total valence electrons for each species. Carbon has 4 valence electrons and hydrogen has 1. For \(CH_3^-\), total electrons = \(4 + (3 \times 1) + 1 = 8\). These form 3 C-H bonds (using 6 electrons), leaving a pair of electrons unbonded on the carbon atom. The other species either have no extra electrons (\(CH_4\), \(CH_3\)) or a deficit (\(CH_3^+\)), so they lack a lone pair.